Explanation:
Okay, so I need to figure out which of these equations has the largest solution for x, given that 0 < r < 3. Let me start by understanding what each equation is saying. All of them are of the form 3 times some base raised to the power of x equals 7. So, if I can solve each equation for x, I can compare the solutions.

First, let me recall that for an equation like A * B^x = C, solving for x would involve taking the logarithm of both sides. Let me write that down. If I have 3*(something)^x = 7, then dividing both sides by 3 gives (something)^x = 7/3. Then taking the natural logarithm (or log base 10, doesn't matter as long as I'm consistent) of both sides gives x * ln(something) = ln(7/3), so x = ln(7/3)/ln(something). Therefore, the value of x depends inversely on the logarithm of the base. So, if the base is larger, then ln(base) is larger, so x would be smaller. Conversely, if the base is smaller, ln(base) is smaller, so x would be larger. Therefore, the equation with the smallest base will have the largest x. Wait, is that correct?

Wait, let me check. Suppose I have two equations: A * B^x = C and A * D^x = C. Then solving for x gives x = ln(C/A)/ln(B) and x = ln(C/A)/ln(D). So, if B < D, then ln(B) < ln(D), so x would be larger for the equation with the smaller base. Therefore, yes, the equation with the smallest base will have the largest x. Therefore, to find which equation has the largest x, I need to find which equation has the smallest base. So, the base is (1 + r), (1 + r/10), (1 + 2r), (1 + sqrt(r)), and (1 + 1/r). So, I need to compare these bases for 0 < r < 3 and find which one is the smallest.

But wait, hold on. Let me make sure. Let me think again. Since x is proportional to 1/ln(base), so if the base is smaller, ln(base) is smaller, so 1/ln(base) is larger. Therefore, x is larger. Therefore, the equation with the smallest base will have the largest x. Therefore, the problem reduces to finding which of these bases is the smallest for 0 < r < 3.

So, let me analyze each base:

A: 1 + r

B: 1 + r/10

C: 1 + 2r

D: 1 + sqrt(r)

E: 1 + 1/r

So, given that 0 < r < 3, let's analyze each base.

First, note that r is between 0 and 3. Let me consider each base:

For base A: 1 + r. Since r is between 0 and 3, this base is between 1 and 4.

For base B: 1 + r/10. Since r is between 0 and 3, r/10 is between 0 and 0.3, so base B is between 1 and 1.3.

For base C: 1 + 2r. Since r is between 0 and 3, 2r is between 0 and 6, so base C is between 1 and 7.

For base D: 1 + sqrt(r). Since r is between 0 and 3, sqrt(r) is between 0 and sqrt(3) ≈ 1.732. Therefore, base D is between 1 and approximately 2.732.

For base E: 1 + 1/r. Since r is between 0 and 3, 1/r is between 1/3 ≈ 0.333 and approaching infinity as r approaches 0. However, since r is greater than 0, 1/r is positive. But since r is less than 3, 1/r is greater than 1/3. Therefore, base E is between 1 + 1/3 ≈ 1.333 and approaching infinity as r approaches 0.

So, now, comparing all these bases:

Base B is between 1 and 1.3.

Base E is between approximately 1.333 and infinity.

Base A is between 1 and 4.

Base D is between 1 and ~2.732.

Base C is between 1 and 7.

So, the smallest base is base B? Wait, but base B is between 1 and 1.3, while base E is between 1.333 and infinity. So, base B is smaller than base E. However, base B is between 1 and 1.3, but base D is between 1 and ~2.732. So, base B is smaller than base D. Similarly, base A is between 1 and 4, so base B is smaller than base A. So, base B is the smallest? Wait, but hold on. Wait, base B is 1 + r/10. Since r is between 0 and 3, then base B is between 1 and 1.3. However, base E is 1 + 1/r. When r approaches 0, 1/r approaches infinity, so base E can be very large. But when r approaches 3, base E approaches 1 + 1/3 ≈ 1.333. So, for r approaching 3, base E is about 1.333, which is larger than base B's upper bound of 1.3. Wait, but base B's upper bound is 1.3 when r=3. So, when r approaches 3, base B approaches 1 + 3/10 = 1.3, and base E approaches 1 + 1/3 ≈ 1.333. Therefore, for r approaching 3, base B is smaller than base E. However, for r approaching 0, base B approaches 1 + 0 = 1, while base E approaches infinity. Therefore, for all r in (0, 3), base B is between 1 and 1.3, and base E is between 1.333 and infinity. Therefore, base B is always smaller than base E. Similarly, base B is smaller than base D, base A, and base C. Therefore, base B is the smallest base. Therefore, equation B would have the largest x? Wait, but hold on. Wait, if base B is the smallest, then x is the largest? Wait, earlier conclusion was that the equation with the smallest base has the largest x. Therefore, equation B would have the largest x? But let me check again.

Wait, let me think again. Let me take an example. Suppose I have two equations:

Equation 1: 3*(1 + r/10)^x = 7

Equation 2: 3*(1 + 1/r)^x = 7

If base B is smaller than base E, then x for equation B is larger than x for equation E. So, if base B is smaller, then x is larger. Therefore, equation B would have larger x. But wait, but maybe there's a case where base B is not the smallest? Wait, for example, if r is very small, say r approaches 0. Then base B approaches 1, and base E approaches infinity. So, base B is smaller. If r is approaching 3, base B approaches 1.3, and base E approaches 1.333. So, base B is still smaller. Therefore, for all r in (0, 3), base B is the smallest. Therefore, equation B would have the largest x. But wait, let me check with another base. For example, base D: 1 + sqrt(r). For r approaching 0, sqrt(r) approaches 0, so base D approaches 1. So, when r approaches 0, base D approaches 1, which is smaller than base B's value of 1 + r/10. Wait, when r approaches 0, base B approaches 1 + 0 = 1, and base D approaches 1 + 0 = 1. So, they are both approaching 1. But which one is smaller? Let me take a specific value of r. Let me take r approaching 0, say r = 0.1. Then base B is 1 + 0.1/10 = 1.01. Base D is 1 + sqrt(0.1) ≈ 1 + 0.316 ≈ 1.316. So, base B is smaller. If r = 0.01, base B is 1 + 0.001 = 1.001, base D is 1 + sqrt(0.01) = 1 + 0.1 = 1.1. So, base B is still smaller. If r approaches 0, base B approaches 1 from above, and base D approaches 1 from above as well, but base B approaches 1 faster? Wait, sqrt(r) approaches 0 slower than r/10. For example, for small r, sqrt(r) is larger than r/10. Let me check. Let me take r = 0.01. sqrt(0.01) = 0.1, and r/10 = 0.001. So, sqrt(r) is larger. Therefore, base D is 1 + sqrt(r) which is larger than base B which is 1 + r/10. Therefore, for small r, base B is smaller. Therefore, base B is smaller than base D. Similarly, for r approaching 3, base B is 1.3, base D is 1 + sqrt(3) ≈ 2.732. So, base B is smaller. Therefore, base B is the smallest base for all r in (0, 3). Therefore, equation B would have the largest x. But wait, let me check equation E. For example, when r approaches 0, base E approaches infinity, so x approaches 0. When r approaches 3, base E approaches 1 + 1/3 ≈ 1.333, which is larger than base B's 1.3. Therefore, equation E's base is larger than base B's base for all r in (0, 3). Therefore, equation B has the smallest base, so equation B has the largest x. Therefore, answer is B? But wait, let me check the answer options again. Wait, the answer options are A, B, C, D, E. Wait, but let me check again. Wait, maybe I made a mistake here. Let me think again. Wait, if base B is the smallest, then x is the largest. But let me check with specific numbers. Let me take a specific value of r, say r = 1. Then compute x for each equation.

Let me take r = 1. Then:

Equation A: 3*(1 + 1)^x = 7 => 3*(2)^x = 7 => 2^x = 7/3 ≈ 2.333. Then x = log2(7/3) ≈ log(2.333)/log(2) ≈ 1.22.

Equation B: 3*(1 + 1/10)^x = 7 => 3*(1.1)^x = 7 => 1.1^x = 7/3 ≈ 2.333. Then x = log(2.333)/log(1.1) ≈ 1.22 / 0.0414 ≈ 29.46? Wait, wait, no. Wait, log(2.333) is approximately 0.367, and log(1.1) is approximately 0.0414. Therefore, x ≈ 0.367 / 0.0414 ≈ 8.86. Wait, that seems high. Wait, let me check with calculator steps. Let me compute ln(7/3) divided by ln(1.1). Let me compute ln(7/3) ≈ ln(2.333) ≈ 0.847. ln(1.1) ≈ 0.0953. Therefore, x ≈ 0.847 / 0.0953 ≈ 8.89. So, x ≈ 8.89.

Equation C: 3*(1 + 2*1)^x = 7 => 3*(3)^x = 7 => 3^x = 7/3 ≈ 2.333. Then x = log3(7/3) ≈ ln(2.333)/ln(3) ≈ 0.847 / 1.0986 ≈ 0.771.

Equation D: 3*(1 + sqrt(1))^x = 7 => 3*(1 + 1)^x = 7 => same as equation A, so x ≈ 1.22.

Equation E: 3*(1 + 1/1)^x = 7 => 3*(2)^x = 7 => same as equation A, so x ≈ 1.22.

So, for r = 1, equation B has x ≈ 8.89, which is the largest. So, that seems to confirm that equation B has the largest x. But let me check another value of r. Let me take r approaching 0, say r = 0.1.

For r = 0.1:

Equation A: 3*(1 + 0.1)^x = 7 => 3*(1.1)^x = 7 => same as equation B when r = 1? Wait, no. Wait, equation B is 3*(1 + r/10)^x = 7. For r = 0.1, equation B is 3*(1 + 0.01)^x = 7. So, base is 1.01. Let me compute x for each equation.

Equation A: 3*(1.1)^x = 7 => 1.1^x = 7/3 ≈ 2.333. So, x ≈ ln(2.333)/ln(1.1) ≈ 0.847 / 0.0953 ≈ 8.89.

Equation B: 3*(1.01)^x = 7 => 1.01^x = 7/3 ≈ 2.333. So, x ≈ ln(2.333)/ln(1.01) ≈ 0.847 / 0.00995 ≈ 85.1. That's way larger.

Equation C: 3*(1 + 2*0.1)^x = 7 => 3*(1.2)^x = 7 => 1.2^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(1.2) ≈ 0.847 / 0.1823 ≈ 4.64.

Equation D: 3*(1 + sqrt(0.1))^x = 7 => sqrt(0.1) ≈ 0.316, so base ≈ 1.316. Then x ≈ ln(2.333)/ln(1.316) ≈ 0.847 / 0.275 ≈ 3.08.

Equation E: 3*(1 + 1/0.1)^x = 7 => 1 + 10 = 11. So, 3*(11)^x = 7 => 11^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(11) ≈ 0.847 / 2.398 ≈ 0.353.

So, for r = 0.1, equation B has x ≈ 85.1, which is the largest. So, again, equation B is the largest. Let me check another value, say r approaching 3, say r = 2.9.

For r = 2.9:

Equation A: 3*(1 + 2.9)^x = 7 => 3*(3.9)^x = 7 => 3.9^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(3.9) ≈ 0.847 / 1.361 ≈ 0.622.

Equation B: 3*(1 + 2.9/10)^x = 7 => 3*(1.29)^x = 7 => 1.29^x = 2.333. x ≈ ln(2.333)/ln(1.29) ≈ 0.847 / 0.255 ≈ 3.32.

Equation C: 3*(1 + 2*2.9)^x = 7 => 3*(1 + 5.8)^x = 7 => 3*(6.8)^x = 7 => 6.8^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(6.8) ≈ 0.847 / 1.916 ≈ 0.442.

Equation D: 3*(1 + sqrt(2.9))^x = 7. sqrt(2.9) ≈ 1.702, so base ≈ 2.702. x ≈ ln(2.333)/ln(2.702) ≈ 0.847 / 1.0 ≈ 0.847.

Equation E: 3*(1 + 1/2.9)^x = 7. 1/2.9 ≈ 0.3448, so base ≈ 1.3448. x ≈ ln(2.333)/ln(1.3448) ≈ 0.847 / 0.298 ≈ 2.84.

