Explanation:
Okay, so I need to figure out what kind of curve is defined by the equation $ r^2 \cos 2\theta = 4 $. The options are line, circle, parabola, ellipse, or hyperbola. Let me start by recalling some polar coordinate equations for conic sections and other curves.

First, I remember that in polar coordinates, conic sections can be represented with equations involving $ r $ and $ \theta $, often involving $ \cos \theta $ or $ \sin \theta $. The general form for a conic section with one focus at the origin is $ r = \frac{ed}{1 + e \cos \theta} $ or something like that, where $ e $ is the eccentricity. But here, the equation is $ r^2 \cos 2\theta = 4 $, which seems different. Maybe I need to convert this equation to Cartesian coordinates to identify the curve?

Let me recall that $ r^2 = x^2 + y^2 $, and $ \cos 2\theta $ can be expressed in terms of $ x $ and $ y $. There's a trigonometric identity for $ \cos 2\theta $: $ \cos 2\theta = \cos^2 \theta - \sin^2 \theta $. Also, $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $. Let me verify that.

Since $ x = r \cos \theta $ and $ y = r \sin \theta $, then $ x^2 = r^2 \cos^2 \theta $ and $ y^2 = r^2 \sin^2 \theta $. Therefore, $ x^2 - y^2 = r^2 (\cos^2 \theta - \sin^2 \theta) = r^2 \cos 2\theta $. So indeed, $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $.

Therefore, substituting back into the original equation $ r^2 \cos 2\theta = 4 $, we get:

$ r^2 \times \frac{x^2 - y^2}{r^2} = 4 $

Simplifying, the $ r^2 $ cancels out, and we get:

$ x^2 - y^2 = 4 $

Wait, so the equation simplifies to $ x^2 - y^2 = 4 $. That looks like the standard form of a hyperbola. The standard form of a hyperbola centered at the origin is $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $. Comparing that with our equation, we can write $ x^2 - y^2 = 4 $ as $ \frac{x^2}{4} - \frac{y^2}{4} = 1 $, so $ a^2 = 4 $ and $ b^2 = 4 $. Therefore, this is a hyperbola centered at the origin, opening along the x-axis.

But let me double-check my steps to make sure I didn't make any mistakes. Starting with the original polar equation:

$ r^2 \cos 2\theta = 4 $.

Expressed in terms of Cartesian coordinates, since $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $, then substituting gives:

$ r^2 \times \frac{x^2 - y^2}{r^2} = 4 $

Which simplifies to $ x^2 - y^2 = 4 $. That seems correct. Therefore, the equation is indeed a hyperbola.

Alternatively, maybe I can think about the original polar equation. Is there another way to interpret $ r^2 \cos 2\theta = 4 $? For example, sometimes polar equations can represent different conic sections or other curves. But I think converting to Cartesian coordinates is the more straightforward approach here, especially since I remember the identity for $ \cos 2\theta $.

Another way to check: Let's consider specific points. For example, when $ \theta = 0 $, $ \cos 2\theta = \cos 0 = 1 $. Then the equation becomes $ r^2 \times 1 = 4 $, so $ r^2 = 4 $, so $ r = \pm 2 $. But in polar coordinates, r is usually taken as non-negative, so r = 2. Therefore, the point (r, θ) = (2, 0) is on the curve. Similarly, when θ = π/4, then 2θ = π/2, so cos(π/2) = 0. Then the equation becomes r² * 0 = 4, which implies 0 = 4, which is impossible. Therefore, there are no points at θ = π/4.

Similarly, when θ = π/2, 2θ = π, cos π = -1. Then equation becomes r²*(-1) = 4 => r² = -4, which is impossible. Therefore, there are no points at θ = π/2. Similarly for θ = 3π/4, 2θ = 3π/2, cos(3π/2) = 0, leading again to 0 = 4, impossible. So this suggests that the curve does not pass through θ = π/4, π/2, etc.

Alternatively, when θ = π/3, then 2θ = 2π/3, cos(2π/3) = -1/2. Then the equation becomes r²*(-1/2) = 4 => r² = -8, which is impossible. Wait, that's not possible. So maybe θ needs to be such that cos 2θ is positive? Because r² is always non-negative, so cos 2θ must be positive, otherwise, the product would be negative or zero, but the right-hand side is positive 4. Therefore, cos 2θ must be positive. So 2θ must lie in the first or fourth quadrants, i.e., 2θ ∈ (-π/2 + 2πk, π/2 + 2πk) for some integer k. Therefore θ ∈ (-π/4 + πk, π/4 + πk). So the curve exists only in those angular regions.

So maybe plotting some points where cos 2θ is positive. For example, θ between -π/4 and π/4, which is the region near the x-axis. Also between 3π/4 and 5π/4, etc. But in those regions, how does the curve look?

But perhaps it's simpler to stick with the Cartesian equation. Since we have x² - y² = 4, which is a hyperbola. Therefore, the answer should be (E) Hyperbola.

But wait, just to be thorough, let me think if there's another possibility. For example, sometimes equations can represent other conic sections if they are manipulated differently. For instance, if the equation was something like r = something involving cos θ or sin θ, but in this case, it's a different form.

Alternatively, could this be a rectangular hyperbola? Yes, because the equation x² - y² = 4 is a rectangular hyperbola since a = b = 2. Rectangular hyperbola is a specific case where a = b, and it has asymptotes y = ±x. Which makes sense here.

So, given that the equation simplifies to x² - y² = 4, which is a hyperbola, the answer should be E.

But let me also recall that the original equation is in polar coordinates. Maybe there's a way to think about it in polar coordinates. For example, the equation is r² cos 2θ = 4. Let me recall that in polar coordinates, equations involving cos 2θ can represent hyperbolas or other conic sections. For example, the general equation for a conic section in polar coordinates is r = ed/(1 + e cos θ), but that's for a conic with one focus at the origin. However, here we have r² cos 2θ = 4, which seems more complicated.

Alternatively, maybe I can think about the equation as r² = 4 / cos 2θ. Which is equivalent to r² = 4 sec 2θ. So, r = ±2 sec^{1/2} 2θ. Not sure if that helps. But if I square both sides, r² = 4 sec 2θ. Then, in Cartesian coordinates, as before, x² + y² = 4 / cos 2θ. But since cos 2θ = (x² - y²)/r², then substituting back, x² + y² = 4 / [(x² - y²)/r²] => x² + y² = 4 r² / (x² - y²). Then, multiplying both sides by (x² - y²):

(x² + y²)(x² - y²) = 4r²

But since r² = x² + y², then:

(x² + y²)(x² - y²) = 4(x² + y²)

Assuming x² + y² ≠ 0, we can divide both sides by x² + y²:

x² - y² = 4

Which is the same equation as before. Therefore, again, confirming that x² - y² = 4 is the Cartesian equation, which is a hyperbola. Therefore, answer E.

Alternatively, if x² + y² = 0, then r = 0, which is just the origin. But substituting into original equation: r² cos 2θ = 4 would be 0 * cos 2θ = 4 => 0 = 4, which is impossible. Therefore, the origin is not on the curve, so we can safely divide by r². Thus, the curve is indeed x² - y² = 4, which is a hyperbola.

Therefore, the answer is E) Hyperbola.

I think that's solid. I don't see any mistakes in my reasoning. So the correct option is E.

**Final Answer**
\boxed{E}

To determine the type of curve described by the polar equation
$$
r^2 \cos 2\theta = 4,
$$
we begin by recognizing that it's not immediately recognizable as a standard conic section in polar form. Therefore, a useful strategy is to **convert the polar equation to Cartesian coordinates**.

---

### Step 1: Use known identities

We recall the identity for the cosine of a double angle:
$$
\cos 2\theta = \frac{x^2 - y^2}{r^2}.
$$
In Cartesian coordinates, $ r^2 = x^2 + y^2 $. Substituting this identity into the original equation:

$$
r^2 \cos 2\theta = 4 \Rightarrow r^2 \cdot \frac{x^2 - y^2}{r^2} = 4.
$$
The $ r^2 $ terms cancel out, leaving:
$$
x^2 - y^2 = 4.
$$

---

### Step 2: Identify the type of curve

The equation $ x^2 - y^2 = 4 $ is a **standard form of a hyperbola** centered at the origin, where the transverse axis lies along the x-axis. Specifically, it can be rewritten as:
$$
\frac{x^2}{4} - \frac{y^2}{4} = 1,
$$
which confirms the equation is a **hyperbola** with $ a^2 = 4 $, $ b^2 = 4 $.

This is a **rectangular hyperbola**, since $ a = b $.

---

### Step 3: Optional Analysis in Polar Coordinates

Let’s consider the original polar form $ r^2 \cos 2\theta = 4 $ again. We can rearrange:
$$
\cos 2\theta = \frac{4}{r^2},
$$
which implies that $ \cos 2\theta > 0 $ for real values of $ r $. That restricts the angle $ \theta $ to certain regions (e.g., between $ -\pi/4 $ and $ \pi/4 $, or $ 3\pi/4 $ to $ 5\pi/4 $, etc.), indicating the curve is symmetric and only defined in those angular regions.

But this is consistent with a hyperbola, which has two branches and is symmetric.

---

### Final Conclusion

After converting the polar equation to Cartesian coordinates and simplifying, we find the curve corresponds to the equation of a hyperbola. Therefore, the correct answer is:

$$
\boxed{E}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what kind of curve is defined by the equation $ r^2 \cos 2\theta = 4 $. The options are line, circle, parabola, ellipse, or hyperbola. Let me start by recalling some polar coordinate equations for conic sections and other curves.

First, I remember that in polar coordinates, conic sections can be represented with equations involving $ r $ and $ \theta $, often involving $ \cos \theta $ or $ \sin \theta $. The general form for a conic section with one focus at the origin is $ r = \frac{ed}{1 + e \cos \theta} $ or something like that, where $ e $ is the eccentricity. But here, the equation is $ r^2 \cos 2\theta = 4 $, which seems different. Maybe I need to convert this equation to Cartesian coordinates to identify the curve?

Let me recall that $ r^2 = x^2 + y^2 $, and $ \cos 2\theta $ can be expressed in terms of $ x $ and $ y $. There's a trigonometric identity for $ \cos 2\theta $: $ \cos 2\theta = \cos^2 \theta - \sin^2 \theta $. Also, $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $. Let me verify that.

Since $ x = r \cos \theta $ and $ y = r \sin \theta $, then $ x^2 = r^2 \cos^2 \theta $ and $ y^2 = r^2 \sin^2 \theta $. Therefore, $ x^2 - y^2 = r^2 (\cos^2 \theta - \sin^2 \theta) = r^2 \cos 2\theta $. So indeed, $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $.

Therefore, substituting back into the original equation $ r^2 \cos 2\theta = 4 $, we get:

$ r^2 \times \frac{x^2 - y^2}{r^2} = 4 $

Simplifying, the $ r^2 $ cancels out, and we get:

$ x^2 - y^2 = 4 $

Wait, so the equation simplifies to $ x^2 - y^2 = 4 $. That looks like the standard form of a hyperbola. The standard form of a hyperbola centered at the origin is $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $. Comparing that with our equation, we can write $ x^2 - y^2 = 4 $ as $ \frac{x^2}{4} - \frac{y^2}{4} = 1 $, so $ a^2 = 4 $ and $ b^2 = 4 $. Therefore, this is a hyperbola centered at the origin, opening along the x-axis.

But let me double-check my steps to make sure I didn't make any mistakes. Starting with the original polar equation:

$ r^2 \cos 2\theta = 4 $.

Expressed in terms of Cartesian coordinates, since $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $, then substituting gives:

$ r^2 \times \frac{x^2 - y^2}{r^2} = 4 $

Which simplifies to $ x^2 - y^2 = 4 $. That seems correct. Therefore, the equation is indeed a hyperbola.

Alternatively, maybe I can think about the original polar equation. Is there another way to interpret $ r^2 \cos 2\theta = 4 $? For example, sometimes polar equations can represent different conic sections or other curves. But I think converting to Cartesian coordinates is the more straightforward approach here, especially since I remember the identity for $ \cos 2\theta $.

Another way to check: Let's consider specific points. For example, when $ \theta = 0 $, $ \cos 2\theta = \cos 0 = 1 $. Then the equation becomes $ r^2 \times 1 = 4 $, so $ r^2 = 4 $, so $ r = \pm 2 $. But in polar coordinates, r is usually taken as non-negative, so r = 2. Therefore, the point (r, θ) = (2, 0) is on the curve. Similarly, when θ = π/4, then 2θ = π/2, so cos(π/2) = 0. Then the equation becomes r² * 0 = 4, which implies 0 = 4, which is impossible. Therefore, there are no points at θ = π/4.

Similarly, when θ = π/2, 2θ = π, cos π = -1. Then equation becomes r²*(-1) = 4 => r² = -4, which is impossible. Therefore, there are no points at θ = π/2. Similarly for θ = 3π/4, 2θ = 3π/2, cos(3π/2) = 0, leading again to 0 = 4, impossible. So this suggests that the curve does not pass through θ = π/4, π/2, etc.

Alternatively, when θ = π/3, then 2θ = 2π/3, cos(2π/3) = -1/2. Then the equation becomes r²*(-1/2) = 4 => r² = -8, which is impossible. Wait, that's not possible. So maybe θ needs to be such that cos 2θ is positive? Because r² is always non-negative, so cos 2θ must be positive, otherwise, the product would be negative or zero, but the right-hand side is positive 4. Therefore, cos 2θ must be positive. So 2θ must lie in the first or fourth quadrants, i.e., 2θ ∈ (-π/2 + 2πk, π/2 + 2πk) for some integer k. Therefore θ ∈ (-π/4 + πk, π/4 + πk). So the curve exists only in those angular regions.

So maybe plotting some points where cos 2θ is positive. For example, θ between -π/4 and π/4, which is the region near the x-axis. Also between 3π/4 and 5π/4, etc. But in those regions, how does the curve look?

But perhaps it's simpler to stick with the Cartesian equation. Since we have x² - y² = 4, which is a hyperbola. Therefore, the answer should be (E) Hyperbola.

But wait, just to be thorough, let me think if there's another possibility. For example, sometimes equations can represent other conic sections if they are manipulated differently. For instance, if the equation was something like r = something involving cos θ or sin θ, but in this case, it's a different form.

Alternatively, could this be a rectangular hyperbola? Yes, because the equation x² - y² = 4 is a rectangular hyperbola since a = b = 2. Rectangular hyperbola is a specific case where a = b, and it has asymptotes y = ±x. Which makes sense here.

So, given that the equation simplifies to x² - y² = 4, which is a hyperbola, the answer should be E.

But let me also recall that the original equation is in polar coordinates. Maybe there's a way to think about it in polar coordinates. For example, the equation is r² cos 2θ = 4. Let me recall that in polar coordinates, equations involving cos 2θ can represent hyperbolas or other conic sections. For example, the general equation for a conic section in polar coordinates is r = ed/(1 + e cos θ), but that's for a conic with one focus at the origin. However, here we have r² cos 2θ = 4, which seems more complicated.

Alternatively, maybe I can think about the equation as r² = 4 / cos 2θ. Which is equivalent to r² = 4 sec 2θ. So, r = ±2 sec^{1/2} 2θ. Not sure if that helps. But if I square both sides, r² = 4 sec 2θ. Then, in Cartesian coordinates, as before, x² + y² = 4 / cos 2θ. But since cos 2θ = (x² - y²)/r², then substituting back, x² + y² = 4 / [(x² - y²)/r²] => x² + y² = 4 r² / (x² - y²). Then, multiplying both sides by (x² - y²):

(x² + y²)(x² - y²) = 4r²

But since r² = x² + y², then:

(x² + y²)(x² - y²) = 4(x² + y²)

Assuming x² + y² ≠ 0, we can divide both sides by x² + y²:

x² - y² = 4

Which is the same equation as before. Therefore, again, confirming that x² - y² = 4 is the Cartesian equation, which is a hyperbola. Therefore, answer E.

Alternatively, if x² + y² = 0, then r = 0, which is just the origin. But substituting into original equation: r² cos 2θ = 4 would be 0 * cos 2θ = 4 => 0 = 4, which is impossible. Therefore, the origin is not on the curve, so we can safely divide by r². Thus, the curve is indeed x² - y² = 4, which is a hyperbola.

Therefore, the answer is E) Hyperbola.

I think that's solid. I don't see any mistakes in my reasoning. So the correct option is E.

**Final Answer**
\boxed{E}

To determine the type of curve described by the polar equation
$$
r^2 \cos 2\theta = 4,
$$
we begin by recognizing that it's not immediately recognizable as a standard conic section in polar form. Therefore, a useful strategy is to **convert the polar equation to Cartesian coordinates**.

---

### Step 1: Use known identities

We recall the identity for the cosine of a double angle:
$$
\cos 2\theta = \frac{x^2 - y^2}{r^2}.
$$
In Cartesian coordinates, $ r^2 = x^2 + y^2 $. Substituting this identity into the original equation:

$$
r^2 \cos 2\theta = 4 \Rightarrow r^2 \cdot \frac{x^2 - y^2}{r^2} = 4.
$$
The $ r^2 $ terms cancel out, leaving:
$$
x^2 - y^2 = 4.
$$

---

### Step 2: Identify the type of curve

The equation $ x^2 - y^2 = 4 $ is a **standard form of a hyperbola** centered at the origin, where the transverse axis lies along the x-axis. Specifically, it can be rewritten as:
$$
\frac{x^2}{4} - \frac{y^2}{4} = 1,
$$
which confirms the equation is a **hyperbola** with $ a^2 = 4 $, $ b^2 = 4 $.

This is a **rectangular hyperbola**, since $ a = b $.

---

### Step 3: Optional Analysis in Polar Coordinates

Let’s consider the original polar form $ r^2 \cos 2\theta = 4 $ again. We can rearrange:
$$
\cos 2\theta = \frac{4}{r^2},
$$
which implies that $ \cos 2\theta > 0 $ for real values of $ r $. That restricts the angle $ \theta $ to certain regions (e.g., between $ -\pi/4 $ and $ \pi/4 $, or $ 3\pi/4 $ to $ 5\pi/4 $, etc.), indicating the curve is symmetric and only defined in those angular regions.

But this is consistent with a hyperbola, which has two branches and is symmetric.

---

### Final Conclusion

After converting the polar equation to Cartesian coordinates and simplifying, we find the curve corresponds to the equation of a hyperbola. Therefore, the correct answer is:

$$
\boxed{E}
$$

Solution:
From $r^2 \cos 2 \theta = 4,$
\[r^2 (\cos^2 \theta - \sin^2 \theta) = r^2 \cos^2 \theta - r^2 \sin^2 \theta = 4.\]Then $x^2 - y^2 = 4,$ or
\[\frac{x^2}{4} - \frac{y^2}{4} = 1.\]Thus, the graph represents a hyperbola.  The answer is $\boxed{\text{(E)}}.$

[asy]
unitsize(0.5 cm);

pair moo (real t) {
  real r = sqrt(4/Cos(2*t));
  return (r*Cos(t), r*Sin(t));
}

path foo = moo(-44);
real t;

for (t = -44; t <= 44; t = t + 0.1) {
  foo = foo--moo(t);
}

draw(foo,red);
draw(reflect((0,0),(0,1))*(foo),red);

draw((-4,0)--(4,0));
draw((0,-4)--(0,4));

limits((-4,-4),(4,4),Crop);

label("$r^2 \cos 2 \theta = 4$", (6.5,1.5), red);
[/asy]


Explanation:
Okay, so I need to find the 2003rd term of the sequence of odd numbers: 1, 3, 5, 7, and so on. Let me think about how sequences like this work.

First, I remember that the sequence of odd numbers is an arithmetic sequence. In an arithmetic sequence, each term increases by a constant difference. For the odd numbers, the difference between each term is 2. Let me verify that.

Starting with 1, the next term is 3. 3 - 1 = 2. Then 5 - 3 = 2, and so on. Yep, the common difference is 2.

Now, I need to recall the formula for the nth term of an arithmetic sequence. I think it's something like a_n = a_1 + (n - 1)d, where a_1 is the first term, d is the common difference, and n is the term number. Let me check that.

If n = 1, then a_1 = a_1 + (1 - 1)d = a_1 + 0 = a_1, which is correct. For n = 2, it would be a_1 + (2 - 1)d = a_1 + d, which matches the second term. So that formula seems right.

In this case, the first term a_1 is 1, and the common difference d is 2. So substituting into the formula, the nth term should be:

a_n = 1 + (n - 1)*2

Let me simplify that.

First, distribute the 2: 1 + 2(n - 1) = 1 + 2n - 2 = 2n - 1.

Wait, so the nth term is 2n - 1? Let me check with the first few terms.

For n = 1: 2*1 - 1 = 2 - 1 = 1. Correct.

n = 2: 2*2 - 1 = 4 - 1 = 3. Correct.

n = 3: 6 - 1 = 5. Yep, that works. So the formula seems to hold.

Therefore, to find the 2003rd term, I just plug n = 2003 into the formula.

So, a_2003 = 2*2003 - 1. Let me compute that.

First, calculate 2*2003. Well, 2000*2 = 4000, and 3*2 = 6. So 4000 + 6 = 4006. Then subtract 1: 4006 - 1 = 4005.

Therefore, the 2003rd term should be 4005.

But let me make sure I didn't make any mistakes here. Maybe I can think of another way to approach this problem.

Another way to think about odd numbers is that they can be represented as 2k - 1, where k is a positive integer. For example, when k = 1, we get 1; k = 2, we get 3, etc. So if this is the case, then the k-th term is 2k - 1. Therefore, if we want the 2003rd term, that would correspond to k = 2003, so 2*2003 -1 = 4005.

That's the same result as before. So that seems to confirm it.

Alternatively, maybe I can think about how many numbers are there between 1 and the term I'm looking for. But that might complicate things. However, let me try.

Each odd number is 1 more than an even number. Wait, maybe not helpful. Alternatively, if I think about the position of the term in the sequence.

For example, the first term is 1, which is 2*1 - 1. The second term is 3, which is 2*2 - 1. So in general, the nth term is 2n - 1.

Alternatively, if I think of the sequence of natural numbers, the odd numbers are the numbers that are not divisible by 2. So the nth odd number is... Well, if you think about it, the nth odd number is equal to 2n - 1.

Another way to see this is that if I have n numbers, each separated by 2, starting at 1. So the total difference from the first term would be (n - 1)*2. So adding that to the first term: 1 + (n - 1)*2 = 2n - 1.

All methods lead to the same formula. Therefore, I feel confident that the formula is correct.

So applying it for n = 2003:

2*2003 = 4006. Then subtract 1 gives 4005.

Let me check with smaller numbers again. Let's take n = 4, which should be 7.

Using the formula: 2*4 -1 = 8 -1 = 7. Correct.

Another example: n = 5. 2*5 -1 = 10 -1 = 9. Which is the fifth odd number. 1, 3, 5, 7, 9. Yes, that's right.

Therefore, seems solid.

Alternatively, maybe I can think about the relationship between odd numbers and squares? Wait, there's a thing that the sum of the first n odd numbers is n squared. But that's a different problem.

But maybe that can help here? Let me think. For example, 1 = 1², 1 + 3 = 4 = 2², 1 + 3 + 5 = 9 = 3², etc.

So the sum of the first n odd numbers is n². However, the question here isn't about the sum, but about the value of the nth term.

But maybe there is a way to connect them? For example, if I know that the sum of the first n odd numbers is n², then if I wanted to find the nth odd number, maybe I can take the difference between the sum up to n and the sum up to n-1.

So sum_{k=1}^n (2k -1) = n². Therefore, the nth term would be sum_{k=1}^n (2k -1) - sum_{k=1}^{n-1} (2k -1) = n² - (n - 1)^2.

Calculating that: n² - (n² - 2n + 1) = n² - n² + 2n -1 = 2n -1.

So, that gives the same formula. Therefore, the nth term is 2n -1. So again, the same result.