So, for r = 2.9, equation B has x ≈ 3.32, which is the largest. So, again, equation B is the largest. Therefore, seems like equation B is the one with the largest x. But wait, let me check equation D for r = 2.9. The base was 2.702, so ln(2.702) ≈ 1.0, so x ≈ 0.847 / 1.0 ≈ 0.847. Which is less than equation B's 3.32. So, equation B is still larger. Therefore, seems like equation B is the answer. But let me check if there is any case where another equation could have a larger x. For example, equation E when r approaches 0. But when r approaches 0, equation E's base approaches infinity, so x approaches 0. So, equation E's x approaches 0. So, equation B's x approaches infinity? Wait, no. Wait, when r approaches 0, equation B's base approaches 1 + 0 = 1. So, the equation becomes 3*(1)^x = 7, which is 3 = 7, which is impossible. Wait, but in reality, as r approaches 0, the base approaches 1 from above. So, for very small r, the base is slightly larger than 1, so the equation 3*(1 + r/10)^x = 7 would have x approaching infinity as r approaches 0? Wait, let me think. If the base is approaching 1 from above, then (1 + ε)^x = 7/3, where ε approaches 0. So, taking logarithms, x = ln(7/3)/ln(1 + ε). As ε approaches 0, ln(1 + ε) ≈ ε. Therefore, x ≈ ln(7/3)/ε. Since ε = r/10, so x ≈ ln(7/3)/(r/10) = 10*ln(7/3)/r. As r approaches 0, x approaches infinity. Therefore, for very small r, equation B's x approaches infinity. Therefore, equation B's x can be made arbitrarily large by taking r approaching 0. Therefore, equation B has the largest x. Therefore, answer is B? But let me check if there is any other equation that can have larger x. For example, equation D. Let me check for very small r. For equation D, base is 1 + sqrt(r). For very small r, sqrt(r) is much larger than r/10. For example, if r approaches 0, sqrt(r) approaches 0, but slower than r/10? Wait, no. Wait, for very small r, sqrt(r) is larger than r/10. For example, take r = 0.01, sqrt(r) = 0.1, and r/10 = 0.001. So, sqrt(r) is larger. Therefore, base D is 1 + sqrt(r) ≈ 1 + 0.1 = 1.1, while base B is 1 + 0.001 = 1.001. Therefore, base B is smaller. Therefore, equation B's x is larger. Therefore, even for very small r, equation B's x is larger. Therefore, equation B is the one with the largest x. Therefore, answer is B. But let me check the answer options again. Wait, the answer options are A, B, C, D, E. So, according to my analysis, answer is B. But let me check again. Wait, but let me think again. Wait, the problem says 0 < r < 3. So, r can't be zero. But as r approaches zero, equation B's x approaches infinity. Therefore, equation B's x is the largest. Therefore, answer is B. However, let me check if there is any mistake in my reasoning. For example, maybe equation E? Wait, equation E is 3*(1 + 1/r)^x = 7. For r approaching 0, 1 + 1/r approaches infinity, so the base approaches infinity. Therefore, solving for x, we have (base)^x = 7/3. Therefore, x = log_{base}(7/3) = ln(7/3)/ln(base). As base approaches infinity, ln(base) approaches infinity, so x approaches 0. Therefore, equation E's x approaches 0. Therefore, equation E's x is small. Therefore, equation B's x is the largest. Therefore, answer is B. But let me check the answer options again. Wait, the answer options are given as A to E. Wait, but let me check if there is a case where equation D could have a larger x. For example, if base D is smaller than base B? Wait, for r approaching 0, base D approaches 1 + 0 = 1, same as base B. But for very small r, sqrt(r) is larger than r/10. Therefore, base D is larger than base B. Therefore, equation B's base is smaller, so equation B's x is larger. Therefore, equation B is the answer. Therefore, I think the answer is B. However, let me check the original problem again to make sure I didn't misread anything. The problem says "which equation has the largest solution x?" and the equations are:

A) 3(1 + r)^x = 7

B) 3(1 + r/10)^x = 7

C) 3(1 + 2r)^x = 7

D) 3(1 + sqrt(r))^x = 7

E) 3(1 + 1/r)^x = 7

Yes, that's correct. So, according to my analysis, answer is B. However, I need to make sure that there isn't a case where another equation could have a larger x. For example, equation D. Let me take r = 0.01. Then base D is 1 + sqrt(0.01) = 1 + 0.1 = 1.1. Base B is 1 + 0.01/10 = 1.001. So, base B is smaller. Therefore, equation B's x is larger. For equation D, solving 3*(1.1)^x = 7. So, x ≈ ln(7/3)/ln(1.1) ≈ 0.847 / 0.0953 ≈ 8.89. For equation B, solving 3*(1.001)^x = 7. So, x ≈ ln(7/3)/ln(1.001) ≈ 0.847 / 0.0009995 ≈ 847. So, way larger. Therefore, equation B is larger. Therefore, answer is B. Therefore, I think the answer is B. However, let me check if there is any other possible mistake. For example, maybe I confused the direction of the inequality? Let me think again. If base is smaller, then ln(base) is smaller, so x = ln(7/3)/ln(base) is larger. Therefore, yes, smaller base gives larger x. Therefore, equation B has the smallest base, so largest x. Therefore, answer is B. Therefore, I think the answer is B. However, let me check the answer options again. Wait, the answer options are labeled A to E. So, answer is B. But let me check once more with another value. Let me take r = 0.5. Then:

Equation A: 3*(1 + 0.5)^x = 7 => 3*(1.5)^x = 7 => 1.5^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(1.5) ≈ 0.847 / 0.405 ≈ 2.09.

Equation B: 3*(1 + 0.5/10)^x = 7 => 3*(1.05)^x = 7 => 1.05^x = 2.333. x ≈ ln(2.333)/ln(1.05) ≈ 0.847 / 0.04879 ≈ 17.37.

Equation C: 3*(1 + 2*0.5)^x = 7 => 3*(2)^x = 7 => 2^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(2) ≈ 0.847 / 0.693 ≈ 1.22.

Equation D: 3*(1 + sqrt(0.5))^x = 7. sqrt(0.5) ≈ 0.707, so base ≈ 1.707. x ≈ ln(2.333)/ln(1.707) ≈ 0.847 / 0.535 ≈ 1.58.

Equation E: 3*(1 + 1/0.5)^x = 7 => 3*(1 + 2)^x = 7 => 3*(3)^x = 7 => 3^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(3) ≈ 0.847 / 1.0986 ≈ 0.771.

So, for r = 0.5, equation B has x ≈ 17.37, which is the largest. Therefore, again, equation B is the largest. Therefore, I think my conclusion is correct. Therefore, the answer is B. However, let me check if the problem says "the largest solution x", and maybe there is a case where another equation could have a larger x? For example, equation E? For r approaching 0, equation E's x approaches 0, so no. Equation D? For r approaching 0, equation D's base approaches 1 + 0 = 1, so similar to equation B. But since base D is larger than base B for small r, equation B's x is larger. Therefore, equation B is the answer. Therefore, I think the answer is B. However, let me check once again if I made a mistake in the initial assumption. For example, if the base is between 0 and 1, then the logarithm is negative, and x would be negative. But since the problem says 0 < r < 3, let me check if any of the bases can be less than 1. For example, base E: 1 + 1/r. Since r is between 0 and 3, 1/r is positive, so base E is always greater than 1. Similarly, base D: 1 + sqrt(r). sqrt(r) is non-negative, so base D is always greater than 1. Base C: 1 + 2r. Since r > 0, base C is greater than 1. Base A: 1 + r, same. Base B: 1 + r/10, same. Therefore, all bases are greater than 1. Therefore, all equations have positive x. Therefore, no issues with negative x. Therefore, my previous analysis holds. Therefore, answer is B. Therefore, I think the correct answer is B.

**Final Answer**
\boxed{B}

To determine which equation has the **largest solution** for $ x $, we analyze the structure of each equation:

All equations are of the form:

$$
3 \cdot (\text{base})^x = 7
$$

We can solve for $ x $ by isolating it:

$$
x = \frac{\ln\left(\frac{7}{3}\right)}{\ln(\text{base})}
$$

This shows that the value of $ x $ is inversely proportional to the natural logarithm of the base. That is, **smaller bases** result in **larger values of $ x $**.

---

### Step 1: Identify the Bases

Let’s extract the base from each equation:

- **(A)** $ 3(1 + r)^x = 7 $ → base = $ 1 + r $
- **(B)** $ 3(1 + r/10)^x = 7 $ → base = $ 1 + \frac{r}{10} $
- **(C)** $ 3(1 + 2r)^x = 7 $ → base = $ 1 + 2r $
- **(D)** $ 3(1 + \sqrt{r})^x = 7 $ → base = $ 1 + \sqrt{r} $
- **(E)** $ 3(1 + 1/r)^x = 7 $ → base = $ 1 + \frac{1}{r} $

---

### Step 2: Compare the Bases

We are given that $ 0 < r < 3 $. Let’s analyze the behavior of each base in this interval:

- **Base A**: $ 1 + r $ → ranges from 1 to 4
- **Base B**: $ 1 + \frac{r}{10} $ → ranges from 1 to 1.3
- **Base C**: $ 1 + 2r $ → ranges from 1 to 7
- **Base D**: $ 1 + \sqrt{r} $ → ranges from 1 to $ 1 + \sqrt{3} \approx 2.732 $
- **Base E**: $ 1 + \frac{1}{r} $ → ranges from $ 1 + \frac{1}{3} \approx 1.333 $ to $ \infty $

---

### Step 3: Determine the Smallest Base

From the above, the **smallest base** is:

$$
\text{Base B: } 1 + \frac{r}{10}
$$

This is the **smallest** among all the bases for all $ r \in (0, 3) $, because:

- For small $ r $, $ \frac{r}{10} $ is much smaller than $ \sqrt{r} $, $ 2r $, or $ \frac{1}{r} $
- For large $ r $, $ \frac{r}{10} $ is still smaller than $ \sqrt{r} $, $ 2r $, or $ \frac{1}{r} $

Thus, **Base B is the smallest**.

---

### Step 4: Conclude the Largest $ x $

Since $ x = \frac{\ln(7/3)}{\ln(\text{base})} $, and the **base is smallest for equation B**, the **value of $ x $ is largest** for equation B.

---

### Final Answer

$$
\boxed{B}
$$

[Assistant]
Explanation:
Okay, so I need to figure out which of these equations has the largest solution for x, given that 0 < r < 3. Let me start by understanding what each equation is saying. All of them are of the form 3 times some base raised to the power of x equals 7. So, if I can solve each equation for x, I can compare the solutions.

First, let me recall that for an equation like A * B^x = C, solving for x would involve taking the logarithm of both sides. Let me write that down. If I have 3*(something)^x = 7, then dividing both sides by 3 gives (something)^x = 7/3. Then taking the natural logarithm (or log base 10, doesn't matter as long as I'm consistent) of both sides gives x * ln(something) = ln(7/3), so x = ln(7/3)/ln(something). Therefore, the value of x depends inversely on the logarithm of the base. So, if the base is larger, then ln(base) is larger, so x would be smaller. Conversely, if the base is smaller, ln(base) is smaller, so x would be larger. Therefore, the equation with the smallest base will have the largest x. Wait, is that correct?

Wait, let me check. Suppose I have two equations: A * B^x = C and A * D^x = C. Then solving for x gives x = ln(C/A)/ln(B) and x = ln(C/A)/ln(D). So, if B < D, then ln(B) < ln(D), so x would be larger for the equation with the smaller base. Therefore, yes, the equation with the smallest base will have the largest x. Therefore, to find which equation has the largest x, I need to find which equation has the smallest base. So, the base is (1 + r), (1 + r/10), (1 + 2r), (1 + sqrt(r)), and (1 + 1/r). So, I need to compare these bases for 0 < r < 3 and find which one is the smallest.

But wait, hold on. Let me make sure. Let me think again. Since x is proportional to 1/ln(base), so if the base is smaller, ln(base) is smaller, so 1/ln(base) is larger. Therefore, x is larger. Therefore, the equation with the smallest base will have the largest x. Therefore, the problem reduces to finding which of these bases is the smallest for 0 < r < 3.

So, let me analyze each base:

A: 1 + r

B: 1 + r/10

C: 1 + 2r

D: 1 + sqrt(r)

E: 1 + 1/r

So, given that 0 < r < 3, let's analyze each base.

First, note that r is between 0 and 3. Let me consider each base:

For base A: 1 + r. Since r is between 0 and 3, this base is between 1 and 4.

For base B: 1 + r/10. Since r is between 0 and 3, r/10 is between 0 and 0.3, so base B is between 1 and 1.3.

For base C: 1 + 2r. Since r is between 0 and 3, 2r is between 0 and 6, so base C is between 1 and 7.

For base D: 1 + sqrt(r). Since r is between 0 and 3, sqrt(r) is between 0 and sqrt(3) ≈ 1.732. Therefore, base D is between 1 and approximately 2.732.

For base E: 1 + 1/r. Since r is between 0 and 3, 1/r is between 1/3 ≈ 0.333 and approaching infinity as r approaches 0. However, since r is greater than 0, 1/r is positive. But since r is less than 3, 1/r is greater than 1/3. Therefore, base E is between 1 + 1/3 ≈ 1.333 and approaching infinity as r approaches 0.

So, now, comparing all these bases:

Base B is between 1 and 1.3.

Base E is between approximately 1.333 and infinity.

Base A is between 1 and 4.

Base D is between 1 and ~2.732.

Base C is between 1 and 7.

So, the smallest base is base B? Wait, but base B is between 1 and 1.3, while base E is between 1.333 and infinity. So, base B is smaller than base E. However, base B is between 1 and 1.3, but base D is between 1 and ~2.732. So, base B is smaller than base D. Similarly, base A is between 1 and 4, so base B is smaller than base A. So, base B is the smallest? Wait, but hold on. Wait, base B is 1 + r/10. Since r is between 0 and 3, then base B is between 1 and 1.3. However, base E is 1 + 1/r. When r approaches 0, 1/r approaches infinity, so base E can be very large. But when r approaches 3, base E approaches 1 + 1/3 ≈ 1.333. So, for r approaching 3, base E is about 1.333, which is larger than base B's upper bound of 1.3. Wait, but base B's upper bound is 1.3 when r=3. So, when r approaches 3, base B approaches 1 + 3/10 = 1.3, and base E approaches 1 + 1/3 ≈ 1.333. Therefore, for r approaching 3, base B is smaller than base E. However, for r approaching 0, base B approaches 1 + 0 = 1, while base E approaches infinity. Therefore, for all r in (0, 3), base B is between 1 and 1.3, and base E is between 1.333 and infinity. Therefore, base B is always smaller than base E. Similarly, base B is smaller than base D, base A, and base C. Therefore, base B is the smallest base. Therefore, equation B would have the largest x? Wait, but hold on. Wait, if base B is the smallest, then x is the largest? Wait, earlier conclusion was that the equation with the smallest base has the largest x. Therefore, equation B would have the largest x? But let me check again.