Therefore, even using this method, we get that the nth term is 2n -1.

So, using this formula for n = 2003, we get 2*2003 -1 = 4005.

Therefore, I think that's the correct answer.

But just to be thorough, let me check with another approach.

Suppose we list the odd numbers. The first term is 1, which is 2*1 -1. The second term is 3, which is 2*2 -1. The third term is 5, which is 2*3 -1. So in general, term number n is 2n -1. Therefore, term 2003 is 2*2003 -1.

Alternatively, if I think about how many odd numbers there are up to a certain number. For example, if I have a number x, the number of odd numbers less than or equal to x is ceil(x/2). But again, not sure if that helps here.

But if I want to find the nth odd number, it's the same as 2n -1. For example, if n is 1, then 2*1 -1 = 1, which is the first odd number. If n is 2, 3, etc.

Alternatively, maybe think of it as the nth odd number is the same as the (n + (n -1))th number in the natural numbers? Wait, no.

Wait, the odd numbers are 1, 3, 5,... So each odd number is 2 less than the next even number. Wait, maybe that's complicating.

Alternatively, think of the nth odd number as the number that is two less than twice n. Wait, 2n - 2? No, that would be 2(n - 1). But the first term is 2(1) -1.

Alternatively, maybe think of it as the numbers congruent to 1 mod 2. So, numbers that are 1 more than an even number.

But again, this seems like going in circles.

Alternatively, think about the position in the sequence. Each term is 2 more than the previous term, starting at 1. So the sequence is 1, 1 + 2, 1 + 2*2, 1 + 2*3,... So general term is 1 + 2(n -1) = 2n -1.

Same formula again.

So all different ways of thinking about it lead me to the same conclusion.

Therefore, unless I made a calculation mistake when computing 2*2003 -1, which is 4005, that should be correct.

Wait, let me check 2*2003. Let me compute 2003*2.

2000*2 = 4000, 3*2 = 6, so 4000 + 6 = 4006. Then subtract 1: 4006 - 1 = 4005. Yep, that seems right.

Alternatively, if I add 1 + 2*(2003 -1). That would be 1 + 2*(2002). 2002*2 = 4004, then 4004 + 1 = 4005. Same result.

So, 4005 is the correct answer.

Is there a possibility that the problem is referring to something else? Like starting the sequence at 0? But no, the sequence is given as 1, 3, 5, 7,... So it's definitely starting at 1.

If for some reason the sequence was starting at 0, then the term would be different, but that's not the case here.

Alternatively, could there be a misunderstanding about the term number? Like, is the first term considered as term 0? But the problem says "the 2003rd term", so that would be term number 2003, which is the 2003rd element in the sequence.

In most mathematical contexts, the first term is term 1, so term 1 is 1, term 2 is 3, etc. Therefore, term number 2003 is 2*2003 -1.

Alternatively, if someone thought that the first term is term 0, then term 2003 would be 2*2004 -1, which would be 4007, but that's not the case here.

Since the problem states the sequence as 1, 3, 5, 7,..., and asks for the 2003rd term, it's safe to assume that the first term is term 1.

Therefore, I think the answer is definitely 4005.

Wait, just to make sure, maybe I can check with a smaller term. For example, term 5 is 9. If I use the formula, 2*5 -1 = 9. Correct.

If I think about term 10, it should be 19. 2*10 -1 = 19. Correct. So the formula works for these.

Alternatively, if I try term 100, it would be 2*100 -1 = 199. Let me count... Well, the 100th odd number is 199. If I think of it as the 100th term, then yes.

Alternatively, if I think of the even numbers. The nth even number is 2n. The odd numbers are the even numbers minus 1. So the nth odd number is 2n -1. Exactly.

Therefore, that also makes sense.

So, given that, term 2003 is 2*2003 -1 = 4005.

Therefore, I think I can be confident that this is the correct answer.

**Final Answer**
The 2003rd term of the sequence is \boxed{4005}.

To determine the **2003rd term** of the sequence of odd numbers, which begins as:

$$
1, 3, 5, 7, \dots
$$

we recognize that this is an **arithmetic sequence** with:

- First term $ a_1 = 1 $
- Common difference $ d = 2 $

---

### Step 1: General Formula for the $ n $-th Term of an Arithmetic Sequence

The formula for the $ n $-th term of an arithmetic sequence is:

$$
a_n = a_1 + (n - 1)d
$$

Substitute the values for our sequence:

$$
a_n = 1 + (n - 1) \cdot 2 = 1 + 2(n - 1)
$$

Simplify:

$$
a_n = 2n - 1
$$

This confirms that the $ n $-th term of the sequence of odd numbers is $ 2n - 1 $.

---

### Step 2: Apply the Formula for the 2003rd Term

Now, substitute $ n = 2003 $ into the simplified formula:

$$
a_{2003} = 2 \cdot 2003 - 1 = 4006 - 1 = 4005
$$

---

### Final Answer

$$
\boxed{4005}
$$

[Assistant]
Explanation:
Okay, so I need to find the 2003rd term of the sequence of odd numbers: 1, 3, 5, 7, and so on. Let me think about how sequences like this work.

First, I remember that the sequence of odd numbers is an arithmetic sequence. In an arithmetic sequence, each term increases by a constant difference. For the odd numbers, the difference between each term is 2. Let me verify that.

Starting with 1, the next term is 3. 3 - 1 = 2. Then 5 - 3 = 2, and so on. Yep, the common difference is 2.

Now, I need to recall the formula for the nth term of an arithmetic sequence. I think it's something like a_n = a_1 + (n - 1)d, where a_1 is the first term, d is the common difference, and n is the term number. Let me check that.

If n = 1, then a_1 = a_1 + (1 - 1)d = a_1 + 0 = a_1, which is correct. For n = 2, it would be a_1 + (2 - 1)d = a_1 + d, which matches the second term. So that formula seems right.

In this case, the first term a_1 is 1, and the common difference d is 2. So substituting into the formula, the nth term should be:

a_n = 1 + (n - 1)*2

Let me simplify that.

First, distribute the 2: 1 + 2(n - 1) = 1 + 2n - 2 = 2n - 1.

Wait, so the nth term is 2n - 1? Let me check with the first few terms.

For n = 1: 2*1 - 1 = 2 - 1 = 1. Correct.

n = 2: 2*2 - 1 = 4 - 1 = 3. Correct.

n = 3: 6 - 1 = 5. Yep, that works. So the formula seems to hold.

Therefore, to find the 2003rd term, I just plug n = 2003 into the formula.

So, a_2003 = 2*2003 - 1. Let me compute that.

First, calculate 2*2003. Well, 2000*2 = 4000, and 3*2 = 6. So 4000 + 6 = 4006. Then subtract 1: 4006 - 1 = 4005.

Therefore, the 2003rd term should be 4005.

But let me make sure I didn't make any mistakes here. Maybe I can think of another way to approach this problem.

Another way to think about odd numbers is that they can be represented as 2k - 1, where k is a positive integer. For example, when k = 1, we get 1; k = 2, we get 3, etc. So if this is the case, then the k-th term is 2k - 1. Therefore, if we want the 2003rd term, that would correspond to k = 2003, so 2*2003 -1 = 4005.

That's the same result as before. So that seems to confirm it.

Alternatively, maybe I can think about how many numbers are there between 1 and the term I'm looking for. But that might complicate things. However, let me try.

Each odd number is 1 more than an even number. Wait, maybe not helpful. Alternatively, if I think about the position of the term in the sequence.

For example, the first term is 1, which is 2*1 - 1. The second term is 3, which is 2*2 - 1. So in general, the nth term is 2n - 1.

Alternatively, if I think of the sequence of natural numbers, the odd numbers are the numbers that are not divisible by 2. So the nth odd number is... Well, if you think about it, the nth odd number is equal to 2n - 1.

Another way to see this is that if I have n numbers, each separated by 2, starting at 1. So the total difference from the first term would be (n - 1)*2. So adding that to the first term: 1 + (n - 1)*2 = 2n - 1.

All methods lead to the same formula. Therefore, I feel confident that the formula is correct.

So applying it for n = 2003:

2*2003 = 4006. Then subtract 1 gives 4005.

Let me check with smaller numbers again. Let's take n = 4, which should be 7.

Using the formula: 2*4 -1 = 8 -1 = 7. Correct.

Another example: n = 5. 2*5 -1 = 10 -1 = 9. Which is the fifth odd number. 1, 3, 5, 7, 9. Yes, that's right.

Therefore, seems solid.

Alternatively, maybe I can think about the relationship between odd numbers and squares? Wait, there's a thing that the sum of the first n odd numbers is n squared. But that's a different problem.

But maybe that can help here? Let me think. For example, 1 = 1², 1 + 3 = 4 = 2², 1 + 3 + 5 = 9 = 3², etc.

So the sum of the first n odd numbers is n². However, the question here isn't about the sum, but about the value of the nth term.

But maybe there is a way to connect them? For example, if I know that the sum of the first n odd numbers is n², then if I wanted to find the nth odd number, maybe I can take the difference between the sum up to n and the sum up to n-1.

So sum_{k=1}^n (2k -1) = n². Therefore, the nth term would be sum_{k=1}^n (2k -1) - sum_{k=1}^{n-1} (2k -1) = n² - (n - 1)^2.

Calculating that: n² - (n² - 2n + 1) = n² - n² + 2n -1 = 2n -1.

So, that gives the same formula. Therefore, the nth term is 2n -1. So again, the same result.

Therefore, even using this method, we get that the nth term is 2n -1.

So, using this formula for n = 2003, we get 2*2003 -1 = 4005.

Therefore, I think that's the correct answer.

But just to be thorough, let me check with another approach.

Suppose we list the odd numbers. The first term is 1, which is 2*1 -1. The second term is 3, which is 2*2 -1. The third term is 5, which is 2*3 -1. So in general, term number n is 2n -1. Therefore, term 2003 is 2*2003 -1.

Alternatively, if I think about how many odd numbers there are up to a certain number. For example, if I have a number x, the number of odd numbers less than or equal to x is ceil(x/2). But again, not sure if that helps here.

But if I want to find the nth odd number, it's the same as 2n -1. For example, if n is 1, then 2*1 -1 = 1, which is the first odd number. If n is 2, 3, etc.

Alternatively, maybe think of it as the nth odd number is the same as the (n + (n -1))th number in the natural numbers? Wait, no.

Wait, the odd numbers are 1, 3, 5,... So each odd number is 2 less than the next even number. Wait, maybe that's complicating.

Alternatively, think of the nth odd number as the number that is two less than twice n. Wait, 2n - 2? No, that would be 2(n - 1). But the first term is 2(1) -1.

Alternatively, maybe think of it as the numbers congruent to 1 mod 2. So, numbers that are 1 more than an even number.

But again, this seems like going in circles.

Alternatively, think about the position in the sequence. Each term is 2 more than the previous term, starting at 1. So the sequence is 1, 1 + 2, 1 + 2*2, 1 + 2*3,... So general term is 1 + 2(n -1) = 2n -1.

Same formula again.

So all different ways of thinking about it lead me to the same conclusion.

Therefore, unless I made a calculation mistake when computing 2*2003 -1, which is 4005, that should be correct.

Wait, let me check 2*2003. Let me compute 2003*2.

2000*2 = 4000, 3*2 = 6, so 4000 + 6 = 4006. Then subtract 1: 4006 - 1 = 4005. Yep, that seems right.

Alternatively, if I add 1 + 2*(2003 -1). That would be 1 + 2*(2002). 2002*2 = 4004, then 4004 + 1 = 4005. Same result.

So, 4005 is the correct answer.

Is there a possibility that the problem is referring to something else? Like starting the sequence at 0? But no, the sequence is given as 1, 3, 5, 7,... So it's definitely starting at 1.

If for some reason the sequence was starting at 0, then the term would be different, but that's not the case here.

Alternatively, could there be a misunderstanding about the term number? Like, is the first term considered as term 0? But the problem says "the 2003rd term", so that would be term number 2003, which is the 2003rd element in the sequence.

In most mathematical contexts, the first term is term 1, so term 1 is 1, term 2 is 3, etc. Therefore, term number 2003 is 2*2003 -1.

Alternatively, if someone thought that the first term is term 0, then term 2003 would be 2*2004 -1, which would be 4007, but that's not the case here.

Since the problem states the sequence as 1, 3, 5, 7,..., and asks for the 2003rd term, it's safe to assume that the first term is term 1.

Therefore, I think the answer is definitely 4005.

Wait, just to make sure, maybe I can check with a smaller term. For example, term 5 is 9. If I use the formula, 2*5 -1 = 9. Correct.

If I think about term 10, it should be 19. 2*10 -1 = 19. Correct. So the formula works for these.

Alternatively, if I try term 100, it would be 2*100 -1 = 199. Let me count... Well, the 100th odd number is 199. If I think of it as the 100th term, then yes.

Alternatively, if I think of the even numbers. The nth even number is 2n. The odd numbers are the even numbers minus 1. So the nth odd number is 2n -1. Exactly.

Therefore, that also makes sense.

So, given that, term 2003 is 2*2003 -1 = 4005.

Therefore, I think I can be confident that this is the correct answer.

**Final Answer**
The 2003rd term of the sequence is \boxed{4005}.

To determine the **2003rd term** of the sequence of odd numbers, which begins as:

$$
1, 3, 5, 7, \dots
$$

we recognize that this is an **arithmetic sequence** with:

- First term $ a_1 = 1 $
- Common difference $ d = 2 $

---

### Step 1: General Formula for the $ n $-th Term of an Arithmetic Sequence

The formula for the $ n $-th term of an arithmetic sequence is:

$$
a_n = a_1 + (n - 1)d
$$

Substitute the values for our sequence:

$$
a_n = 1 + (n - 1) \cdot 2 = 1 + 2(n - 1)
$$

Simplify:

$$
a_n = 2n - 1
$$

This confirms that the $ n $-th term of the sequence of odd numbers is $ 2n - 1 $.

---

### Step 2: Apply the Formula for the 2003rd Term

Now, substitute $ n = 2003 $ into the simplified formula:

$$
a_{2003} = 2 \cdot 2003 - 1 = 4006 - 1 = 4005
$$

---

### Final Answer

$$
\boxed{4005}
$$

Solution: The sequence of odd numbers 1, 3, 5, 7, and so on, is an arithmetic sequence, with common difference 2.  Therefore, the $2003^{\text{rd}}$ term is $1+2002\cdot2=\boxed{4005}$.


Explanation:
Okay, so I need to find the value of g(f(5) - 1). Let me think about how to approach this. First, I know that f(x) is 2x - 3 and g(x) is x + 1. The problem is asking for g evaluated at (f(5) - 1). So, I guess I need to start by finding f(5), then subtract 1 from that result, and finally plug that into the function g. Let me write that down step by step.

First, let me compute f(5). Since f(x) = 2x - 3, substituting x with 5 should give me f(5) = 2*(5) - 3. Let me calculate that. 2 times 5 is 10, right? And then subtract 3, so 10 - 3 equals 7. So, f(5) is 7. Let me double-check that. If x is 5, then 2*5 is 10, minus 3 is 7. Yep, that seems right.

Now, the next part is to compute f(5) - 1. Since I already found that f(5) is 7, subtracting 1 from that would be 7 - 1. That equals 6. So, f(5) - 1 is 6. Let me confirm once more: 7 minus 1 is indeed 6. Okay, so that part checks out.

Now, I need to find g of that result, which is g(6). Since g(x) is defined as x + 1, substituting 6 into g would be g(6) = 6 + 1. That's straightforward. 6 + 1 is 7. Wait, so g(6) is 7? That seems simple enough. Let me check again. If x is 6, then adding 1 gives 7. Yep, that's correct.

So putting it all together, g(f(5) - 1) equals 7. Hmm. Let me make sure I didn't mix up any steps. Let me go through it again.

Start with f(5): 2*5 - 3 = 10 - 3 = 7. Then subtract 1: 7 - 1 = 6. Then apply g to 6: 6 + 1 = 7. Yep, same result. So, the answer is 7? That seems straightforward, but maybe I should check if there's a different interpretation of the question. For example, is it possible that the problem is asking for g(f(5 - 1))? But no, the original problem says g(f(5) - 1), which is different from g(f(5 - 1)). The parentheses are important here. So according to order of operations, we need to compute f(5) first, then subtract 1, then apply g. So, my steps are correct.

Alternatively, if someone misreads it as g(f(5 - 1)), then they would compute 5 - 1 first, which is 4, then f(4) would be 2*4 - 3 = 8 - 3 = 5, then g(5) = 5 + 1 = 6. But that's a different answer. However, the problem specifically says g(f(5) - 1), so the correct interpretation is the first one. So I think my answer is correct.

Wait, but let me make sure that there isn't any other ambiguity. The expression is g(f(5) - 1). Since f(5) is a single value, then subtract 1, so it's definitely g evaluated at (f(5) - 1). So, no confusion there.

Another way to think about it is to write it as g(f(5) - 1) = g(7 - 1) = g(6) = 6 + 1 = 7. So, yes, that's all consistent.

Therefore, I think the answer is 7. I don't see any mistakes in my calculations. Let me just verify each step once more:

1. f(5) = 2*5 - 3 = 10 - 3 = 7 ✔
2. f(5) - 1 = 7 - 1 = 6 ✔
3. g(6) = 6 + 1 = 7 ✔

All steps are correct. Therefore, the value of g(f(5) - 1) is indeed 7. I think that's the final answer.

**Final Answer**
The value of $ g(f(5) - 1) $ is \boxed{7}.

To solve the problem, we are asked to evaluate the expression:

$$
g(f(5) - 1)
$$

We are given the functions:

- $ f(x) = 2x - 3 $
- $ g(x) = x + 1 $

---

### Step 1: Evaluate $ f(5) $

Use the definition of $ f(x) $:

$$
f(5) = 2(5) - 3 = 10 - 3 = 7
$$

---

### Step 2: Compute $ f(5) - 1 $

$$
f(5) - 1 = 7 - 1 = 6
$$

---

### Step 3: Apply $ g $ to the result from Step 2

$$
g(6) = 6 + 1 = 7
$$

---

### Final Answer

$$
g(f(5) - 1) = \boxed{7}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of g(f(5) - 1). Let me think about how to approach this. First, I know that f(x) is 2x - 3 and g(x) is x + 1. The problem is asking for g evaluated at (f(5) - 1). So, I guess I need to start by finding f(5), then subtract 1 from that result, and finally plug that into the function g. Let me write that down step by step.

First, let me compute f(5). Since f(x) = 2x - 3, substituting x with 5 should give me f(5) = 2*(5) - 3. Let me calculate that. 2 times 5 is 10, right? And then subtract 3, so 10 - 3 equals 7. So, f(5) is 7. Let me double-check that. If x is 5, then 2*5 is 10, minus 3 is 7. Yep, that seems right.

Now, the next part is to compute f(5) - 1. Since I already found that f(5) is 7, subtracting 1 from that would be 7 - 1. That equals 6. So, f(5) - 1 is 6. Let me confirm once more: 7 minus 1 is indeed 6. Okay, so that part checks out.

Now, I need to find g of that result, which is g(6). Since g(x) is defined as x + 1, substituting 6 into g would be g(6) = 6 + 1. That's straightforward. 6 + 1 is 7. Wait, so g(6) is 7? That seems simple enough. Let me check again. If x is 6, then adding 1 gives 7. Yep, that's correct.

So putting it all together, g(f(5) - 1) equals 7. Hmm. Let me make sure I didn't mix up any steps. Let me go through it again.

Start with f(5): 2*5 - 3 = 10 - 3 = 7. Then subtract 1: 7 - 1 = 6. Then apply g to 6: 6 + 1 = 7. Yep, same result. So, the answer is 7? That seems straightforward, but maybe I should check if there's a different interpretation of the question. For example, is it possible that the problem is asking for g(f(5 - 1))? But no, the original problem says g(f(5) - 1), which is different from g(f(5 - 1)). The parentheses are important here. So according to order of operations, we need to compute f(5) first, then subtract 1, then apply g. So, my steps are correct.

Alternatively, if someone misreads it as g(f(5 - 1)), then they would compute 5 - 1 first, which is 4, then f(4) would be 2*4 - 3 = 8 - 3 = 5, then g(5) = 5 + 1 = 6. But that's a different answer. However, the problem specifically says g(f(5) - 1), so the correct interpretation is the first one. So I think my answer is correct.

Wait, but let me make sure that there isn't any other ambiguity. The expression is g(f(5) - 1). Since f(5) is a single value, then subtract 1, so it's definitely g evaluated at (f(5) - 1). So, no confusion there.

Another way to think about it is to write it as g(f(5) - 1) = g(7 - 1) = g(6) = 6 + 1 = 7. So, yes, that's all consistent.

Therefore, I think the answer is 7. I don't see any mistakes in my calculations. Let me just verify each step once more:

1. f(5) = 2*5 - 3 = 10 - 3 = 7 ✔
2. f(5) - 1 = 7 - 1 = 6 ✔
3. g(6) = 6 + 1 = 7 ✔

All steps are correct. Therefore, the value of g(f(5) - 1) is indeed 7. I think that's the final answer.

**Final Answer**
The value of $ g(f(5) - 1) $ is \boxed{7}.

To solve the problem, we are asked to evaluate the expression:

$$
g(f(5) - 1)
$$

We are given the functions:

- $ f(x) = 2x - 3 $
- $ g(x) = x + 1 $

---

### Step 1: Evaluate $ f(5) $

Use the definition of $ f(x) $:

$$
f(5) = 2(5) - 3 = 10 - 3 = 7
$$

---

### Step 2: Compute $ f(5) - 1 $

$$
f(5) - 1 = 7 - 1 = 6
$$

---

### Step 3: Apply $ g $ to the result from Step 2

$$
g(6) = 6 + 1 = 7
$$

---

### Final Answer

$$
g(f(5) - 1) = \boxed{7}
$$

Solution: We have $f(5) = 2(5) -3 = 7$, so $g(f(5)-1) = g(7-1) = g(6) = 6+1 = \boxed{7}$.


Explanation:
Okay, so I need to figure out which of these equations has the largest solution for x, given that 0 < r < 3. All the equations are of the form 3 times something raised to the power of x equals 7. Let me write them down again to make sure I have them right:

A) 3(1 + r)^x = 7
B) 3(1 + r/10)^x = 7
C) 3(1 + 2r)^x = 7
D) 3(1 + √r)^x = 7
E) 3(1 + 1/r)^x = 7

Since all equations are similar, I can probably solve for x in each case and then compare them. Let me recall how to solve an equation like this. If I have an equation like a*b^x = c, then solving for x would involve taking logarithms. Let me try that with each equation.

Starting with equation A:
3(1 + r)^x = 7
Divide both sides by 3:
(1 + r)^x = 7/3
Take natural logarithm on both sides:
x * ln(1 + r) = ln(7/3)
Therefore:
x = ln(7/3) / ln(1 + r)

Similarly for equation B:
3(1 + r/10)^x = 7
Divide by 3:
(1 + r/10)^x = 7/3
Take ln:
x * ln(1 + r/10) = ln(7/3)
x = ln(7/3) / ln(1 + r/10)

Same for equation C:
x = ln(7/3) / ln(1 + 2r)

Equation D:
x = ln(7/3) / ln(1 + √r)

Equation E:
x = ln(7/3) / ln(1 + 1/r)

So, in all cases, x is equal to ln(7/3) divided by the natural log of some base that depends on r. Since ln(7/3) is a constant (positive, since 7/3 > 1), the value of x is inversely proportional to the logarithm of the base. Therefore, to find which equation has the largest x, we need to find which equation has the smallest denominator, i.e., the smallest value of ln(base). Because if the denominator is smaller, then the overall x is larger.