Wait, let me think again. Let me take an example. Suppose I have two equations:

Equation 1: 3*(1 + r/10)^x = 7

Equation 2: 3*(1 + 1/r)^x = 7

If base B is smaller than base E, then x for equation B is larger than x for equation E. So, if base B is smaller, then x is larger. Therefore, equation B would have larger x. But wait, but maybe there's a case where base B is not the smallest? Wait, for example, if r is very small, say r approaches 0. Then base B approaches 1, and base E approaches infinity. So, base B is smaller. If r is approaching 3, base B approaches 1.3, and base E approaches 1.333. So, base B is still smaller. Therefore, for all r in (0, 3), base B is the smallest. Therefore, equation B would have the largest x. But wait, let me check with another base. For example, base D: 1 + sqrt(r). For r approaching 0, sqrt(r) approaches 0, so base D approaches 1. So, when r approaches 0, base D approaches 1, which is smaller than base B's value of 1 + r/10. Wait, when r approaches 0, base B approaches 1 + 0 = 1, and base D approaches 1 + 0 = 1. So, they are both approaching 1. But which one is smaller? Let me take a specific value of r. Let me take r approaching 0, say r = 0.1. Then base B is 1 + 0.1/10 = 1.01. Base D is 1 + sqrt(0.1) ≈ 1 + 0.316 ≈ 1.316. So, base B is smaller. If r = 0.01, base B is 1 + 0.001 = 1.001, base D is 1 + sqrt(0.01) = 1 + 0.1 = 1.1. So, base B is still smaller. If r approaches 0, base B approaches 1 from above, and base D approaches 1 from above as well, but base B approaches 1 faster? Wait, sqrt(r) approaches 0 slower than r/10. For example, for small r, sqrt(r) is larger than r/10. Let me check. Let me take r = 0.01. sqrt(0.01) = 0.1, and r/10 = 0.001. So, sqrt(r) is larger. Therefore, base D is 1 + sqrt(r) which is larger than base B which is 1 + r/10. Therefore, for small r, base B is smaller. Therefore, base B is smaller than base D. Similarly, for r approaching 3, base B is 1.3, base D is 1 + sqrt(3) ≈ 2.732. So, base B is smaller. Therefore, base B is the smallest base for all r in (0, 3). Therefore, equation B would have the largest x. But wait, let me check equation E. For example, when r approaches 0, base E approaches infinity, so x approaches 0. When r approaches 3, base E approaches 1 + 1/3 ≈ 1.333, which is larger than base B's 1.3. Therefore, equation E's base is larger than base B's base for all r in (0, 3). Therefore, equation B has the smallest base, so equation B has the largest x. Therefore, answer is B? But wait, let me check the answer options again. Wait, the answer options are A, B, C, D, E. Wait, but let me check again. Wait, maybe I made a mistake here. Let me think again. Wait, if base B is the smallest, then x is the largest. But let me check with specific numbers. Let me take a specific value of r, say r = 1. Then compute x for each equation.

Let me take r = 1. Then:

Equation A: 3*(1 + 1)^x = 7 => 3*(2)^x = 7 => 2^x = 7/3 ≈ 2.333. Then x = log2(7/3) ≈ log(2.333)/log(2) ≈ 1.22.

Equation B: 3*(1 + 1/10)^x = 7 => 3*(1.1)^x = 7 => 1.1^x = 7/3 ≈ 2.333. Then x = log(2.333)/log(1.1) ≈ 1.22 / 0.0414 ≈ 29.46? Wait, wait, no. Wait, log(2.333) is approximately 0.367, and log(1.1) is approximately 0.0414. Therefore, x ≈ 0.367 / 0.0414 ≈ 8.86. Wait, that seems high. Wait, let me check with calculator steps. Let me compute ln(7/3) divided by ln(1.1). Let me compute ln(7/3) ≈ ln(2.333) ≈ 0.847. ln(1.1) ≈ 0.0953. Therefore, x ≈ 0.847 / 0.0953 ≈ 8.89. So, x ≈ 8.89.

Equation C: 3*(1 + 2*1)^x = 7 => 3*(3)^x = 7 => 3^x = 7/3 ≈ 2.333. Then x = log3(7/3) ≈ ln(2.333)/ln(3) ≈ 0.847 / 1.0986 ≈ 0.771.

Equation D: 3*(1 + sqrt(1))^x = 7 => 3*(1 + 1)^x = 7 => same as equation A, so x ≈ 1.22.

Equation E: 3*(1 + 1/1)^x = 7 => 3*(2)^x = 7 => same as equation A, so x ≈ 1.22.

So, for r = 1, equation B has x ≈ 8.89, which is the largest. So, that seems to confirm that equation B has the largest x. But let me check another value of r. Let me take r approaching 0, say r = 0.1.

For r = 0.1:

Equation A: 3*(1 + 0.1)^x = 7 => 3*(1.1)^x = 7 => same as equation B when r = 1? Wait, no. Wait, equation B is 3*(1 + r/10)^x = 7. For r = 0.1, equation B is 3*(1 + 0.01)^x = 7. So, base is 1.01. Let me compute x for each equation.

Equation A: 3*(1.1)^x = 7 => 1.1^x = 7/3 ≈ 2.333. So, x ≈ ln(2.333)/ln(1.1) ≈ 0.847 / 0.0953 ≈ 8.89.

Equation B: 3*(1.01)^x = 7 => 1.01^x = 7/3 ≈ 2.333. So, x ≈ ln(2.333)/ln(1.01) ≈ 0.847 / 0.00995 ≈ 85.1. That's way larger.

Equation C: 3*(1 + 2*0.1)^x = 7 => 3*(1.2)^x = 7 => 1.2^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(1.2) ≈ 0.847 / 0.1823 ≈ 4.64.

Equation D: 3*(1 + sqrt(0.1))^x = 7 => sqrt(0.1) ≈ 0.316, so base ≈ 1.316. Then x ≈ ln(2.333)/ln(1.316) ≈ 0.847 / 0.275 ≈ 3.08.

Equation E: 3*(1 + 1/0.1)^x = 7 => 1 + 10 = 11. So, 3*(11)^x = 7 => 11^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(11) ≈ 0.847 / 2.398 ≈ 0.353.

So, for r = 0.1, equation B has x ≈ 85.1, which is the largest. So, again, equation B is the largest. Let me check another value, say r approaching 3, say r = 2.9.

For r = 2.9:

Equation A: 3*(1 + 2.9)^x = 7 => 3*(3.9)^x = 7 => 3.9^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(3.9) ≈ 0.847 / 1.361 ≈ 0.622.

Equation B: 3*(1 + 2.9/10)^x = 7 => 3*(1.29)^x = 7 => 1.29^x = 2.333. x ≈ ln(2.333)/ln(1.29) ≈ 0.847 / 0.255 ≈ 3.32.

Equation C: 3*(1 + 2*2.9)^x = 7 => 3*(1 + 5.8)^x = 7 => 3*(6.8)^x = 7 => 6.8^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(6.8) ≈ 0.847 / 1.916 ≈ 0.442.

Equation D: 3*(1 + sqrt(2.9))^x = 7. sqrt(2.9) ≈ 1.702, so base ≈ 2.702. x ≈ ln(2.333)/ln(2.702) ≈ 0.847 / 1.0 ≈ 0.847.

Equation E: 3*(1 + 1/2.9)^x = 7. 1/2.9 ≈ 0.3448, so base ≈ 1.3448. x ≈ ln(2.333)/ln(1.3448) ≈ 0.847 / 0.298 ≈ 2.84.

So, for r = 2.9, equation B has x ≈ 3.32, which is the largest. So, again, equation B is the largest. Therefore, seems like equation B is the one with the largest x. But wait, let me check equation D for r = 2.9. The base was 2.702, so ln(2.702) ≈ 1.0, so x ≈ 0.847 / 1.0 ≈ 0.847. Which is less than equation B's 3.32. So, equation B is still larger. Therefore, seems like equation B is the answer. But let me check if there is any case where another equation could have a larger x. For example, equation E when r approaches 0. But when r approaches 0, equation E's base approaches infinity, so x approaches 0. So, equation E's x approaches 0. So, equation B's x approaches infinity? Wait, no. Wait, when r approaches 0, equation B's base approaches 1 + 0 = 1. So, the equation becomes 3*(1)^x = 7, which is 3 = 7, which is impossible. Wait, but in reality, as r approaches 0, the base approaches 1 from above. So, for very small r, the base is slightly larger than 1, so the equation 3*(1 + r/10)^x = 7 would have x approaching infinity as r approaches 0? Wait, let me think. If the base is approaching 1 from above, then (1 + ε)^x = 7/3, where ε approaches 0. So, taking logarithms, x = ln(7/3)/ln(1 + ε). As ε approaches 0, ln(1 + ε) ≈ ε. Therefore, x ≈ ln(7/3)/ε. Since ε = r/10, so x ≈ ln(7/3)/(r/10) = 10*ln(7/3)/r. As r approaches 0, x approaches infinity. Therefore, for very small r, equation B's x approaches infinity. Therefore, equation B's x can be made arbitrarily large by taking r approaching 0. Therefore, equation B has the largest x. Therefore, answer is B? But let me check if there is any other equation that can have larger x. For example, equation D. Let me check for very small r. For equation D, base is 1 + sqrt(r). For very small r, sqrt(r) is much larger than r/10. For example, if r approaches 0, sqrt(r) approaches 0, but slower than r/10? Wait, no. Wait, for very small r, sqrt(r) is larger than r/10. For example, take r = 0.01, sqrt(r) = 0.1, and r/10 = 0.001. So, sqrt(r) is larger. Therefore, base D is 1 + sqrt(r) ≈ 1 + 0.1 = 1.1, while base B is 1 + 0.001 = 1.001. Therefore, base B is smaller. Therefore, equation B's x is larger. Therefore, even for very small r, equation B's x is larger. Therefore, equation B is the one with the largest x. Therefore, answer is B. But let me check the answer options again. Wait, the answer options are A, B, C, D, E. So, according to my analysis, answer is B. But let me check again. Wait, but let me think again. Wait, the problem says 0 < r < 3. So, r can't be zero. But as r approaches zero, equation B's x approaches infinity. Therefore, equation B's x is the largest. Therefore, answer is B. However, let me check if there is any mistake in my reasoning. For example, maybe equation E? Wait, equation E is 3*(1 + 1/r)^x = 7. For r approaching 0, 1 + 1/r approaches infinity, so the base approaches infinity. Therefore, solving for x, we have (base)^x = 7/3. Therefore, x = log_{base}(7/3) = ln(7/3)/ln(base). As base approaches infinity, ln(base) approaches infinity, so x approaches 0. Therefore, equation E's x approaches 0. Therefore, equation E's x is small. Therefore, equation B's x is the largest. Therefore, answer is B. But let me check the answer options again. Wait, the answer options are given as A to E. Wait, but let me check if there is a case where equation D could have a larger x. For example, if base D is smaller than base B? Wait, for r approaching 0, base D approaches 1 + 0 = 1, same as base B. But for very small r, sqrt(r) is larger than r/10. Therefore, base D is larger than base B. Therefore, equation B's base is smaller, so equation B's x is larger. Therefore, equation B is the answer. Therefore, I think the answer is B. However, let me check the original problem again to make sure I didn't misread anything. The problem says "which equation has the largest solution x?" and the equations are:

A) 3(1 + r)^x = 7

B) 3(1 + r/10)^x = 7

C) 3(1 + 2r)^x = 7

D) 3(1 + sqrt(r))^x = 7

E) 3(1 + 1/r)^x = 7

Yes, that's correct. So, according to my analysis, answer is B. However, I need to make sure that there isn't a case where another equation could have a larger x. For example, equation D. Let me take r = 0.01. Then base D is 1 + sqrt(0.01) = 1 + 0.1 = 1.1. Base B is 1 + 0.01/10 = 1.001. So, base B is smaller. Therefore, equation B's x is larger. For equation D, solving 3*(1.1)^x = 7. So, x ≈ ln(7/3)/ln(1.1) ≈ 0.847 / 0.0953 ≈ 8.89. For equation B, solving 3*(1.001)^x = 7. So, x ≈ ln(7/3)/ln(1.001) ≈ 0.847 / 0.0009995 ≈ 847. So, way larger. Therefore, equation B is larger. Therefore, answer is B. Therefore, I think the answer is B. However, let me check if there is any other possible mistake. For example, maybe I confused the direction of the inequality? Let me think again. If base is smaller, then ln(base) is smaller, so x = ln(7/3)/ln(base) is larger. Therefore, yes, smaller base gives larger x. Therefore, equation B has the smallest base, so largest x. Therefore, answer is B. Therefore, I think the answer is B. However, let me check the answer options again. Wait, the answer options are labeled A to E. So, answer is B. But let me check once more with another value. Let me take r = 0.5. Then:

Equation A: 3*(1 + 0.5)^x = 7 => 3*(1.5)^x = 7 => 1.5^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(1.5) ≈ 0.847 / 0.405 ≈ 2.09.

Equation B: 3*(1 + 0.5/10)^x = 7 => 3*(1.05)^x = 7 => 1.05^x = 2.333. x ≈ ln(2.333)/ln(1.05) ≈ 0.847 / 0.04879 ≈ 17.37.