So, the problem reduces to comparing the bases of the exponents in each equation. Since ln is an increasing function, the base with the smallest value will have the smallest ln(base), hence making x the largest. Wait, but hold on: the base is inside the log, so if the base is smaller, then ln(base) is smaller. So, if we have x = constant / ln(base), then smaller ln(base) would lead to larger x. Therefore, we need to find the equation where the base (which is (1 + r), (1 + r/10), (1 + 2r), (1 + √r), or (1 + 1/r)) is the smallest. Because the smaller the base, the smaller the ln(base), leading to larger x. Therefore, our task is to determine which of these bases is the smallest for 0 < r < 3.

But we have to be careful here. Wait, but maybe not? Let me think again. Let me check: For example, suppose two equations: 3*(base1)^x = 7 and 3*(base2)^x = 7. Then, solving for x, we get x = ln(7/3)/ln(base1) and x = ln(7/3)/ln(base2). So, if base1 < base2, then ln(base1) < ln(base2), so 1/ln(base1) > 1/ln(base2), meaning that x1 > x2. So yes, the smaller base gives a larger x. Therefore, to find the equation with the largest x, we need the equation with the smallest base. Therefore, the problem reduces to finding which of the five bases is the smallest, given that 0 < r < 3.

Therefore, let me analyze each base:

Base for A: 1 + r
Base for B: 1 + r/10
Base for C: 1 + 2r
Base for D: 1 + √r
Base for E: 1 + 1/r

We need to compare these bases for 0 < r < 3. Let me note that r is between 0 and 3, not including the endpoints.

So first, let's analyze each base:

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

Base B: 1 + r/10. Since 0 < r < 3, then r/10 is between 0 and 0.3, so 1 + r/10 is between 1 and 1.3.

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

Base D: 1 + √r. Since 0 < r < 3, √r is between 0 and √3 ≈ 1.732. Therefore, 1 + √r is between 1 and approx 2.732.

Base E: 1 + 1/r. Since 0 < r < 3, 1/r is greater than 1/3. However, since r approaches 0 from the right, 1/r approaches infinity. When r approaches 3 from the left, 1/r approaches 1/3. So, 1 + 1/r is between 1 + 1/3 ≈ 1.333 and infinity.

So, the bases:

A: between 1 and 4
B: between 1 and 1.3
C: between 1 and 7
D: between 1 and approx 2.73
E: between approx 1.333 and infinity

So, if we want the smallest base, which one is the smallest? Well, looking at the ranges: B is between 1 and 1.3, E is above 1.333, D is up to about 2.73, A up to 4, C up to 7. So, B is the smallest base? Wait, but hold on. Wait, base B is 1 + r/10. But r is between 0 and 3, so r/10 is between 0 and 0.3. So, base B is between 1 and 1.3. However, base E is 1 + 1/r. For r approaching 0, 1/r becomes very large, so base E can be as large as we want, but for r approaching 3, 1/r approaches 1/3, so base E approaches 1 + 1/3 ≈ 1.333. So, base E is between approximately 1.333 and infinity, while base B is between 1 and 1.3. Therefore, base B is definitely smaller than base E because even when r is at its maximum (r approaching 3), base B approaches 1 + 3/10 = 1.3, while base E approaches 1 + 1/3 ≈ 1.333. Therefore, base B is smaller than base E for all 0 < r < 3. Similarly, base B is between 1 and 1.3, while base D is between 1 and approx 2.73, so base B is smaller than D. Similarly, base B is smaller than A and C as well.

Therefore, base B is the smallest among all. Therefore, according to our previous reasoning, equation B would have the largest x? Wait, but hold on, because if base is smaller, then ln(base) is smaller, hence x is larger. Therefore, equation B would have the largest x? Wait, but wait, the answer might not be B. Wait, but hold on. Wait, hold on! Wait, let me check again. Because equation E: base is 1 + 1/r. Let me check for specific values. Let me pick a value for r and compute the bases for each equation to compare. Maybe my previous conclusion is wrong. Let me take r = 1, for example. Then:

For r = 1:

A: 1 + 1 = 2
B: 1 + 1/10 = 1.1
C: 1 + 2*1 = 3
D: 1 + √1 = 2
E: 1 + 1/1 = 2

So, bases are:

A: 2
B: 1.1
C: 3
D: 2
E: 2

So, here, base B is smallest, so according to our logic, equation B would have the largest x. However, what if we take another value of r. Let me try r = 2:

For r = 2:

A: 1 + 2 = 3
B: 1 + 2/10 = 1.2
C: 1 + 2*2 = 5
D: 1 + √2 ≈ 1 + 1.414 ≈ 2.414
E: 1 + 1/2 = 1.5

So, bases here:

A: 3
B: 1.2
C: 5
D: ≈2.414
E: 1.5

So, again base B is 1.2 which is smaller than E's 1.5. So, still base B is the smallest.

But let me check for r approaching 0. Suppose r approaches 0, say r = 0.1:

For r = 0.1:

A: 1 + 0.1 = 1.1
B: 1 + 0.1/10 = 1 + 0.01 = 1.01
C: 1 + 2*0.1 = 1.2
D: 1 + √0.1 ≈ 1 + 0.316 ≈ 1.316
E: 1 + 1/0.1 = 1 + 10 = 11

So, base B is 1.01, which is smaller than base A (1.1), so again base B is the smallest. However, if r approaches 0, base E becomes very large (like 1 + 1/r approaches infinity), so base B is still the smallest. So, in all these examples, base B is the smallest. Therefore, according to that, equation B would have the largest x. But wait, but wait, let me check if maybe for some r, another base is smaller than base B?

Wait, for example, can the base in E ever be smaller than base B?

Base E is 1 + 1/r, base B is 1 + r/10. Let me see if 1 + 1/r < 1 + r/10 for some r in (0, 3). Let's solve 1 + 1/r < 1 + r/10. Subtract 1 from both sides: 1/r < r/10. Multiply both sides by 10r (since r > 0): 10 < r^2. Therefore, r^2 > 10. So, r > sqrt(10) ≈ 3.16. However, our range is 0 < r < 3, so sqrt(10) ≈ 3.16 is outside of this range. Therefore, in the entire range 0 < r < 3, we have 1/r > 1/3, so 1 + 1/r > 1 + 1/3 ≈ 1.333, while base B is 1 + r/10. The maximum value of base B is when r approaches 3, so 1 + 3/10 = 1.3. Therefore, base B approaches 1.3, while base E is always greater than 1.333. Hence, in the entire range, base B is less than base E. So, base B is the smallest. Similarly, base D is 1 + sqrt(r). Let's compare base D with base B. For example, when r is 1, base D is 2, base B is 1.1. For r approaching 0, base D approaches 1 + 0 = 1, while base B approaches 1 + 0 = 1. But for r approaching 0, base B approaches 1 + 0 = 1, but base D approaches 1 + 0 = 1? Wait, no: for r approaching 0, sqrt(r) approaches 0, so base D approaches 1 + 0 = 1. Similarly, base B approaches 1 + 0 = 1. But when r is very small, say r = 0.01:

Base B: 1 + 0.01/10 = 1 + 0.001 = 1.001
Base D: 1 + sqrt(0.01) = 1 + 0.1 = 1.1

So in this case, base B is smaller. What if r is such that sqrt(r) is smaller than r/10? Let me see when sqrt(r) < r/10. Let me solve sqrt(r) < r/10. Multiply both sides by 10: 10 sqrt(r) < r. Let me square both sides (since both sides are positive): 100 r < r^2 => r^2 - 100 r > 0 => r(r - 100) > 0. So, this inequality holds when r > 100 or r < 0. But since r is positive, this would require r > 100. However, our domain is 0 < r < 3, so this inequality never holds. Therefore, for all 0 < r < 3, sqrt(r) > r/10. Therefore, 1 + sqrt(r) > 1 + r/10. Therefore, base D is greater than base B. Therefore, base B is still the smallest.

Similarly, for base A, 1 + r. Since r > 0, 1 + r > 1, and since base B is 1 + r/10, which is less than 1 + r (since r/10 < r for positive r). Therefore, base B is smaller than base A. Similarly, base C is 1 + 2r, which for r > 0 is greater than 1 + r, so again base B is smaller. Therefore, in conclusion, base B is the smallest among all the bases for 0 < r < 3. Therefore, according to our previous reasoning, equation B has the largest x. However, wait, let me check with another example where perhaps base E could be smaller? Wait, earlier analysis showed that base E is always larger than base B in the given range. Therefore, equation B is the one with the smallest base, so equation B's x is the largest. However, let me check for some specific value again. Let's take r = 0.5:

For r = 0.5:

A: 1 + 0.5 = 1.5
B: 1 + 0.5/10 = 1.05
C: 1 + 2*0.5 = 2
D: 1 + sqrt(0.5) ≈ 1 + 0.707 ≈ 1.707
E: 1 + 1/0.5 = 1 + 2 = 3

So, here, base B is 1.05, which is the smallest. Therefore, equation B has the smallest base, so x is the largest. Therefore, according to this, the answer is B. However, wait, the answer options are A, B, C, D, E. But let me check again. Wait, in the problem statement, the answer options are:

(A) 3(1 + r)^x = 7
(B) 3(1 + r/10)^x = 7
(C) 3(1 + 2r)^x = 7
(D) 3(1 + √r)^x = 7
(E) 3(1 + 1/r)^x = 7

But according to my analysis, equation B has the smallest base, so equation B would have the largest x. However, the answer might be different? Let me check again. Wait, wait, perhaps I made a mistake here. Wait, the question is asking for the equation with the largest solution x. If the base is smaller, then ln(base) is smaller, so x is larger. Therefore, equation B would have larger x than the others. However, let me test with numbers. Let me choose r = 1, then compute x for each equation.

Let me compute x for each equation when r = 1.

For equation A:
x = ln(7/3) / ln(1 + 1) = ln(7/3)/ln(2) ≈ (0.8473)/0.6931 ≈ 1.222

For equation B:
x = ln(7/3)/ln(1 + 1/10) = ln(7/3)/ln(1.1) ≈ 0.8473 / 0.09531 ≈ 8.89

For equation C:
x = ln(7/3)/ln(1 + 2*1) = ln(7/3)/ln(3) ≈ 0.8473 / 1.0986 ≈ 0.771

For equation D:
x = ln(7/3)/ln(1 + √1) = ln(7/3)/ln(2) ≈ same as A, so ≈ 1.222

For equation E:
x = ln(7/3)/ln(1 + 1/1) = ln(7/3)/ln(2) ≈ same as A and D, so ≈ 1.222

So here, equation B gives x ≈ 8.89, which is much larger than others. So in this case, equation B has the largest x.

But wait, let me test with another value of r, say r = 2. Let me compute x for each equation.

For r = 2:

Equation A:
x = ln(7/3)/ln(1 + 2) = ln(7/3)/ln(3) ≈ 0.8473 / 1.0986 ≈ 0.771

Equation B:
x = ln(7/3)/ln(1 + 2/10) = ln(7/3)/ln(1.2) ≈ 0.8473 / 0.1823 ≈ 4.647

Equation C:
x = ln(7/3)/ln(1 + 4) = ln(7/3)/ln(5) ≈ 0.8473 / 1.6094 ≈ 0.526

Equation D:
x = ln(7/3)/ln(1 + √2) ≈ 0.8473 / ln(2.4142) ≈ 0.8473 / 0.8813 ≈ 0.961

Equation E:
x = ln(7/3)/ln(1 + 1/2) = ln(7/3)/ln(1.5) ≈ 0.8473 / 0.4055 ≈ 2.09

So again, equation B gives x ≈ 4.647, which is the largest. So even here, equation B is the largest. Let me try with r approaching zero, say r = 0.1:

Equation A:
x = ln(7/3)/ln(1 + 0.1) ≈ 0.8473 / 0.09531 ≈ 8.89

Equation B:
x = ln(7/3)/ln(1 + 0.01) ≈ 0.8473 / 0.00995 ≈ 85.15

Equation C:
x = ln(7/3)/ln(1 + 0.2) ≈ 0.8473 / 0.1823 ≈ 4.647

Equation D:
x = ln(7/3)/ln(1 + sqrt(0.1)) ≈ 0.8473 / ln(1.316) ≈ 0.8473 / 0.2755 ≈ 3.075

Equation E:
x = ln(7/3)/ln(1 + 1/0.1) = ln(7/3)/ln(11) ≈ 0.8473 / 2.3979 ≈ 0.354

So here, equation B gives x ≈85.15, which is the largest. Therefore, seems like equation B is giving the largest x for different values of r. However, let me check for another value where maybe another equation might have larger x. Let me check for r approaching 3, say r = 2.9:

For r = 2.9:

Equation A:
x = ln(7/3)/ln(1 + 2.9) ≈ ln(7/3)/ln(3.9) ≈ 0.8473 / 1.3609 ≈ 0.622

Equation B:
x = ln(7/3)/ln(1 + 2.9/10) = ln(7/3)/ln(1 + 0.29) ≈ 0.8473 / ln(1.29) ≈ 0.8473 / 0.2553 ≈ 3.32

Equation C:
x = ln(7/3)/ln(1 + 2*2.9) = ln(7/3)/ln(1 + 5.8) ≈ 0.8473 / ln(6.8) ≈ 0.8473 / 1.916 ≈ 0.442

Equation D:
x = ln(7/3)/ln(1 + sqrt(2.9)) ≈ sqrt(2.9) ≈ 1.702, so ln(1 + 1.702) = ln(2.702) ≈ 0.996, so x ≈ 0.8473 / 0.996 ≈ 0.851

Equation E:
x = ln(7/3)/ln(1 + 1/2.9) ≈ ln(7/3)/ln(1 + 0.3448) ≈ ln(7/3)/ln(1.3448) ≈ 0.8473 / 0.298 ≈ 2.84

So here, equation B gives x ≈3.32, which is the largest. So again, equation B is the largest. Therefore, seems like equation B has the largest x.

But hold on, wait, the answer options don't include B as the answer? Wait no, the answer options are A to E. However, let me check if there's a mistake in my reasoning. Because according to the problem statement, 0 < r < 3. But when I considered r approaching 0, equation B gives very large x. But maybe there is a case where another equation has a larger x? Wait, let me think again. Suppose for some r, the base for equation E is smaller than base B? Earlier, we saw that it's impossible because 1 + 1/r > 1 + r/10 for all r in (0, 3). Wait, but maybe equation D? Wait, equation D's base is 1 + sqrt(r). For some r, maybe base D is smaller than base B?

Let me check when is 1 + sqrt(r) < 1 + r/10? That is, sqrt(r) < r/10. As before, this is equivalent to r^2 > 100r => r > 100. But our domain is 0 < r < 3, so no solution. Therefore, equation D's base is always larger than equation B's base. So equation B's base is always the smallest. Therefore, equation B has the largest x. So the answer should be B. However, I need to check the answer options again. Wait, the problem says "Which equation has the largest solution x?" So according to all my analysis, it should be equation B. But let me check original problem again to make sure I read it correctly.

Wait, the problem says "Assume 0 < r < 3. Below are five equations for x. Which equation has the largest solution x?" Then equations:

A) 3(1 + r)^x = 7
B) 3(1 + r/10)^x = 7
C) 3(1 + 2r)^x = 7
D) 3(1 + √r)^x = 7
E) 3(1 + 1/r)^x = 7

Yes, that's what I considered. So according to my analysis, equation B has the largest x. But let me check if perhaps there's an error in assuming that smaller base leads to larger x. Let me recall that x = ln(7/3)/ln(base). So, if base is smaller, then ln(base) is smaller (since base > 1, because 1 + something positive). Therefore, 1/ln(base) is larger, so x is larger. Therefore, equation with smallest base has largest x. Therefore, equation B is the answer. However, let me check if maybe the question is designed to have a different answer. Wait, let me check again for r = 1, equation B gives x ≈8.89, which is significantly larger than others. Let me check equation E for r =1: 1 + 1/r = 2, same as equation A and D. So equation E gives same x as A and D. So equation B is the only one with larger x.

Alternatively, maybe there's a case where equation E has a smaller base? For example, if 1 + 1/r is smaller than 1 + r/10. But as we saw earlier, that would require r > sqrt(10) which is outside the domain. Hence, no. Therefore, equation B is the answer.

But wait, let me check the answer choices again. Wait, maybe I made a mistake in the problem statement? Wait, the original problem is:

"Assume 0 < r < 3. Below are five equations for x. Which equation has the largest solution x?

(A) 3(1 + r)^x = 7
(B) 3(1 + r/10)^x = 7
(C) 3(1 + 2r)^x = 7
(D) 3(1 + √r)^x = 7
(E) 3(1 + 1/r)^x = 7"

Yes, that's correct. So, according to this, equation B is correct. But let me check again with r approaching 0. Let me take r approaching 0, say r = 0.0001:

Equation B: base = 1 + 0.0001/10 = 1 + 0.00001 = 1.00001. Then ln(1.00001) ≈ 0.00001. Therefore, x ≈ ln(7/3)/0.00001 ≈ 0.8473 / 0.00001 ≈ 84730. Which is huge.

Equation E: base = 1 + 1/r = 1 + 1/0.0001 = 1 + 10000 = 10001. So ln(10001) ≈ 9.21. Therefore, x ≈ 0.8473 / 9.21 ≈ 0.092, which is way smaller. Therefore, equation B is way larger. So, seems like equation B is the answer.

But maybe the answer is E? Wait, no. Wait, let me check the problem again. Wait, the problem says "Assume 0 < r < 3". For example, if r is close to 0, then 1/r is very large. However, in equation E, the base is 1 + 1/r, which is 1 + large number. Therefore, the base is large, so ln(base) is large, hence x is small. Whereas, in equation B, the base is 1 + r/10, which is very slightly larger than 1, so ln(base) is very small, so x is large. Therefore, equation B is the correct answer. So why do I feel uncertain? Maybe because sometimes in these problems, there is a trick. Wait, but all my calculations seem to confirm that equation B has the largest x. However, maybe there is a case where another equation has a smaller base?

Wait, let me check the base for equation D: 1 + sqrt(r). For r approaching 0, sqrt(r) approaches 0, so base approaches 1. Similarly, equation B's base approaches 1 as r approaches 0. But which one approaches 1 faster? For very small r, sqrt(r) is larger than r/10. For example, take r = 0.01:

sqrt(r) = 0.1, while r/10 = 0.001. Therefore, base D is 1 + 0.1 = 1.1, while base B is 1 + 0.001 = 1.001. Therefore, base B is smaller. So for any r > 0, base B is smaller than base D. Therefore, equation B's base is always the smallest. Therefore, equation B is the answer.

Therefore, the answer is B. However, maybe I need to check the answer options again. Wait, the original problem says "Which equation has the largest solution x?" and the answer choices are A through E. So according to my analysis, it's B. However, maybe the answer is E? Wait, let me check again when r is approaching 3.

Wait, when r approaches 3 from the left, let's take r = 2.99:

Equation E's base: 1 + 1/r ≈ 1 + 1/2.99 ≈ 1 + 0.334 ≈ 1.334

Equation B's base: 1 + r/10 ≈ 1 + 0.299 ≈ 1.299

Therefore, base B is smaller than base E. Therefore, equation B's x is larger. Therefore, even when r approaches 3, equation B is still better. Therefore, my conclusion is that equation B is the correct answer. Therefore, the answer should be \boxed{B}.

But wait, looking back at the original problem statement, I need to make sure that I didn't misread any of the equations. For example, equation E is 3(1 + 1/r)^x = 7. So, if r is between 0 and 3, then 1/r is greater than 1/3, so 1 + 1/r is greater than 4/3 ≈ 1.333. Therefore, equation E's base is always larger than equation B's base (which can be as low as approaching 1). So, equation B is definitely the one with the minimal base. Therefore, equation B has the largest x.

Therefore, I think the correct answer is B. However, to be 100% sure, let me check with another approach. Let me consider that for all equations, since the right-hand side is 7, the value of x depends on how "fast" the base grows. If the base grows faster, then x is smaller. If the base grows slower, x is larger. Therefore, to have the largest x, you want the base to grow as slowly as possible. Since for the equations, the base is 1 + something. So, the smaller the something, the slower the growth, so the larger x. Therefore, which equation has the smallest "something"? Let's see:

For equation A: something = r
For equation B: something = r/10
For equation C: something = 2r
For equation D: something = sqrt(r)
For equation E: something = 1/r

So, the question is, which of these "something" is the smallest for 0 < r < 3?

Let me analyze:

For 0 < r < 3, which of the following expressions is the smallest:

r, r/10, 2r, sqrt(r), 1/r.

We need to compare these expressions.

First, note that for r in (0, 3):

- r is between 0 and 3.
- r/10 is between 0 and 0.3.
- 2r is between 0 and 6.
- sqrt(r) is between 0 and sqrt(3) ≈ 1.732.
- 1/r is between 1/3 ≈ 0.333 and infinity.

Therefore, the smallest among these is r/10, since it's between 0 and 0.3, while others are either larger or unbounded. Therefore, equation B has the smallest "something", meaning the base grows the slowest, hence gives the largest x. Therefore, equation B is the answer. Therefore, my conclusion is consistent. Therefore, the correct answer is B.

**Final Answer**
\boxed{B}

To determine which equation has the **largest solution** for $ x $, we analyze the general structure of the given equations. All equations have the form:

$$
3 \cdot (1 + \text{something})^x = 7
$$

Our goal is to find the value of $ x $ for each equation and identify which one has the **maximum** value of $ x $ under the condition $ 0 < r < 3 $. Solving for $ x $ in each equation yields:

$$
x = \frac{\ln\left(\frac{7}{3}\right)}{\ln(1 + \text{something})}
$$

---

### Key Insight

Since the numerator $ \ln\left(\frac{7}{3}\right) $ is a **positive constant**, the value of $ x $ is **inversely proportional** to $ \ln(\text{base}) $, where the base is:

$$
1 + \text{something}
$$

Because the natural log function $ \ln(x) $ is **increasing**, a smaller base results in a smaller $ \ln(\text{base}) $, and thus a **larger** value of $ x $.

Hence, our task reduces to comparing the **bases** of each equation under the condition $ 0 < r < 3 $.

---

### Step 1: Analyze Each Base

Let's write down the base for each equation:

- (A): $ 1 + r $
- (B): $ 1 + \frac{r}{10} $
- (C): $ 1 + 2r $
- (D): $ 1 + \sqrt{r} $
- (E): $ 1 + \frac{1}{r} $

Now consider the possible values of the **something** added to 1:

- For $ r \in (0,3) $:
  - $ r \in (0, 3) $
  - $ \frac{r}{10} \in (0, 0.3) $
  - $ 2r \in (0, 6) $
  - $ \sqrt{r} \in (0, \sqrt{3}) \approx (0, 1.732) $
  - $ \frac{1}{r} \in (\frac{1}{3}, \infty) $

---

### Step 2: Compare the Expressions

We examine which expression is **smallest**:

- $ \frac{r}{10} $: Smallest value across the range due to division by 10
- $ r $: Larger than $ r/10 $
- $ 2r $: Much larger
- $ \sqrt{r} $: Larger than $ \frac{r}{10} $
- $ \frac{1}{r} $: Increases as $ r \to 0 $, but always larger than $ \frac{r}{10} $

So, the smallest value is:

$$
\frac{r}{10}
$$

Therefore, **Equation B** has the **smallest base**, and thus gives the **largest $ x $**.

---

### Final Conclusion

All other bases are greater than $ 1 + \frac{r}{10} $, leading to larger $ \ln(\text{base}) $, and thus smaller $ x $. This confirms that Equation B produces the **largest** value of $ x $ for the given constraint $ 0 < r < 3 $.

---

### 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. All the equations are of the form 3 times something raised to the power of x equals 7. Let me write them down again to make sure I have them right:

A) 3(1 + r)^x = 7
B) 3(1 + r/10)^x = 7
C) 3(1 + 2r)^x = 7
D) 3(1 + √r)^x = 7
E) 3(1 + 1/r)^x = 7

Since all equations are similar, I can probably solve for x in each case and then compare them. Let me recall how to solve an equation like this. If I have an equation like a*b^x = c, then solving for x would involve taking logarithms. Let me try that with each equation.