Equation C: 3*(1 + 2*0.5)^x = 7 => 3*(2)^x = 7 => 2^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(2) ≈ 0.847 / 0.693 ≈ 1.22.

Equation D: 3*(1 + sqrt(0.5))^x = 7. sqrt(0.5) ≈ 0.707, so base ≈ 1.707. x ≈ ln(2.333)/ln(1.707) ≈ 0.847 / 0.535 ≈ 1.58.

Equation E: 3*(1 + 1/0.5)^x = 7 => 3*(1 + 2)^x = 7 => 3*(3)^x = 7 => 3^x = 7/3 ≈ 2.333. x ≈ ln(2.333)/ln(3) ≈ 0.847 / 1.0986 ≈ 0.771.

So, for r = 0.5, equation B has x ≈ 17.37, which is the largest. Therefore, again, equation B is the largest. Therefore, I think my conclusion is correct. Therefore, the answer is B. However, let me check if the problem says "the largest solution x", and maybe there is a case where another equation could have a larger x? For example, equation E? For r approaching 0, equation E's x approaches 0, so no. Equation D? For r approaching 0, equation D's base approaches 1 + 0 = 1, so similar to equation B. But since base D is larger than base B for small r, equation B's x is larger. Therefore, equation B is the answer. Therefore, I think the answer is B. However, let me check once again if I made a mistake in the initial assumption. For example, if the base is between 0 and 1, then the logarithm is negative, and x would be negative. But since the problem says 0 < r < 3, let me check if any of the bases can be less than 1. For example, base E: 1 + 1/r. Since r is between 0 and 3, 1/r is positive, so base E is always greater than 1. Similarly, base D: 1 + sqrt(r). sqrt(r) is non-negative, so base D is always greater than 1. Base C: 1 + 2r. Since r > 0, base C is greater than 1. Base A: 1 + r, same. Base B: 1 + r/10, same. Therefore, all bases are greater than 1. Therefore, all equations have positive x. Therefore, no issues with negative x. Therefore, my previous analysis holds. Therefore, answer is B. Therefore, I think the correct answer is B.

**Final Answer**
\boxed{B}

To determine which equation has the **largest solution** for $ x $, we analyze the structure of each equation:

All equations are of the form:

$$
3 \cdot (\text{base})^x = 7
$$

We can solve for $ x $ by isolating it:

$$
x = \frac{\ln\left(\frac{7}{3}\right)}{\ln(\text{base})}
$$

This shows that the value of $ x $ is inversely proportional to the natural logarithm of the base. That is, **smaller bases** result in **larger values of $ x $**.

---

### Step 1: Identify the Bases

Let’s extract the base from each equation:

- **(A)** $ 3(1 + r)^x = 7 $ → base = $ 1 + r $
- **(B)** $ 3(1 + r/10)^x = 7 $ → base = $ 1 + \frac{r}{10} $
- **(C)** $ 3(1 + 2r)^x = 7 $ → base = $ 1 + 2r $
- **(D)** $ 3(1 + \sqrt{r})^x = 7 $ → base = $ 1 + \sqrt{r} $
- **(E)** $ 3(1 + 1/r)^x = 7 $ → base = $ 1 + \frac{1}{r} $

---

### Step 2: Compare the Bases

We are given that $ 0 < r < 3 $. Let’s analyze the behavior of each base in this interval:

- **Base A**: $ 1 + r $ → ranges from 1 to 4
- **Base B**: $ 1 + \frac{r}{10} $ → ranges from 1 to 1.3
- **Base C**: $ 1 + 2r $ → ranges from 1 to 7
- **Base D**: $ 1 + \sqrt{r} $ → ranges from 1 to $ 1 + \sqrt{3} \approx 2.732 $
- **Base E**: $ 1 + \frac{1}{r} $ → ranges from $ 1 + \frac{1}{3} \approx 1.333 $ to $ \infty $

---

### Step 3: Determine the Smallest Base

From the above, the **smallest base** is:

$$
\text{Base B: } 1 + \frac{r}{10}
$$

This is the **smallest** among all the bases for all $ r \in (0, 3) $, because:

- For small $ r $, $ \frac{r}{10} $ is much smaller than $ \sqrt{r} $, $ 2r $, or $ \frac{1}{r} $
- For large $ r $, $ \frac{r}{10} $ is still smaller than $ \sqrt{r} $, $ 2r $, or $ \frac{1}{r} $

Thus, **Base B is the smallest**.

---

### Step 4: Conclude the Largest $ x $

Since $ x = \frac{\ln(7/3)}{\ln(\text{base})} $, and the **base is smallest for equation B**, the **value of $ x $ is largest** for equation B.

---

### Final Answer

$$
\boxed{B}
$$

Solution:
Intuitively, $x$ will be largest for the option for which the value in the parentheses is smallest.

Formally, first note that each of the values in parentheses is larger than $1$. Now, each of the options is of the form $3f(r)^x = 7$. This can be rewritten as $x\log f(r) = \log\frac 73$. As $f(r)>1$, we have $\log f(r)>0$. Thus $x$ is the largest for the option for which $\log f(r)$ is smallest. As $\log f(r)$ is an increasing function, this is the option for which $f(r)$ is smallest.

We now get the following easier problem: Given that $0<r<3$, find the smallest value in the set $\{ 1+r, 1+r/10, 1+2r, 1+\sqrt r, 1+1/r\}$.

Clearly, $1+r/10$ is smaller than the first and the third option.

We have $r^2 < 10$, so dividing both sides by $10r$, we get $r/10 < 1/r$.

Finally, $r/100 < 1$, therefore $r^2/100 < r$. Since both sides are positive, we can take the square root and get $r/10 < \sqrt r$.

Thus the answer is $\boxed{\text{(B)}} 3(1 + r/10)^x = 7$.


Explanation:
Okay, so I need to find the range of the function y = log base 2 of the square root of sin x, where x is between 0 and 180 degrees. Let me start by recalling some basic properties of logarithmic functions and trigonometric functions to figure this out.

First, the function is y = log₂(√(sin x)). Let me rewrite that to make it easier to handle. Since the square root is the same as raising to the power of 1/2, I can write this as y = log₂((sin x)^(1/2)). Then, using the logarithm power rule, which says log_b(a^c) = c*log_b(a), I can bring the exponent 1/2 in front. So that becomes y = (1/2) * log₂(sin x). So, y = (1/2) log₂(sin x). That might be easier to work with.

Now, the next step is to figure out the domain of the original function, but the problem already specifies that x is between 0° and 180°, so I don't need to worry about that. However, I should check if there are any restrictions within that interval. The argument of the logarithm must be positive, so √(sin x) must be positive. Since square root of something is positive as long as the inside is positive, and sin x is positive in the interval 0° < x < 180°, except at the endpoints where sin x is zero. But since x is strictly between 0° and 180°, sin x is positive throughout. Therefore, the domain is indeed 0° < x < 180°, and the function is defined there.

Now, to find the range of y, I need to figure out all possible values that y can take as x varies between 0° and 180°. Let me think about how sin x behaves in this interval. The sine function starts at 0 when x approaches 0°, increases to 1 at x = 90°, and then decreases back to 0 at x approaching 180°. So sin x is in the interval (0, 1] for 0° < x < 180°. Therefore, sin x can take any value between 0 and 1, but not including 0 and 1? Wait, at x = 90°, sin x is exactly 1, so the maximum value of sin x is 1, and the minimum approaches 0. So sin x ∈ (0, 1].

Therefore, √(sin x) would be the square root of something in (0, 1]. The square root of numbers between 0 and 1 is also between 0 and 1, but since sin x is strictly greater than 0, √(sin x) is in (0, 1]. Therefore, √(sin x) ∈ (0, 1].

Now, taking the logarithm base 2 of that. Remember that log₂(z) is defined for z > 0. Since √(sin x) is in (0, 1], log₂(√(sin x)) is defined. Now, log₂(z) when z is between 0 and 1 is negative because log₂(1) = 0, and as z approaches 0 from the right, log₂(z) approaches negative infinity. So log₂(z) ∈ (-∞, 0] when z ∈ (0, 1]. Therefore, since √(sin x) ∈ (0, 1], then log₂(√(sin x)) ∈ (-∞, 0]. But wait, let me check that again.

Wait, if z approaches 0, log₂(z) approaches -infty, and when z = 1, log₂(1) = 0. Therefore, log₂(z) can take any value from -infty up to 0. Therefore, log₂(√(sin x)) ∈ (-∞, 0]. However, since √(sin x) can be equal to 1 only when sin x = 1, which occurs at x = 90°, so the maximum value of log₂(√(sin x)) is 0, achieved at x = 90°, and as x approaches 0° or 180°, sin x approaches 0, so √(sin x) approaches 0, and log₂(√(sin x)) approaches -infty. Therefore, the range of y is (-∞, 0].

But wait, the original function was y = (1/2) log₂(sin x). Let me check if that's equivalent. Let me verify:

Original function: y = log₂(√(sin x)) = log₂((sin x)^{1/2}) = (1/2) log₂(sin x). Yes, that's correct. So, if I consider sin x ∈ (0, 1], then log₂(sin x) is in (-infty, 0], so multiplying by 1/2, the range of y is also (-infty, 0]. Therefore, the range is all real numbers less than or equal to 0. But let me check if there are any restrictions here.

Wait, but maybe I need to check if there are any other constraints. For example, when sin x approaches 0, log₂(sin x) approaches -infty, so multiplying by 1/2, it still approaches -infty. When sin x is 1, log₂(1) = 0, so y = 0. Therefore, the range is (-infty, 0]. But let me think again if there's a mistake here.

Alternatively, maybe I can approach this by considering substitution. Let me set t = sin x, where t ∈ (0, 1]. Then, y = (1/2) log₂ t. So, since t ∈ (0, 1], log₂ t ∈ (-infty, 0], so multiplying by 1/2, y ∈ (-infty, 0]. Therefore, the range is (-infty, 0]. Therefore, the answer should be all real numbers y such that y ≤ 0. But let me check with specific values.

For example, when x = 90°, sin x = 1, so y = log₂(√1) = log₂(1) = 0. That's correct. When x approaches 0°, sin x approaches 0, so √(sin x) approaches 0, so log₂(√(sin x)) approaches log₂(0+) which is -infty. Similarly, when x approaches 180°, sin x approaches 0 again, so same result. Therefore, the function y can take any value from negative infinity up to 0. Therefore, the range is (-∞, 0].

But let me check if there's a possibility that the function might not reach all values in that interval. For example, is there a value in (-∞, 0] that isn't achieved? Let me think. Since sin x can take any value between 0 and 1, then t = sin x can be any value in (0, 1]. Therefore, log₂ t can take any value in (-infty, 0], so multiplying by 1/2, y can take any value in (-infty, 0]. Therefore, the range is indeed all real numbers less than or equal to 0.

But wait, let me check with another example. Suppose I want y = -1. Is there an x such that log₂(√(sin x)) = -1? Let's solve for sin x. Let me set log₂(√(sin x)) = -1. Then, √(sin x) = 2^{-1} = 1/2. Therefore, sin x = (1/2)^2 = 1/4. Since 1/4 is between 0 and 1, there are x values in (0°, 180°) where sin x = 1/4. For example, x = arcsin(1/4) which is approximately 14.5°, and also 180° - 14.5° = 165.5°, both within the interval. Therefore, y = -1 is attainable. Similarly, for any y ≤ 0, we can solve for sin x. Let me take an arbitrary y ≤ 0. Let me set y = (1/2) log₂(sin x). Then, multiplying both sides by 2: 2y = log₂(sin x). Then, exponentiating both sides with base 2: sin x = 2^{2y}. Since y ≤ 0, 2y ≤ 0, so 2^{2y} is between 0 and 1. Therefore, sin x = 2^{2y} ∈ (0, 1], which is valid. Therefore, for every y ≤ 0, there exists an x in (0°, 180°) such that y is achieved. Therefore, the range is indeed (-∞, 0].

But let me check if there's a mistake in the transformation. Let me go back again. Original function: y = log₂(√(sin x)). Let me think about the steps again. Since √(sin x) is positive, and sin x is between 0 and 1, then √(sin x) is between 0 and 1. Therefore, log₂ of something between 0 and 1 is negative. So y is negative or zero. So the range is all real numbers less than or equal to zero. Therefore, (-∞, 0].

But maybe the problem expects the answer in terms of intervals with specific notation? Let me check if there's any possible mistake here. For example, sometimes when dealing with logarithms, people might confuse the range, but I think my reasoning is correct.

Alternatively, maybe I can think of the function in terms of transformations. Let me consider the function step by step. Let me define f(x) = sin x, which has range (0, 1] for x in (0°, 180°). Then, g(t) = √t, which maps (0, 1] to (0, 1]. Then, h(z) = log₂ z, which maps (0, 1] to (-∞, 0]. Therefore, composing these functions, the range is (-∞, 0]. Therefore, the range of y is (-∞, 0].

Alternatively, maybe I can think of the function y = log₂(√(sin x)) as y = (1/2) log₂(sin x). So, if I let u = log₂(sin x), then y = u/2. Since sin x ∈ (0, 1], log₂(sin x) ∈ (-∞, 0], so u ∈ (-∞, 0], so y ∈ (-∞, 0]. Therefore, same conclusion.

Therefore, I think the range is all real numbers less than or equal to zero. Therefore, the answer is (-∞, 0], which in boxed form is \boxed{(-\infty, 0]}.

But let me check once more for errors. Suppose someone might think that since sin x is between 0 and 1, then √(sin x) is between 0 and 1, so log₂(√(sin x)) is between -infty and 0. But is there a case where log₂(√(sin x)) can be positive? For example, if √(sin x) > 1, then log₂(√(sin x)) would be positive. But since sin x is at most 1, √(sin x) is at most 1, so √(sin x) cannot be greater than 1. Therefore, log₂(√(sin x)) cannot be positive. Therefore, the maximum value is 0, and it goes to negative infinity. Therefore, the range is indeed (-∞, 0].

Another way to think about it: Let me consider the function y = log₂(√(sin x)). Let me write this as y = (1/2) log₂(sin x). Let me consider the function log₂(sin x). Since sin x is between 0 and 1, log₂(sin x) is between -infty and 0. Therefore, multiplying by 1/2, the range is still between -infty and 0. Therefore, same conclusion.

Alternatively, maybe I can use calculus to find the maximum and minimum values. Let me try that approach. Let me consider y = (1/2) log₂(sin x). To find the range, I can find the maximum and minimum values of y. Since sin x reaches its maximum at x = 90°, which is 1, so y = (1/2) log₂(1) = 0. That's the maximum value. As x approaches 0° or 180°, sin x approaches 0, so log₂(sin x) approaches -infty, so y approaches -infty. Therefore, the range is from -infty up to 0. Therefore, (-∞, 0].

Therefore, I think this confirms the same result. Therefore, I can be confident that the range is (-∞, 0].

But let me check if there's a possible mistake in assuming that all values between -infty and 0 are achieved. For example, suppose I pick a value y = -1000. Is there an x such that y = -1000? Let me solve for sin x. Starting with y = (1/2) log₂(sin x) = -1000. Multiply both sides by 2: log₂(sin x) = -2000. Then, sin x = 2^{-2000}. Since 2^{-2000} is a very small positive number, which is between 0 and 1, so there exists x in (0°, 180°) such that sin x = 2^{-2000}. Therefore, such x exists. Therefore, y can be as small as desired, approaching -infty, and can reach 0. Therefore, the range is indeed all real numbers less than or equal to 0.

Therefore, I think my answer is correct. The range is (-∞, 0].

**Final Answer**
The range of the function is \boxed{(-\infty, 0]}.

To determine the **range** of the function
$$
y = \log_2(\sqrt{\sin x}) \quad \text{for } 0^\circ < x < 180^\circ,
$$
we proceed step by step, analyzing the behavior of the function and its components.

---

### Step 1: Simplify the Function

We begin by simplifying the expression using logarithmic and exponent rules:

$$
y = \log_2(\sqrt{\sin x}) = \log_2((\sin x)^{1/2}) = \frac{1}{2} \log_2(\sin x)
$$

So, the function becomes:
$$
y = \frac{1}{2} \log_2(\sin x)
$$

---

### Step 2: Analyze the Domain of the Function

The function is defined only when the argument of the logarithm is positive. Since $ \sin x > 0 $ for $ 0^\circ < x < 180^\circ $, the function is well-defined in this interval.

---

### Step 3: Determine the Range of $ \sin x $

In the interval $ 0^\circ < x < 180^\circ $, the sine function behaves as follows:

- $ \sin x $ increases from 0 to 1 as $ x $ goes from $ 0^\circ $ to $ 90^\circ $
- $ \sin x $ decreases back to 0 as $ x $ goes from $ 90^\circ $ to $ 180^\circ $

Thus, the range of $ \sin x $ in this interval is:
$$
\sin x \in (0, 1]
$$

---

### Step 4: Analyze the Behavior of $ \log_2(\sin x) $

Since $ \sin x \in (0, 1] $, we consider the logarithm base 2 of values in this interval:

- $ \log_2(1) = 0 $
- $ \log_2(z) \to -\infty $ as $ z \to 0^+ $

Therefore:
$$
\log_2(\sin x) \in (-\infty, 0]
$$

---

### Step 5: Apply the Factor of $ \frac{1}{2} $

Now, since:
$$
y = \frac{1}{2} \log_2(\sin x)
$$
and $ \log_2(\sin x) \in (-\infty, 0] $, multiplying by $ \frac{1}{2} $ scales the interval:

$$
y \in (-\infty, 0]
$$

---

### Step 6: Confirm the Range

- The **maximum** value of $ y $ occurs when $ \sin x = 1 $, which gives $ y = \frac{1}{2} \log_2(1) = 0 $
- The **minimum** value of $ y $ approaches $ -\infty $ as $ \sin x \to 0^+ $

Thus, the function $ y = \log_2(\sqrt{\sin x}) $ can take **any real value less than or equal to 0**.

---

### Final Answer

$$
\boxed{(-\infty, 0]}
$$

[Assistant]
Explanation:
Okay, so I need to find the range of the function y = log base 2 of the square root of sin x, where x is between 0 and 180 degrees. Let me start by recalling some basic properties of logarithmic functions and trigonometric functions to figure this out.

First, the function is y = log₂(√(sin x)). Let me rewrite that to make it easier to handle. Since the square root is the same as raising to the power of 1/2, I can write this as y = log₂((sin x)^(1/2)). Then, using the logarithm power rule, which says log_b(a^c) = c*log_b(a), I can bring the exponent 1/2 in front. So that becomes y = (1/2) * log₂(sin x). So, y = (1/2) log₂(sin x). That might be easier to work with.

Now, the next step is to figure out the domain of the original function, but the problem already specifies that x is between 0° and 180°, so I don't need to worry about that. However, I should check if there are any restrictions within that interval. The argument of the logarithm must be positive, so √(sin x) must be positive. Since square root of something is positive as long as the inside is positive, and sin x is positive in the interval 0° < x < 180°, except at the endpoints where sin x is zero. But since x is strictly between 0° and 180°, sin x is positive throughout. Therefore, the domain is indeed 0° < x < 180°, and the function is defined there.

Now, to find the range of y, I need to figure out all possible values that y can take as x varies between 0° and 180°. Let me think about how sin x behaves in this interval. The sine function starts at 0 when x approaches 0°, increases to 1 at x = 90°, and then decreases back to 0 at x approaching 180°. So sin x is in the interval (0, 1] for 0° < x < 180°. Therefore, sin x can take any value between 0 and 1, but not including 0 and 1? Wait, at x = 90°, sin x is exactly 1, so the maximum value of sin x is 1, and the minimum approaches 0. So sin x ∈ (0, 1].

Therefore, √(sin x) would be the square root of something in (0, 1]. The square root of numbers between 0 and 1 is also between 0 and 1, but since sin x is strictly greater than 0, √(sin x) is in (0, 1]. Therefore, √(sin x) ∈ (0, 1].

Now, taking the logarithm base 2 of that. Remember that log₂(z) is defined for z > 0. Since √(sin x) is in (0, 1], log₂(√(sin x)) is defined. Now, log₂(z) when z is between 0 and 1 is negative because log₂(1) = 0, and as z approaches 0 from the right, log₂(z) approaches negative infinity. So log₂(z) ∈ (-∞, 0] when z ∈ (0, 1]. Therefore, since √(sin x) ∈ (0, 1], then log₂(√(sin x)) ∈ (-∞, 0]. But wait, let me check that again.

Wait, if z approaches 0, log₂(z) approaches -infty, and when z = 1, log₂(1) = 0. Therefore, log₂(z) can take any value from -infty up to 0. Therefore, log₂(√(sin x)) ∈ (-∞, 0]. However, since √(sin x) can be equal to 1 only when sin x = 1, which occurs at x = 90°, so the maximum value of log₂(√(sin x)) is 0, achieved at x = 90°, and as x approaches 0° or 180°, sin x approaches 0, so √(sin x) approaches 0, and log₂(√(sin x)) approaches -infty. Therefore, the range of y is (-∞, 0].

But wait, the original function was y = (1/2) log₂(sin x). Let me check if that's equivalent. Let me verify:

Original function: y = log₂(√(sin x)) = log₂((sin x)^{1/2}) = (1/2) log₂(sin x). Yes, that's correct. So, if I consider sin x ∈ (0, 1], then log₂(sin x) is in (-infty, 0], so multiplying by 1/2, the range of y is also (-infty, 0]. Therefore, the range is all real numbers less than or equal to 0. But let me check if there are any restrictions here.

Wait, but maybe I need to check if there are any other constraints. For example, when sin x approaches 0, log₂(sin x) approaches -infty, so multiplying by 1/2, it still approaches -infty. When sin x is 1, log₂(1) = 0, so y = 0. Therefore, the range is (-infty, 0]. But let me think again if there's a mistake here.

Alternatively, maybe I can approach this by considering substitution. Let me set t = sin x, where t ∈ (0, 1]. Then, y = (1/2) log₂ t. So, since t ∈ (0, 1], log₂ t ∈ (-infty, 0], so multiplying by 1/2, y ∈ (-infty, 0]. Therefore, the range is (-infty, 0]. Therefore, the answer should be all real numbers y such that y ≤ 0. But let me check with specific values.

For example, when x = 90°, sin x = 1, so y = log₂(√1) = log₂(1) = 0. That's correct. When x approaches 0°, sin x approaches 0, so √(sin x) approaches 0, so log₂(√(sin x)) approaches log₂(0+) which is -infty. Similarly, when x approaches 180°, sin x approaches 0 again, so same result. Therefore, the function y can take any value from negative infinity up to 0. Therefore, the range is (-∞, 0].

But let me check if there's a possibility that the function might not reach all values in that interval. For example, is there a value in (-∞, 0] that isn't achieved? Let me think. Since sin x can take any value between 0 and 1, then t = sin x can be any value in (0, 1]. Therefore, log₂ t can take any value in (-infty, 0], so multiplying by 1/2, y can take any value in (-infty, 0]. Therefore, the range is indeed all real numbers less than or equal to 0.

But wait, let me check with another example. Suppose I want y = -1. Is there an x such that log₂(√(sin x)) = -1? Let's solve for sin x. Let me set log₂(√(sin x)) = -1. Then, √(sin x) = 2^{-1} = 1/2. Therefore, sin x = (1/2)^2 = 1/4. Since 1/4 is between 0 and 1, there are x values in (0°, 180°) where sin x = 1/4. For example, x = arcsin(1/4) which is approximately 14.5°, and also 180° - 14.5° = 165.5°, both within the interval. Therefore, y = -1 is attainable. Similarly, for any y ≤ 0, we can solve for sin x. Let me take an arbitrary y ≤ 0. Let me set y = (1/2) log₂(sin x). Then, multiplying both sides by 2: 2y = log₂(sin x). Then, exponentiating both sides with base 2: sin x = 2^{2y}. Since y ≤ 0, 2y ≤ 0, so 2^{2y} is between 0 and 1. Therefore, sin x = 2^{2y} ∈ (0, 1], which is valid. Therefore, for every y ≤ 0, there exists an x in (0°, 180°) such that y is achieved. Therefore, the range is indeed (-∞, 0].

But let me check if there's a mistake in the transformation. Let me go back again. Original function: y = log₂(√(sin x)). Let me think about the steps again. Since √(sin x) is positive, and sin x is between 0 and 1, then √(sin x) is between 0 and 1. Therefore, log₂ of something between 0 and 1 is negative. So y is negative or zero. So the range is all real numbers less than or equal to zero. Therefore, (-∞, 0].

But maybe the problem expects the answer in terms of intervals with specific notation? Let me check if there's any possible mistake here. For example, sometimes when dealing with logarithms, people might confuse the range, but I think my reasoning is correct.

Alternatively, maybe I can think of the function in terms of transformations. Let me consider the function step by step. Let me define f(x) = sin x, which has range (0, 1] for x in (0°, 180°). Then, g(t) = √t, which maps (0, 1] to (0, 1]. Then, h(z) = log₂ z, which maps (0, 1] to (-∞, 0]. Therefore, composing these functions, the range is (-∞, 0]. Therefore, the range of y is (-∞, 0].

Alternatively, maybe I can think of the function y = log₂(√(sin x)) as y = (1/2) log₂(sin x). So, if I let u = log₂(sin x), then y = u/2. Since sin x ∈ (0, 1], log₂(sin x) ∈ (-∞, 0], so u ∈ (-∞, 0], so y ∈ (-∞, 0]. Therefore, same conclusion.

Therefore, I think the range is all real numbers less than or equal to zero. Therefore, the answer is (-∞, 0], which in boxed form is \boxed{(-\infty, 0]}.

But let me check once more for errors. Suppose someone might think that since sin x is between 0 and 1, then √(sin x) is between 0 and 1, so log₂(√(sin x)) is between -infty and 0. But is there a case where log₂(√(sin x)) can be positive? For example, if √(sin x) > 1, then log₂(√(sin x)) would be positive. But since sin x is at most 1, √(sin x) is at most 1, so √(sin x) cannot be greater than 1. Therefore, log₂(√(sin x)) cannot be positive. Therefore, the maximum value is 0, and it goes to negative infinity. Therefore, the range is indeed (-∞, 0].

Another way to think about it: Let me consider the function y = log₂(√(sin x)). Let me write this as y = (1/2) log₂(sin x). Let me consider the function log₂(sin x). Since sin x is between 0 and 1, log₂(sin x) is between -infty and 0. Therefore, multiplying by 1/2, the range is still between -infty and 0. Therefore, same conclusion.