Starting with equation A:
3(1 + r)^x = 7
Divide both sides by 3:
(1 + r)^x = 7/3
Take natural logarithm on both sides:
x * ln(1 + r) = ln(7/3)
Therefore:
x = ln(7/3) / ln(1 + r)

Similarly for equation B:
3(1 + r/10)^x = 7
Divide by 3:
(1 + r/10)^x = 7/3
Take ln:
x * ln(1 + r/10) = ln(7/3)
x = ln(7/3) / ln(1 + r/10)

Same for equation C:
x = ln(7/3) / ln(1 + 2r)

Equation D:
x = ln(7/3) / ln(1 + √r)

Equation E:
x = ln(7/3) / ln(1 + 1/r)

So, in all cases, x is equal to ln(7/3) divided by the natural log of some base that depends on r. Since ln(7/3) is a constant (positive, since 7/3 > 1), the value of x is inversely proportional to the logarithm of the base. Therefore, to find which equation has the largest x, we need to find which equation has the smallest denominator, i.e., the smallest value of ln(base). Because if the denominator is smaller, then the overall x is larger.

So, the problem reduces to comparing the bases of the exponents in each equation. Since ln is an increasing function, the base with the smallest value will have the smallest ln(base), hence making x the largest. Wait, but hold on: the base is inside the log, so if the base is smaller, then ln(base) is smaller. So, if we have x = constant / ln(base), then smaller ln(base) would lead to larger x. Therefore, we need to find the equation where the base (which is (1 + r), (1 + r/10), (1 + 2r), (1 + √r), or (1 + 1/r)) is the smallest. Because the smaller the base, the smaller the ln(base), leading to larger x. Therefore, our task is to determine which of these bases is the smallest for 0 < r < 3.

But we have to be careful here. Wait, but maybe not? Let me think again. Let me check: For example, suppose two equations: 3*(base1)^x = 7 and 3*(base2)^x = 7. Then, solving for x, we get x = ln(7/3)/ln(base1) and x = ln(7/3)/ln(base2). So, if base1 < base2, then ln(base1) < ln(base2), so 1/ln(base1) > 1/ln(base2), meaning that x1 > x2. So yes, the smaller base gives a larger x. Therefore, to find the equation with the largest x, we need the equation with the smallest base. Therefore, the problem reduces to finding which of the five bases is the smallest, given that 0 < r < 3.

Therefore, let me analyze each base:

Base for A: 1 + r
Base for B: 1 + r/10
Base for C: 1 + 2r
Base for D: 1 + √r
Base for E: 1 + 1/r

We need to compare these bases for 0 < r < 3. Let me note that r is between 0 and 3, not including the endpoints.

So first, let's analyze each base:

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

Base B: 1 + r/10. Since 0 < r < 3, then r/10 is between 0 and 0.3, so 1 + r/10 is between 1 and 1.3.

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

Base D: 1 + √r. Since 0 < r < 3, √r is between 0 and √3 ≈ 1.732. Therefore, 1 + √r is between 1 and approx 2.732.

Base E: 1 + 1/r. Since 0 < r < 3, 1/r is greater than 1/3. However, since r approaches 0 from the right, 1/r approaches infinity. When r approaches 3 from the left, 1/r approaches 1/3. So, 1 + 1/r is between 1 + 1/3 ≈ 1.333 and infinity.

So, the bases:

A: between 1 and 4
B: between 1 and 1.3
C: between 1 and 7
D: between 1 and approx 2.73
E: between approx 1.333 and infinity

So, if we want the smallest base, which one is the smallest? Well, looking at the ranges: B is between 1 and 1.3, E is above 1.333, D is up to about 2.73, A up to 4, C up to 7. So, B is the smallest base? Wait, but hold on. Wait, base B is 1 + r/10. But r is between 0 and 3, so r/10 is between 0 and 0.3. So, base B is between 1 and 1.3. However, base E is 1 + 1/r. For r approaching 0, 1/r becomes very large, so base E can be as large as we want, but for r approaching 3, 1/r approaches 1/3, so base E approaches 1 + 1/3 ≈ 1.333. So, base E is between approximately 1.333 and infinity, while base B is between 1 and 1.3. Therefore, base B is definitely smaller than base E because even when r is at its maximum (r approaching 3), base B approaches 1 + 3/10 = 1.3, while base E approaches 1 + 1/3 ≈ 1.333. Therefore, base B is smaller than base E for all 0 < r < 3. Similarly, base B is between 1 and 1.3, while base D is between 1 and approx 2.73, so base B is smaller than D. Similarly, base B is smaller than A and C as well.

Therefore, base B is the smallest among all. Therefore, according to our previous reasoning, equation B would have the largest x? Wait, but hold on, because if base is smaller, then ln(base) is smaller, hence x is larger. Therefore, equation B would have the largest x? Wait, but wait, the answer might not be B. Wait, but hold on. Wait, hold on! Wait, let me check again. Because equation E: base is 1 + 1/r. Let me check for specific values. Let me pick a value for r and compute the bases for each equation to compare. Maybe my previous conclusion is wrong. Let me take r = 1, for example. Then:

For r = 1:

A: 1 + 1 = 2
B: 1 + 1/10 = 1.1
C: 1 + 2*1 = 3
D: 1 + √1 = 2
E: 1 + 1/1 = 2

So, bases are:

A: 2
B: 1.1
C: 3
D: 2
E: 2

So, here, base B is smallest, so according to our logic, equation B would have the largest x. However, what if we take another value of r. Let me try r = 2:

For r = 2:

A: 1 + 2 = 3
B: 1 + 2/10 = 1.2
C: 1 + 2*2 = 5
D: 1 + √2 ≈ 1 + 1.414 ≈ 2.414
E: 1 + 1/2 = 1.5

So, bases here:

A: 3
B: 1.2
C: 5
D: ≈2.414
E: 1.5

So, again base B is 1.2 which is smaller than E's 1.5. So, still base B is the smallest.

But let me check for r approaching 0. Suppose r approaches 0, say r = 0.1:

For r = 0.1:

A: 1 + 0.1 = 1.1
B: 1 + 0.1/10 = 1 + 0.01 = 1.01
C: 1 + 2*0.1 = 1.2
D: 1 + √0.1 ≈ 1 + 0.316 ≈ 1.316
E: 1 + 1/0.1 = 1 + 10 = 11

So, base B is 1.01, which is smaller than base A (1.1), so again base B is the smallest. However, if r approaches 0, base E becomes very large (like 1 + 1/r approaches infinity), so base B is still the smallest. So, in all these examples, base B is the smallest. Therefore, according to that, equation B would have the largest x. But wait, but wait, let me check if maybe for some r, another base is smaller than base B?

Wait, for example, can the base in E ever be smaller than base B?

Base E is 1 + 1/r, base B is 1 + r/10. Let me see if 1 + 1/r < 1 + r/10 for some r in (0, 3). Let's solve 1 + 1/r < 1 + r/10. Subtract 1 from both sides: 1/r < r/10. Multiply both sides by 10r (since r > 0): 10 < r^2. Therefore, r^2 > 10. So, r > sqrt(10) ≈ 3.16. However, our range is 0 < r < 3, so sqrt(10) ≈ 3.16 is outside of this range. Therefore, in the entire range 0 < r < 3, we have 1/r > 1/3, so 1 + 1/r > 1 + 1/3 ≈ 1.333, while base B is 1 + r/10. The maximum value of base B is when r approaches 3, so 1 + 3/10 = 1.3. Therefore, base B approaches 1.3, while base E is always greater than 1.333. Hence, in the entire range, base B is less than base E. So, base B is the smallest. Similarly, base D is 1 + sqrt(r). Let's compare base D with base B. For example, when r is 1, base D is 2, base B is 1.1. For r approaching 0, base D approaches 1 + 0 = 1, while base B approaches 1 + 0 = 1. But for r approaching 0, base B approaches 1 + 0 = 1, but base D approaches 1 + 0 = 1? Wait, no: for r approaching 0, sqrt(r) approaches 0, so base D approaches 1 + 0 = 1. Similarly, base B approaches 1 + 0 = 1. But when r is very small, say r = 0.01:

Base B: 1 + 0.01/10 = 1 + 0.001 = 1.001
Base D: 1 + sqrt(0.01) = 1 + 0.1 = 1.1

So in this case, base B is smaller. What if r is such that sqrt(r) is smaller than r/10? Let me see when sqrt(r) < r/10. Let me solve sqrt(r) < r/10. Multiply both sides by 10: 10 sqrt(r) < r. Let me square both sides (since both sides are positive): 100 r < r^2 => r^2 - 100 r > 0 => r(r - 100) > 0. So, this inequality holds when r > 100 or r < 0. But since r is positive, this would require r > 100. However, our domain is 0 < r < 3, so this inequality never holds. Therefore, for all 0 < r < 3, sqrt(r) > r/10. Therefore, 1 + sqrt(r) > 1 + r/10. Therefore, base D is greater than base B. Therefore, base B is still the smallest.

Similarly, for base A, 1 + r. Since r > 0, 1 + r > 1, and since base B is 1 + r/10, which is less than 1 + r (since r/10 < r for positive r). Therefore, base B is smaller than base A. Similarly, base C is 1 + 2r, which for r > 0 is greater than 1 + r, so again base B is smaller. Therefore, in conclusion, base B is the smallest among all the bases for 0 < r < 3. Therefore, according to our previous reasoning, equation B has the largest x. However, wait, let me check with another example where perhaps base E could be smaller? Wait, earlier analysis showed that base E is always larger than base B in the given range. Therefore, equation B is the one with the smallest base, so equation B's x is the largest. However, let me check for some specific value again. Let's take r = 0.5:

For r = 0.5:

A: 1 + 0.5 = 1.5
B: 1 + 0.5/10 = 1.05
C: 1 + 2*0.5 = 2
D: 1 + sqrt(0.5) ≈ 1 + 0.707 ≈ 1.707
E: 1 + 1/0.5 = 1 + 2 = 3

So, here, base B is 1.05, which is the smallest. Therefore, equation B has the smallest base, so x is the largest. Therefore, according to this, the answer is B. However, wait, the answer options are A, B, C, D, E. But let me check again. Wait, in the problem statement, the answer options are:

(A) 3(1 + r)^x = 7
(B) 3(1 + r/10)^x = 7
(C) 3(1 + 2r)^x = 7
(D) 3(1 + √r)^x = 7
(E) 3(1 + 1/r)^x = 7

But according to my analysis, equation B has the smallest base, so equation B would have the largest x. However, the answer might be different? Let me check again. Wait, wait, perhaps I made a mistake here. Wait, the question is asking for the equation with the largest solution x. If the base is smaller, then ln(base) is smaller, so x is larger. Therefore, equation B would have larger x than the others. However, let me test with numbers. Let me choose r = 1, then compute x for each equation.

Let me compute x for each equation when r = 1.

For equation A:
x = ln(7/3) / ln(1 + 1) = ln(7/3)/ln(2) ≈ (0.8473)/0.6931 ≈ 1.222

For equation B:
x = ln(7/3)/ln(1 + 1/10) = ln(7/3)/ln(1.1) ≈ 0.8473 / 0.09531 ≈ 8.89

For equation C:
x = ln(7/3)/ln(1 + 2*1) = ln(7/3)/ln(3) ≈ 0.8473 / 1.0986 ≈ 0.771

For equation D:
x = ln(7/3)/ln(1 + √1) = ln(7/3)/ln(2) ≈ same as A, so ≈ 1.222

For equation E:
x = ln(7/3)/ln(1 + 1/1) = ln(7/3)/ln(2) ≈ same as A and D, so ≈ 1.222

So here, equation B gives x ≈ 8.89, which is much larger than others. So in this case, equation B has the largest x.

But wait, let me test with another value of r, say r = 2. Let me compute x for each equation.