Alternatively, maybe I can use calculus to find the maximum and minimum values. Let me try that approach. Let me consider y = (1/2) log₂(sin x). To find the range, I can find the maximum and minimum values of y. Since sin x reaches its maximum at x = 90°, which is 1, so y = (1/2) log₂(1) = 0. That's the maximum value. As x approaches 0° or 180°, sin x approaches 0, so log₂(sin x) approaches -infty, so y approaches -infty. Therefore, the range is from -infty up to 0. Therefore, (-∞, 0].

Therefore, I think this confirms the same result. Therefore, I can be confident that the range is (-∞, 0].

But let me check if there's a possible mistake in assuming that all values between -infty and 0 are achieved. For example, suppose I pick a value y = -1000. Is there an x such that y = -1000? Let me solve for sin x. Starting with y = (1/2) log₂(sin x) = -1000. Multiply both sides by 2: log₂(sin x) = -2000. Then, sin x = 2^{-2000}. Since 2^{-2000} is a very small positive number, which is between 0 and 1, so there exists x in (0°, 180°) such that sin x = 2^{-2000}. Therefore, such x exists. Therefore, y can be as small as desired, approaching -infty, and can reach 0. Therefore, the range is indeed all real numbers less than or equal to 0.

Therefore, I think my answer is correct. The range is (-∞, 0].

**Final Answer**
The range of the function is \boxed{(-\infty, 0]}.

To determine the **range** of the function
$$
y = \log_2(\sqrt{\sin x}) \quad \text{for } 0^\circ < x < 180^\circ,
$$
we proceed step by step, analyzing the behavior of the function and its components.

---

### Step 1: Simplify the Function

We begin by simplifying the expression using logarithmic and exponent rules:

$$
y = \log_2(\sqrt{\sin x}) = \log_2((\sin x)^{1/2}) = \frac{1}{2} \log_2(\sin x)
$$

So, the function becomes:
$$
y = \frac{1}{2} \log_2(\sin x)
$$

---

### Step 2: Analyze the Domain of the Function

The function is defined only when the argument of the logarithm is positive. Since $ \sin x > 0 $ for $ 0^\circ < x < 180^\circ $, the function is well-defined in this interval.

---

### Step 3: Determine the Range of $ \sin x $

In the interval $ 0^\circ < x < 180^\circ $, the sine function behaves as follows:

- $ \sin x $ increases from 0 to 1 as $ x $ goes from $ 0^\circ $ to $ 90^\circ $
- $ \sin x $ decreases back to 0 as $ x $ goes from $ 90^\circ $ to $ 180^\circ $

Thus, the range of $ \sin x $ in this interval is:
$$
\sin x \in (0, 1]
$$

---

### Step 4: Analyze the Behavior of $ \log_2(\sin x) $

Since $ \sin x \in (0, 1] $, we consider the logarithm base 2 of values in this interval:

- $ \log_2(1) = 0 $
- $ \log_2(z) \to -\infty $ as $ z \to 0^+ $

Therefore:
$$
\log_2(\sin x) \in (-\infty, 0]
$$

---

### Step 5: Apply the Factor of $ \frac{1}{2} $

Now, since:
$$
y = \frac{1}{2} \log_2(\sin x)
$$
and $ \log_2(\sin x) \in (-\infty, 0] $, multiplying by $ \frac{1}{2} $ scales the interval:

$$
y \in (-\infty, 0]
$$

---

### Step 6: Confirm the Range

- The **maximum** value of $ y $ occurs when $ \sin x = 1 $, which gives $ y = \frac{1}{2} \log_2(1) = 0 $
- The **minimum** value of $ y $ approaches $ -\infty $ as $ \sin x \to 0^+ $

Thus, the function $ y = \log_2(\sqrt{\sin x}) $ can take **any real value less than or equal to 0**.

---

### Final Answer

$$
\boxed{(-\infty, 0]}
$$

Solution: Between 0 and 180 degrees, the value of $\sin x$ is between 0 (exclusive) and 1 (inclusive). Thus, the value of $\sqrt{\sin x}$ is between 0 (exclusive) and 1 (inclusive). Since the range of $\log_2 x$ for $0<x\le1$ is all non-positive numbers, the range of the entire function is all non-positive numbers, or $x \in \boxed{(-\infty, 0]}$.


Explanation:
Okay, so I need to figure out the value of this operation @ between two fractions: 7/30 and 10/21. The operation is defined as (m/n) @ (p/q) = m*p*(q/n). Let me make sure I understand the definition correctly.

First, let me parse the definition again. If I have two fractions, the first one is m/n and the second one is p/q. Then the operation @ takes these two fractions and multiplies m by p, then multiplies that by q divided by n. So, in symbols, it's m*p*(q/n). Let me write that down to visualize it better:

(m/n) @ (p/q) = m * p * (q / n)

So, substituting the given fractions, which are 7/30 and 10/21, into this formula. Let me identify m, n, p, q.

For the first fraction, 7/30, m is 7 and n is 30. For the second fraction, 10/21, p is 10 and q is 21. So substituting these into the formula:

= 7 * 10 * (21 / 30)

Okay, so that simplifies to 7*10*(21/30). Let me compute this step by step.

First, let me compute 7*10. That's straightforward: 7*10 = 70.

Then, I have to multiply that by 21/30. So, 70*(21/30). Let me write that as (70*21)/30.

Now, I can simplify this fraction before multiplying. Let me see if 70 and 30 have a common factor. 70 divided by 10 is 7, and 30 divided by 10 is 3. So, 70/30 simplifies to 7/3. Therefore, (70*21)/30 = (7/3)*21.

Wait, let me check that again. If I have (70*21)/30, I can divide 70 by 30 first. But 70 divided by 30 is 7/3, so then multiplying by 21 gives (7/3)*21. Alternatively, I can divide 21 by 30 first? Wait, maybe I should factor the numbers to see if there are common factors.

Alternatively, maybe I can simplify 70 and 30 with 21? Let me think. Let me factor all numbers:

70 = 7 * 10 = 7 * 2 * 5

21 = 3 * 7

30 = 3 * 10 = 3 * 2 * 5

So, the numerator is 70*21 = (7*2*5)*(3*7) = 7^2 * 2 * 3 * 5

Denominator is 30 = 3 * 2 * 5

So, when we divide numerator by denominator, we have:

(7^2 * 2 * 3 * 5) / (3 * 2 * 5) = 7^2

Because the 3, 2, and 5 in the denominator cancel out with the same factors in the numerator. So, that leaves 7^2, which is 49.

Wait, so that would mean the entire expression simplifies to 49? Let me check that again because that seems straightforward, but let me verify with another approach.

Alternatively, let me compute 70*(21/30). Let me compute 21 divided by 30 first. 21/30 simplifies to 7/10. So, 70*(7/10). Then, 70 divided by 10 is 7, so 7*7 = 49. Yes, that's the same result.

So, both methods give me 49. Therefore, the value of (7/30) @ (10/21) is 49.

But wait, the problem says "for simplified fractions p/q". Wait, does that mean that the fractions need to be simplified before applying the operation? Let me check the original problem again.

"The operation @ is defined as (m/n) @ (p/q) = (m)(p)(q/n) for simplified fractions p/q."

Wait, the wording is a bit confusing. It says "for simplified fractions p/q". Maybe that means that the fractions p/q are simplified? Or maybe that the operation is defined for simplified fractions? Let me check again.

Original problem: "The operation @ is defined as (m/n) @ (p/q) = (m)(p)(q/n) for simplified fractions p/q."

Hmm. Maybe it's saying that the operation is defined for simplified fractions, meaning that m/n and p/q are simplified fractions? Or maybe that p/q is simplified? The wording is a bit unclear. But since the problem gives us two fractions, 7/30 and 10/21, and asks for the operation, maybe we need to check if they are simplified.

Let me check if 7/30 is simplified. The numerator is 7, denominator 30. The GCD of 7 and 30 is 1, so yes, it's simplified. Similarly, 10/21: GCD of 10 and 21 is 1, so that's simplified. So, since both fractions are already simplified, we can proceed with the operation as defined.

Therefore, my calculation seems correct. Let me recap:

(7/30) @ (10/21) = 7 * 10 * (21 / 30) = 70 * (7/10) = 49.

Alternatively, simplifying before multiplying:

70*(21/30) = (70/30)*21 = (7/3)*21 = 7*7 = 49.

Yes, that's correct. Therefore, the simplified value is 49.

But just to make sure I didn't misinterpret the operation. Let me check the definition again. The operation is (m/n) @ (p/q) = m*p*(q/n). So, m is numerator of first fraction, n is denominator, p is numerator of second fraction, q is denominator of second fraction. So, yes, substituting m=7, n=30, p=10, q=21. So, 7*10*(21/30).

Another way to think about it: maybe the operation is m*p*(q/n). So, m*p is 7*10=70, and then multiplied by q/n which is 21/30. So, 70*(21/30). Which is same as before.

Alternatively, maybe there's a different interpretation? For example, is the operation (m/n) @ (p/q) = m * p * (q/n) or is there a different grouping? Like, maybe m * (p * (q/n))? But that's the same as m*p*(q/n).

Alternatively, could it be m * (p * q) / n? Which would be same as m*p*q / n. Wait, but the original definition is written as (m)(p)(q/n). So, that is m*p*(q/n). Which is equal to (m*p*q)/n.

So, if I compute (m*p*q)/n, that would be (7*10*21)/30. Let me compute that.

7*10 is 70, 70*21 is... Let me compute 70*20=1400, plus 70*1=70, so total 1470. Then divide by 30. 1470 divided by 30. Let me compute that.

Divide numerator and denominator by 10: 147 / 3 = 49. Yes, same result.

So, regardless of the method, I get 49. Therefore, I think that is the correct answer.

But just to make sure, let me check if there's any step where I could have made a mistake. For example, maybe I misread the operation? Let me check again.

Original definition: (m/n) @ (p/q) = m*p*(q/n). So, m is numerator of first fraction, p is numerator of second fraction, q is denominator of second fraction, n is denominator of first fraction.

Yes, so substituting m=7, n=30, p=10, q=21. So, 7*10*(21/30).

Alternatively, if someone misreads the operation as m*p*q/n, which is same as above.

Alternatively, if someone thought that the operation is m/n * p/q * something? But no, the definition is explicit.

Alternatively, maybe the operation is (m/n) @ (p/q) = m * p * (q/n). So, m*p is 7*10=70, and then multiplied by q/n which is 21/30.

Yes, so 70*(21/30).

Alternatively, if someone thought that the operation is m/n * p/q * something else? But no, the definition is clear.

Alternatively, maybe the problem says "simplified fractions p/q" meaning that the result should be simplified? But in this case, the result is 49, which is an integer, so it's already simplified.

Alternatively, maybe there's a different interpretation of the operation? For example, if the operation was (m/n) @ (p/q) = m * p * (q/n) where the fractions are simplified? But in this case, the fractions are already simplified.

Alternatively, maybe the operation is defined for simplified fractions, so if the fractions weren't simplified, we need to simplify them first? But in this case, both fractions are already simplified.

Therefore, I think my calculation is correct.

But just to make sure, let me do the calculation once again:

First, compute m*p*q/n:

m = 7, p = 10, q = 21, n = 30.

So, 7 * 10 * 21 / 30.

Compute numerator: 7*10=70; 70*21=1470.

Divide by 30: 1470 / 30.

Divide numerator and denominator by 10: 147 / 3 = 49.

Yes, same result.

Alternatively, simplifying before multiplying:

Note that 21 and 30 have a common factor of 3. So, 21/30 = 7/10. Then, 7*10*(7/10).

Wait, that's another way to look at it. Let me see:

Original expression: 7*10*(21/30).

Note that 21/30 can be simplified to 7/10. Therefore, substituting that in:

7*10*(7/10).

Now, 10 in the numerator and denominator cancels out:

7*(7) = 49.

Yes, that's another way to see it. So, that's even simpler.

Therefore, regardless of the method, the answer is 49.

Therefore, I think the answer is 49.

**Final Answer**
\boxed{49}

To evaluate the expression $\frac{7}{30}@\frac{10}{21}$ using the defined operation:

$$
\frac{m}{n}@\frac{p}{q} = m \cdot p \cdot \left(\frac{q}{n}\right)
$$

we first identify the components from the given fractions:

- $m = 7$, $n = 30$
- $p = 10$, $q = 21$

Substituting these into the operation:

$$
\frac{7}{30}@\frac{10}{21} = 7 \cdot 10 \cdot \left(\frac{21}{30}\right)
$$

---

### Step 1: Multiply the numerators

$$
7 \cdot 10 = 70
$$

---

### Step 2: Multiply by the fraction $\frac{21}{30}$

$$
70 \cdot \left(\frac{21}{30}\right) = \frac{70 \cdot 21}{30}
$$

---

### Step 3: Simplify the fraction

We simplify $\frac{70 \cdot 21}{30}$ by factoring:

- $70 = 7 \cdot 10$
- $21 = 3 \cdot 7$
- $30 = 3 \cdot 10$

So:

$$
\frac{70 \cdot 21}{30} = \frac{(7 \cdot 10) \cdot (3 \cdot 7)}{3 \cdot 10}
$$

Cancel common factors:

- $10$ in numerator and denominator cancels
- $3$ in numerator and denominator cancels

Leaving:

$$
\frac{7 \cdot 7}{1} = 49
$$

---

### Final Answer

$$
\boxed{49}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the value of this operation @ between two fractions: 7/30 and 10/21. The operation is defined as (m/n) @ (p/q) = m*p*(q/n). Let me make sure I understand the definition correctly.

First, let me parse the definition again. If I have two fractions, the first one is m/n and the second one is p/q. Then the operation @ takes these two fractions and multiplies m by p, then multiplies that by q divided by n. So, in symbols, it's m*p*(q/n). Let me write that down to visualize it better:

(m/n) @ (p/q) = m * p * (q / n)

So, substituting the given fractions, which are 7/30 and 10/21, into this formula. Let me identify m, n, p, q.

For the first fraction, 7/30, m is 7 and n is 30. For the second fraction, 10/21, p is 10 and q is 21. So substituting these into the formula:

= 7 * 10 * (21 / 30)

Okay, so that simplifies to 7*10*(21/30). Let me compute this step by step.

First, let me compute 7*10. That's straightforward: 7*10 = 70.

Then, I have to multiply that by 21/30. So, 70*(21/30). Let me write that as (70*21)/30.

Now, I can simplify this fraction before multiplying. Let me see if 70 and 30 have a common factor. 70 divided by 10 is 7, and 30 divided by 10 is 3. So, 70/30 simplifies to 7/3. Therefore, (70*21)/30 = (7/3)*21.

Wait, let me check that again. If I have (70*21)/30, I can divide 70 by 30 first. But 70 divided by 30 is 7/3, so then multiplying by 21 gives (7/3)*21. Alternatively, I can divide 21 by 30 first? Wait, maybe I should factor the numbers to see if there are common factors.

Alternatively, maybe I can simplify 70 and 30 with 21? Let me think. Let me factor all numbers:

70 = 7 * 10 = 7 * 2 * 5

21 = 3 * 7

30 = 3 * 10 = 3 * 2 * 5

So, the numerator is 70*21 = (7*2*5)*(3*7) = 7^2 * 2 * 3 * 5

Denominator is 30 = 3 * 2 * 5

So, when we divide numerator by denominator, we have:

(7^2 * 2 * 3 * 5) / (3 * 2 * 5) = 7^2

Because the 3, 2, and 5 in the denominator cancel out with the same factors in the numerator. So, that leaves 7^2, which is 49.

Wait, so that would mean the entire expression simplifies to 49? Let me check that again because that seems straightforward, but let me verify with another approach.

Alternatively, let me compute 70*(21/30). Let me compute 21 divided by 30 first. 21/30 simplifies to 7/10. So, 70*(7/10). Then, 70 divided by 10 is 7, so 7*7 = 49. Yes, that's the same result.

So, both methods give me 49. Therefore, the value of (7/30) @ (10/21) is 49.

But wait, the problem says "for simplified fractions p/q". Wait, does that mean that the fractions need to be simplified before applying the operation? Let me check the original problem again.

"The operation @ is defined as (m/n) @ (p/q) = (m)(p)(q/n) for simplified fractions p/q."

Wait, the wording is a bit confusing. It says "for simplified fractions p/q". Maybe that means that the fractions p/q are simplified? Or maybe that the operation is defined for simplified fractions? Let me check again.

Original problem: "The operation @ is defined as (m/n) @ (p/q) = (m)(p)(q/n) for simplified fractions p/q."

Hmm. Maybe it's saying that the operation is defined for simplified fractions, meaning that m/n and p/q are simplified fractions? Or maybe that p/q is simplified? The wording is a bit unclear. But since the problem gives us two fractions, 7/30 and 10/21, and asks for the operation, maybe we need to check if they are simplified.

Let me check if 7/30 is simplified. The numerator is 7, denominator 30. The GCD of 7 and 30 is 1, so yes, it's simplified. Similarly, 10/21: GCD of 10 and 21 is 1, so that's simplified. So, since both fractions are already simplified, we can proceed with the operation as defined.

Therefore, my calculation seems correct. Let me recap:

(7/30) @ (10/21) = 7 * 10 * (21 / 30) = 70 * (7/10) = 49.

Alternatively, simplifying before multiplying:

70*(21/30) = (70/30)*21 = (7/3)*21 = 7*7 = 49.

Yes, that's correct. Therefore, the simplified value is 49.

But just to make sure I didn't misinterpret the operation. Let me check the definition again. The operation is (m/n) @ (p/q) = m*p*(q/n). So, m is numerator of first fraction, n is denominator, p is numerator of second fraction, q is denominator of second fraction. So, yes, substituting m=7, n=30, p=10, q=21. So, 7*10*(21/30).

Another way to think about it: maybe the operation is m*p*(q/n). So, m*p is 7*10=70, and then multiplied by q/n which is 21/30. So, 70*(21/30). Which is same as before.

Alternatively, maybe there's a different interpretation? For example, is the operation (m/n) @ (p/q) = m * p * (q/n) or is there a different grouping? Like, maybe m * (p * (q/n))? But that's the same as m*p*(q/n).

Alternatively, could it be m * (p * q) / n? Which would be same as m*p*q / n. Wait, but the original definition is written as (m)(p)(q/n). So, that is m*p*(q/n). Which is equal to (m*p*q)/n.

So, if I compute (m*p*q)/n, that would be (7*10*21)/30. Let me compute that.

7*10 is 70, 70*21 is... Let me compute 70*20=1400, plus 70*1=70, so total 1470. Then divide by 30. 1470 divided by 30. Let me compute that.

Divide numerator and denominator by 10: 147 / 3 = 49. Yes, same result.

So, regardless of the method, I get 49. Therefore, I think that is the correct answer.

But just to make sure, let me check if there's any step where I could have made a mistake. For example, maybe I misread the operation? Let me check again.

Original definition: (m/n) @ (p/q) = m*p*(q/n). So, m is numerator of first fraction, p is numerator of second fraction, q is denominator of second fraction, n is denominator of first fraction.

Yes, so substituting m=7, n=30, p=10, q=21. So, 7*10*(21/30).

Alternatively, if someone misreads the operation as m*p*q/n, which is same as above.

Alternatively, if someone thought that the operation is m/n * p/q * something? But no, the definition is explicit.

Alternatively, maybe the operation is (m/n) @ (p/q) = m * p * (q/n). So, m*p is 7*10=70, and then multiplied by q/n which is 21/30.

Yes, so 70*(21/30).

Alternatively, if someone thought that the operation is m/n * p/q * something else? But no, the definition is clear.

Alternatively, maybe the problem says "simplified fractions p/q" meaning that the result should be simplified? But in this case, the result is 49, which is an integer, so it's already simplified.

Alternatively, maybe there's a different interpretation of the operation? For example, if the operation was (m/n) @ (p/q) = m * p * (q/n) where the fractions are simplified? But in this case, the fractions are already simplified.

Alternatively, maybe the operation is defined for simplified fractions, so if the fractions weren't simplified, we need to simplify them first? But in this case, both fractions are already simplified.

Therefore, I think my calculation is correct.

But just to make sure, let me do the calculation once again:

First, compute m*p*q/n:

m = 7, p = 10, q = 21, n = 30.

So, 7 * 10 * 21 / 30.

Compute numerator: 7*10=70; 70*21=1470.

Divide by 30: 1470 / 30.

Divide numerator and denominator by 10: 147 / 3 = 49.

Yes, same result.

Alternatively, simplifying before multiplying:

Note that 21 and 30 have a common factor of 3. So, 21/30 = 7/10. Then, 7*10*(7/10).

Wait, that's another way to look at it. Let me see:

Original expression: 7*10*(21/30).

Note that 21/30 can be simplified to 7/10. Therefore, substituting that in:

7*10*(7/10).

Now, 10 in the numerator and denominator cancels out:

7*(7) = 49.

Yes, that's another way to see it. So, that's even simpler.

Therefore, regardless of the method, the answer is 49.

Therefore, I think the answer is 49.

**Final Answer**
\boxed{49}

To evaluate the expression $\frac{7}{30}@\frac{10}{21}$ using the defined operation:

$$
\frac{m}{n}@\frac{p}{q} = m \cdot p \cdot \left(\frac{q}{n}\right)
$$

we first identify the components from the given fractions:

- $m = 7$, $n = 30$
- $p = 10$, $q = 21$

Substituting these into the operation:

$$
\frac{7}{30}@\frac{10}{21} = 7 \cdot 10 \cdot \left(\frac{21}{30}\right)
$$

---

### Step 1: Multiply the numerators

$$
7 \cdot 10 = 70
$$

---

### Step 2: Multiply by the fraction $\frac{21}{30}$

$$
70 \cdot \left(\frac{21}{30}\right) = \frac{70 \cdot 21}{30}
$$

---

### Step 3: Simplify the fraction

We simplify $\frac{70 \cdot 21}{30}$ by factoring:

- $70 = 7 \cdot 10$
- $21 = 3 \cdot 7$
- $30 = 3 \cdot 10$

So:

$$
\frac{70 \cdot 21}{30} = \frac{(7 \cdot 10) \cdot (3 \cdot 7)}{3 \cdot 10}
$$

Cancel common factors:

- $10$ in numerator and denominator cancels
- $3$ in numerator and denominator cancels

Leaving:

$$
\frac{7 \cdot 7}{1} = 49
$$

---

### Final Answer

$$
\boxed{49}
$$

Solution: We have $\frac{7}{30}@\frac{10}{21}=(7)(10)\left(\frac{21}{30}\right)=\boxed{49}$.


Explanation:
Okay, so I need to figure out where the point (0,0) ends up after being reflected over the vertical line x=1 and then over the line y=2. Let me start by recalling what reflection over a vertical line does.

First, reflecting a point over a vertical line x = a. I think the formula for that is something like if you have a point (x, y), its reflection over x = a would be (2a - x, y). Let me check that. If the original point is to the left of the line, then the reflection should be the same distance on the other side. For example, if I reflect (0,0) over x=1, then the distance from 0 to 1 is 1 unit, so the reflection should be 1 unit to the right of x=1, which would be x=2. So the reflected point would be (2, 0). That seems right. So applying that formula, 2a - x where a is 1 and x is 0, so 2*1 - 0 = 2. Yep, that gives (2, 0).

Okay, so first reflection takes (0,0) to (2, 0). Now, the next step is reflecting that image over the line y=2. So now I need to figure out how to reflect a point over a horizontal line y = b. I think the formula is similar, but for the y-coordinate. If you have a point (x, y), its reflection over y = b would be (x, 2b - y). Let me verify. If the original point is below the line, then the reflection should be the same distance above. For example, reflecting (0,0) over y=2 would be 2 units up from y=2, which is y=4. So the reflected point would be (0, 4). Using the formula, 2*2 - 0 = 4, so (0, 4). That checks out.

So applying that to the point (2, 0) which we got after the first reflection. So reflecting over y=2, the new y-coordinate would be 2*2 - 0 = 4. Therefore, the resulting point is (2, 4).

Wait, let me make sure I didn't mix up anything. Let me go through each step again.

Original point: (0, 0).

First reflection over x=1. The formula for reflection over vertical line x=a is (2a - x, y). So substituting a=1, x=0, y=0: 2*1 - 0 = 2, so (2, 0). That seems correct.

Then reflecting (2, 0) over y=2. The formula for reflection over horizontal line y=b is (x, 2b - y). So substituting b=2, x=2, y=0: 2*2 - 0 = 4. So the new point is (2, 4).

Hmm, that seems straightforward. But maybe I should visualize it or check with another method.

Alternatively, thinking about reflections: reflecting over x=1. The point (0,0) is 1 unit to the left of x=1, so the reflection should be 1 unit to the right, which is (2, 0). Then reflecting (2, 0) over y=2. The point (2, 0) is 2 units below y=2, so the reflection should be 2 units above y=2, which is y=4. Therefore, (2, 4).

Yes, that matches. So maybe that's the answer.

But let me think if there's another way to approach this. Maybe using coordinate geometry formulas.

Alternatively, maybe using matrices or something? But reflections are usually straightforward with formulas.

Alternatively, think about the composition of reflections. But since the first reflection is over a vertical line and the second over a horizontal line, they are perpendicular. But maybe that's complicating.

Alternatively, think about the distance from the line. For reflection over x=1, the distance from (0,0) to x=1 is |0 - 1| = 1. So the reflection is on the other side, same distance, so 1 + 1 = 2. So x-coordinate is 2.

Similarly, for reflecting (2, 0) over y=2. The distance from (2, 0) to y=2 is |0 - 2| = 2. So the reflection is 2 units above y=2, so y-coordinate is 2 + 2 = 4.

Therefore, the final point is (2, 4).

Wait, so all methods lead to (2, 4). Is there any chance I made a mistake here? Let me check with coordinates.

Let me draw a rough sketch mentally. Original point is at the origin. Reflecting over x=1, which is a vertical line one unit to the right. So the reflection is at (2, 0). Then reflecting that over y=2, which is a horizontal line two units up. The point (2, 0) is two units below y=2, so reflecting over y=2 would take it to (2, 4).

Yes, that seems correct.

Alternatively, if I had messed up the order of reflections? Wait, the problem says first reflect over x=1, then over y=2. So the order is important. If I had done the other way around, reflecting over y=2 first, then over x=1, the result would be different. But here, the order is correct.

Alternatively, maybe I can check with another point. Let me take a different point to see if my reflection formulas are correct.

Suppose I reflect (3, 5) over x=1. According to the formula, it should be (2*1 - 3, 5) = (2 - 3, 5) = (-1, 5). Let me check. The distance from x=1 to x=3 is 2 units. So reflection should be 2 units to the left of x=1, which is x=1 - 2 = -1. So yes, (-1, 5). Correct.

Similarly, reflecting (-1, 5) over y=2. The formula would be ( -1, 2*2 - 5 ) = (-1, 4 - 5) = (-1, -1). Let me check. The distance from y=5 to y=2 is 3 units. So reflection should be 3 units below y=2, which is y=2 - 3 = -1. So yes, (-1, -1). Correct.

So the formulas are working. Therefore, applying them to the original problem, the answer is (2, 4).

Alternatively, maybe I can think of the composition of two reflections. But since reflections over vertical and horizontal lines are not the same as a single reflection, but rather a combination. However, since they are over different axes, the composition is a reflection over some line? Wait, no. Wait, reflecting over two lines that are perpendicular would result in a rotation. Wait, but here we have two reflections: first over vertical, then over horizontal.

But maybe that's more complicated. However, since we already have the answer through direct computation, maybe it's not necessary. But just for my understanding, let me think.

If you reflect over x=1 and then over y=2, what is the composition?

In general, reflecting over two lines that are not parallel results in a rotation. Since x=1 and y=2 are perpendicular, the composition is a rotation by 180 degrees around the point which is the intersection of the two lines? Wait, but the two lines x=1 and y=2 are perpendicular, but they don't intersect each other? Wait, no, they do intersect at (1, 2). Wait, x=1 is a vertical line, y=2 is a horizontal line, so they intersect at (1, 2).

So reflecting over x=1 and then over y=2 would be equivalent to a rotation of 180 degrees around the point (1, 2). Let me check that.

If that's the case, then applying a 180-degree rotation around (1, 2) to the original point (0, 0) would give us the final point.

How does a 180-degree rotation around a point work? The formula is: for a point (x, y), rotating 180 degrees around (h, k) gives (2h - x, 2k - y).

So applying that here, h=1, k=2. So the rotated point would be (2*1 - 0, 2*2 - 0) = (2, 4). Which is the same result as before!

So that's another way to see it. Therefore, the answer is (2, 4).

Therefore, regardless of the method, the answer is (2, 4).

But just to make sure, let me think again. If I have a 180-degree rotation around (1, 2), then the original point (0, 0) is being rotated. The vector from (1, 2) to (0, 0) is (-1, -2). Rotating that 180 degrees would give (1, 2) + (-1, -2) rotated 180, which is (1, 2) + (1, 2) = (2, 4). Yep, that works.

So that's another confirmation.

Alternatively, if I didn't know about the composition of reflections, but just did the two reflections step by step, I still get the same answer.

Therefore, I think I can be confident that the answer is (2, 4).

**Final Answer**
The resulting point is \boxed{(2, 4)}.

To determine the final position of the point $(0, 0)$ after two successive reflections, we proceed step by step, applying the rules of reflection over vertical and horizontal lines.

---

### **Step 1: Reflect over the vertical line $x = 1$**

The formula for reflecting a point $(x, y)$ over a vertical line $x = a$ is:

$$
(x', y') = (2a - x, y)
$$

Applying this to the point $(0, 0)$ with $a = 1$:

$$
x' = 2(1) - 0 = 2, \quad y' = 0
$$

So, the image after the first reflection is:

$$
(2, 0)
$$

---

### **Step 2: Reflect over the horizontal line $y = 2$**

The formula for reflecting a point $(x, y)$ over a horizontal line $y = b$ is:

$$
(x', y') = (x, 2b - y)
$$

Applying this to the point $(2, 0)$ with $b = 2$:

$$
x' = 2, \quad y' = 2(2) - 0 = 4
$$

So, the image after the second reflection is:

$$
(2, 4)
$$

---

### **Alternative Interpretation: Composition of Reflections**

Reflecting over two perpendicular lines (a vertical and a horizontal) results in a **180° rotation** about the point of intersection of the two lines. The lines $x = 1$ and $y = 2$ intersect at the point $(1, 2)$.

A 180° rotation about a point $(h, k)$ transforms a point $(x, y)$ to:

$$
(2h - x, 2k - y)
$$

Applying this to the original point $(0, 0)$ with $(h, k) = (1, 2)$:

$$
x' = 2(1) - 0 = 2, \quad y' = 2(2) - 0 = 4
$$

This again gives the final point:

$$
(2, 4)
$$

---

### **Conclusion**

Both the step-by-step reflection method and the geometric interpretation using rotation confirm that the final position of the point after the two reflections is:

$$
\boxed{(2, 4)}
$$

[Assistant]
Explanation:
Okay, so I need to figure out where the point (0,0) ends up after being reflected over the vertical line x=1 and then over the line y=2. Let me start by recalling what reflection over a vertical line does.

First, reflecting a point over a vertical line x = a. I think the formula for that is something like if you have a point (x, y), its reflection over x = a would be (2a - x, y). Let me check that. If the original point is to the left of the line, then the reflection should be the same distance on the other side. For example, if I reflect (0,0) over x=1, then the distance from 0 to 1 is 1 unit, so the reflection should be 1 unit to the right of x=1, which would be x=2. So the reflected point would be (2, 0). That seems right. So applying that formula, 2a - x where a is 1 and x is 0, so 2*1 - 0 = 2. Yep, that gives (2, 0).

Okay, so first reflection takes (0,0) to (2, 0). Now, the next step is reflecting that image over the line y=2. So now I need to figure out how to reflect a point over a horizontal line y = b. I think the formula is similar, but for the y-coordinate. If you have a point (x, y), its reflection over y = b would be (x, 2b - y). Let me verify. If the original point is below the line, then the reflection should be the same distance above. For example, reflecting (0,0) over y=2 would be 2 units up from y=2, which is y=4. So the reflected point would be (0, 4). Using the formula, 2*2 - 0 = 4, so (0, 4). That checks out.

So applying that to the point (2, 0) which we got after the first reflection. So reflecting over y=2, the new y-coordinate would be 2*2 - 0 = 4. Therefore, the resulting point is (2, 4).

Wait, let me make sure I didn't mix up anything. Let me go through each step again.

Original point: (0, 0).

First reflection over x=1. The formula for reflection over vertical line x=a is (2a - x, y). So substituting a=1, x=0, y=0: 2*1 - 0 = 2, so (2, 0). That seems correct.

Then reflecting (2, 0) over y=2. The formula for reflection over horizontal line y=b is (x, 2b - y). So substituting b=2, x=2, y=0: 2*2 - 0 = 4. So the new point is (2, 4).

Hmm, that seems straightforward. But maybe I should visualize it or check with another method.

Alternatively, thinking about reflections: reflecting over x=1. The point (0,0) is 1 unit to the left of x=1, so the reflection should be 1 unit to the right, which is (2, 0). Then reflecting (2, 0) over y=2. The point (2, 0) is 2 units below y=2, so the reflection should be 2 units above y=2, which is y=4. Therefore, (2, 4).

Yes, that matches. So maybe that's the answer.

But let me think if there's another way to approach this. Maybe using coordinate geometry formulas.

Alternatively, maybe using matrices or something? But reflections are usually straightforward with formulas.

Alternatively, think about the composition of reflections. But since the first reflection is over a vertical line and the second over a horizontal line, they are perpendicular. But maybe that's complicating.

Alternatively, think about the distance from the line. For reflection over x=1, the distance from (0,0) to x=1 is |0 - 1| = 1. So the reflection is on the other side, same distance, so 1 + 1 = 2. So x-coordinate is 2.

Similarly, for reflecting (2, 0) over y=2. The distance from (2, 0) to y=2 is |0 - 2| = 2. So the reflection is 2 units above y=2, so y-coordinate is 2 + 2 = 4.

Therefore, the final point is (2, 4).

Wait, so all methods lead to (2, 4). Is there any chance I made a mistake here? Let me check with coordinates.

Let me draw a rough sketch mentally. Original point is at the origin. Reflecting over x=1, which is a vertical line one unit to the right. So the reflection is at (2, 0). Then reflecting that over y=2, which is a horizontal line two units up. The point (2, 0) is two units below y=2, so reflecting over y=2 would take it to (2, 4).

Yes, that seems correct.

Alternatively, if I had messed up the order of reflections? Wait, the problem says first reflect over x=1, then over y=2. So the order is important. If I had done the other way around, reflecting over y=2 first, then over x=1, the result would be different. But here, the order is correct.

Alternatively, maybe I can check with another point. Let me take a different point to see if my reflection formulas are correct.

Suppose I reflect (3, 5) over x=1. According to the formula, it should be (2*1 - 3, 5) = (2 - 3, 5) = (-1, 5). Let me check. The distance from x=1 to x=3 is 2 units. So reflection should be 2 units to the left of x=1, which is x=1 - 2 = -1. So yes, (-1, 5). Correct.

Similarly, reflecting (-1, 5) over y=2. The formula would be ( -1, 2*2 - 5 ) = (-1, 4 - 5) = (-1, -1). Let me check. The distance from y=5 to y=2 is 3 units. So reflection should be 3 units below y=2, which is y=2 - 3 = -1. So yes, (-1, -1). Correct.

So the formulas are working. Therefore, applying them to the original problem, the answer is (2, 4).

Alternatively, maybe I can think of the composition of two reflections. But since reflections over vertical and horizontal lines are not the same as a single reflection, but rather a combination. However, since they are over different axes, the composition is a reflection over some line? Wait, no. Wait, reflecting over two lines that are perpendicular would result in a rotation. Wait, but here we have two reflections: first over vertical, then over horizontal.

But maybe that's more complicated. However, since we already have the answer through direct computation, maybe it's not necessary. But just for my understanding, let me think.

If you reflect over x=1 and then over y=2, what is the composition?

In general, reflecting over two lines that are not parallel results in a rotation. Since x=1 and y=2 are perpendicular, the composition is a rotation by 180 degrees around the point which is the intersection of the two lines? Wait, but the two lines x=1 and y=2 are perpendicular, but they don't intersect each other? Wait, no, they do intersect at (1, 2). Wait, x=1 is a vertical line, y=2 is a horizontal line, so they intersect at (1, 2).

So reflecting over x=1 and then over y=2 would be equivalent to a rotation of 180 degrees around the point (1, 2). Let me check that.

If that's the case, then applying a 180-degree rotation around (1, 2) to the original point (0, 0) would give us the final point.

How does a 180-degree rotation around a point work? The formula is: for a point (x, y), rotating 180 degrees around (h, k) gives (2h - x, 2k - y).

So applying that here, h=1, k=2. So the rotated point would be (2*1 - 0, 2*2 - 0) = (2, 4). Which is the same result as before!

So that's another way to see it. Therefore, the answer is (2, 4).

Therefore, regardless of the method, the answer is (2, 4).

But just to make sure, let me think again. If I have a 180-degree rotation around (1, 2), then the original point (0, 0) is being rotated. The vector from (1, 2) to (0, 0) is (-1, -2). Rotating that 180 degrees would give (1, 2) + (-1, -2) rotated 180, which is (1, 2) + (1, 2) = (2, 4). Yep, that works.

So that's another confirmation.

Alternatively, if I didn't know about the composition of reflections, but just did the two reflections step by step, I still get the same answer.

Therefore, I think I can be confident that the answer is (2, 4).

**Final Answer**
The resulting point is \boxed{(2, 4)}.

To determine the final position of the point $(0, 0)$ after two successive reflections, we proceed step by step, applying the rules of reflection over vertical and horizontal lines.

---

### **Step 1: Reflect over the vertical line $x = 1$**

The formula for reflecting a point $(x, y)$ over a vertical line $x = a$ is:

$$
(x', y') = (2a - x, y)
$$

Applying this to the point $(0, 0)$ with $a = 1$:

$$
x' = 2(1) - 0 = 2, \quad y' = 0
$$

So, the image after the first reflection is:

$$
(2, 0)
$$

---

### **Step 2: Reflect over the horizontal line $y = 2$**

The formula for reflecting a point $(x, y)$ over a horizontal line $y = b$ is:

$$
(x', y') = (x, 2b - y)
$$

Applying this to the point $(2, 0)$ with $b = 2$:

$$
x' = 2, \quad y' = 2(2) - 0 = 4
$$

So, the image after the second reflection is:

$$
(2, 4)
$$

---

### **Alternative Interpretation: Composition of Reflections**

Reflecting over two perpendicular lines (a vertical and a horizontal) results in a **180° rotation** about the point of intersection of the two lines. The lines $x = 1$ and $y = 2$ intersect at the point $(1, 2)$.

A 180° rotation about a point $(h, k)$ transforms a point $(x, y)$ to:

$$
(2h - x, 2k - y)
$$

Applying this to the original point $(0, 0)$ with $(h, k) = (1, 2)$:

$$
x' = 2(1) - 0 = 2, \quad y' = 2(2) - 0 = 4
$$

This again gives the final point:

$$
(2, 4)
$$

---

### **Conclusion**

Both the step-by-step reflection method and the geometric interpretation using rotation confirm that the final position of the point after the two reflections is:

$$
\boxed{(2, 4)}
$$

Solution:
When $(0,0)$ is reflected over the line $x=1$, the image is $(2,0)$. [asy]
draw((-2, 0)--(6, 0), Arrow); draw((0, -2)--(0, 6), Arrow);
label("$x$", (6, 0), E); label("$y$", (0, 6), N);
label("$(0, 0)$", (0, 0), SW); label("$(2, 0)$", (2, 0), SE);
label("$(2, 4)$", (2, 4), NE);
label("$y = 2$", (6, 2), E); label("$x = 1$", (1, -2), S);
draw((-2, 2)--(6, 2), dashed); draw((1, -2)--(1, 6), dashed);
dot((0, 0)); dot((2, 0)); dot((2, 4));
[/asy] When $(2,0)$ is reflected over the line $y=2$, the image is $\boxed{(2,4)}$.