For r = 2:

Equation A:
x = ln(7/3)/ln(1 + 2) = ln(7/3)/ln(3) ≈ 0.8473 / 1.0986 ≈ 0.771

Equation B:
x = ln(7/3)/ln(1 + 2/10) = ln(7/3)/ln(1.2) ≈ 0.8473 / 0.1823 ≈ 4.647

Equation C:
x = ln(7/3)/ln(1 + 4) = ln(7/3)/ln(5) ≈ 0.8473 / 1.6094 ≈ 0.526

Equation D:
x = ln(7/3)/ln(1 + √2) ≈ 0.8473 / ln(2.4142) ≈ 0.8473 / 0.8813 ≈ 0.961

Equation E:
x = ln(7/3)/ln(1 + 1/2) = ln(7/3)/ln(1.5) ≈ 0.8473 / 0.4055 ≈ 2.09

So again, equation B gives x ≈ 4.647, which is the largest. So even here, equation B is the largest. Let me try with r approaching zero, say r = 0.1:

Equation A:
x = ln(7/3)/ln(1 + 0.1) ≈ 0.8473 / 0.09531 ≈ 8.89

Equation B:
x = ln(7/3)/ln(1 + 0.01) ≈ 0.8473 / 0.00995 ≈ 85.15

Equation C:
x = ln(7/3)/ln(1 + 0.2) ≈ 0.8473 / 0.1823 ≈ 4.647

Equation D:
x = ln(7/3)/ln(1 + sqrt(0.1)) ≈ 0.8473 / ln(1.316) ≈ 0.8473 / 0.2755 ≈ 3.075

Equation E:
x = ln(7/3)/ln(1 + 1/0.1) = ln(7/3)/ln(11) ≈ 0.8473 / 2.3979 ≈ 0.354

So here, equation B gives x ≈85.15, which is the largest. Therefore, seems like equation B is giving the largest x for different values of r. However, let me check for another value where maybe another equation might have larger x. Let me check for r approaching 3, say r = 2.9:

For r = 2.9:

Equation A:
x = ln(7/3)/ln(1 + 2.9) ≈ ln(7/3)/ln(3.9) ≈ 0.8473 / 1.3609 ≈ 0.622

Equation B:
x = ln(7/3)/ln(1 + 2.9/10) = ln(7/3)/ln(1 + 0.29) ≈ 0.8473 / ln(1.29) ≈ 0.8473 / 0.2553 ≈ 3.32

Equation C:
x = ln(7/3)/ln(1 + 2*2.9) = ln(7/3)/ln(1 + 5.8) ≈ 0.8473 / ln(6.8) ≈ 0.8473 / 1.916 ≈ 0.442

Equation D:
x = ln(7/3)/ln(1 + sqrt(2.9)) ≈ sqrt(2.9) ≈ 1.702, so ln(1 + 1.702) = ln(2.702) ≈ 0.996, so x ≈ 0.8473 / 0.996 ≈ 0.851

Equation E:
x = ln(7/3)/ln(1 + 1/2.9) ≈ ln(7/3)/ln(1 + 0.3448) ≈ ln(7/3)/ln(1.3448) ≈ 0.8473 / 0.298 ≈ 2.84

So here, equation B gives x ≈3.32, which is the largest. So again, equation B is the largest. Therefore, seems like equation B has the largest x.

But hold on, wait, the answer options don't include B as the answer? Wait no, the answer options are A to E. However, let me check if there's a mistake in my reasoning. Because according to the problem statement, 0 < r < 3. But when I considered r approaching 0, equation B gives very large x. But maybe there is a case where another equation has a larger x? Wait, let me think again. Suppose for some r, the base for equation E is smaller than base B? Earlier, we saw that it's impossible because 1 + 1/r > 1 + r/10 for all r in (0, 3). Wait, but maybe equation D? Wait, equation D's base is 1 + sqrt(r). For some r, maybe base D is smaller than base B?

Let me check when is 1 + sqrt(r) < 1 + r/10? That is, sqrt(r) < r/10. As before, this is equivalent to r^2 > 100r => r > 100. But our domain is 0 < r < 3, so no solution. Therefore, equation D's base is always larger than equation B's base. So equation B's base is always the smallest. Therefore, equation B has the largest x. So the answer should be B. However, I need to check the answer options again. Wait, the problem says "Which equation has the largest solution x?" So according to all my analysis, it should be equation B. But let me check original problem again to make sure I read it correctly.

Wait, the problem says "Assume 0 < r < 3. Below are five equations for x. Which equation has the largest solution x?" Then equations:

A) 3(1 + r)^x = 7
B) 3(1 + r/10)^x = 7
C) 3(1 + 2r)^x = 7
D) 3(1 + √r)^x = 7
E) 3(1 + 1/r)^x = 7

Yes, that's what I considered. So according to my analysis, equation B has the largest x. But let me check if perhaps there's an error in assuming that smaller base leads to larger x. Let me recall that x = ln(7/3)/ln(base). So, if base is smaller, then ln(base) is smaller (since base > 1, because 1 + something positive). Therefore, 1/ln(base) is larger, so x is larger. Therefore, equation with smallest base has largest x. Therefore, equation B is the answer. However, let me check if maybe the question is designed to have a different answer. Wait, let me check again for r = 1, equation B gives x ≈8.89, which is significantly larger than others. Let me check equation E for r =1: 1 + 1/r = 2, same as equation A and D. So equation E gives same x as A and D. So equation B is the only one with larger x.

Alternatively, maybe there's a case where equation E has a smaller base? For example, if 1 + 1/r is smaller than 1 + r/10. But as we saw earlier, that would require r > sqrt(10) which is outside the domain. Hence, no. Therefore, equation B is the answer.

But wait, let me check the answer choices again. Wait, maybe I made a mistake in the problem statement? Wait, the original problem is:

"Assume 0 < r < 3. Below are five equations for x. Which equation has the largest solution x?

(A) 3(1 + r)^x = 7
(B) 3(1 + r/10)^x = 7
(C) 3(1 + 2r)^x = 7
(D) 3(1 + √r)^x = 7
(E) 3(1 + 1/r)^x = 7"

Yes, that's correct. So, according to this, equation B is correct. But let me check again with r approaching 0. Let me take r approaching 0, say r = 0.0001:

Equation B: base = 1 + 0.0001/10 = 1 + 0.00001 = 1.00001. Then ln(1.00001) ≈ 0.00001. Therefore, x ≈ ln(7/3)/0.00001 ≈ 0.8473 / 0.00001 ≈ 84730. Which is huge.

Equation E: base = 1 + 1/r = 1 + 1/0.0001 = 1 + 10000 = 10001. So ln(10001) ≈ 9.21. Therefore, x ≈ 0.8473 / 9.21 ≈ 0.092, which is way smaller. Therefore, equation B is way larger. So, seems like equation B is the answer.

But maybe the answer is E? Wait, no. Wait, let me check the problem again. Wait, the problem says "Assume 0 < r < 3". For example, if r is close to 0, then 1/r is very large. However, in equation E, the base is 1 + 1/r, which is 1 + large number. Therefore, the base is large, so ln(base) is large, hence x is small. Whereas, in equation B, the base is 1 + r/10, which is very slightly larger than 1, so ln(base) is very small, so x is large. Therefore, equation B is the correct answer. So why do I feel uncertain? Maybe because sometimes in these problems, there is a trick. Wait, but all my calculations seem to confirm that equation B has the largest x. However, maybe there is a case where another equation has a smaller base?

Wait, let me check the base for equation D: 1 + sqrt(r). For r approaching 0, sqrt(r) approaches 0, so base approaches 1. Similarly, equation B's base approaches 1 as r approaches 0. But which one approaches 1 faster? For very small r, sqrt(r) is larger than r/10. For example, take r = 0.01:

sqrt(r) = 0.1, while r/10 = 0.001. Therefore, base D is 1 + 0.1 = 1.1, while base B is 1 + 0.001 = 1.001. Therefore, base B is smaller. So for any r > 0, base B is smaller than base D. Therefore, equation B's base is always the smallest. Therefore, equation B is the answer.

Therefore, the answer is B. However, maybe I need to check the answer options again. Wait, the original problem says "Which equation has the largest solution x?" and the answer choices are A through E. So according to my analysis, it's B. However, maybe the answer is E? Wait, let me check again when r is approaching 3.

Wait, when r approaches 3 from the left, let's take r = 2.99:

Equation E's base: 1 + 1/r ≈ 1 + 1/2.99 ≈ 1 + 0.334 ≈ 1.334

Equation B's base: 1 + r/10 ≈ 1 + 0.299 ≈ 1.299

Therefore, base B is smaller than base E. Therefore, equation B's x is larger. Therefore, even when r approaches 3, equation B is still better. Therefore, my conclusion is that equation B is the correct answer. Therefore, the answer should be \boxed{B}.

But wait, looking back at the original problem statement, I need to make sure that I didn't misread any of the equations. For example, equation E is 3(1 + 1/r)^x = 7. So, if r is between 0 and 3, then 1/r is greater than 1/3, so 1 + 1/r is greater than 4/3 ≈ 1.333. Therefore, equation E's base is always larger than equation B's base (which can be as low as approaching 1). So, equation B is definitely the one with the minimal base. Therefore, equation B has the largest x.

Therefore, I think the correct answer is B. However, to be 100% sure, let me check with another approach. Let me consider that for all equations, since the right-hand side is 7, the value of x depends on how "fast" the base grows. If the base grows faster, then x is smaller. If the base grows slower, x is larger. Therefore, to have the largest x, you want the base to grow as slowly as possible. Since for the equations, the base is 1 + something. So, the smaller the something, the slower the growth, so the larger x. Therefore, which equation has the smallest "something"? Let's see:

For equation A: something = r
For equation B: something = r/10
For equation C: something = 2r
For equation D: something = sqrt(r)
For equation E: something = 1/r

So, the question is, which of these "something" is the smallest for 0 < r < 3?

Let me analyze:

For 0 < r < 3, which of the following expressions is the smallest:

r, r/10, 2r, sqrt(r), 1/r.

We need to compare these expressions.

First, note that for r in (0, 3):

- r is between 0 and 3.
- r/10 is between 0 and 0.3.
- 2r is between 0 and 6.
- sqrt(r) is between 0 and sqrt(3) ≈ 1.732.
- 1/r is between 1/3 ≈ 0.333 and infinity.

Therefore, the smallest among these is r/10, since it's between 0 and 0.3, while others are either larger or unbounded. Therefore, equation B has the smallest "something", meaning the base grows the slowest, hence gives the largest x. Therefore, equation B is the answer. Therefore, my conclusion is consistent. Therefore, the correct answer is B.

**Final Answer**
\boxed{B}

To determine which equation has the **largest solution** for $ x $, we analyze the general structure of the given equations. All equations have the form:

$$
3 \cdot (1 + \text{something})^x = 7
$$

Our goal is to find the value of $ x $ for each equation and identify which one has the **maximum** value of $ x $ under the condition $ 0 < r < 3 $. Solving for $ x $ in each equation yields:

$$
x = \frac{\ln\left(\frac{7}{3}\right)}{\ln(1 + \text{something})}
$$

---

### Key Insight

Since the numerator $ \ln\left(\frac{7}{3}\right) $ is a **positive constant**, the value of $ x $ is **inversely proportional** to $ \ln(\text{base}) $, where the base is:

$$
1 + \text{something}
$$

Because the natural log function $ \ln(x) $ is **increasing**, a smaller base results in a smaller $ \ln(\text{base}) $, and thus a **larger** value of $ x $.

Hence, our task reduces to comparing the **bases** of each equation under the condition $ 0 < r < 3 $.

---

### Step 1: Analyze Each Base

Let's write down the base for each equation:

- (A): $ 1 + r $
- (B): $ 1 + \frac{r}{10} $
- (C): $ 1 + 2r $
- (D): $ 1 + \sqrt{r} $
- (E): $ 1 + \frac{1}{r} $

Now consider the possible values of the **something** added to 1:

- For $ r \in (0,3) $:
  - $ r \in (0, 3) $
  - $ \frac{r}{10} \in (0, 0.3) $
  - $ 2r \in (0, 6) $
  - $ \sqrt{r} \in (0, \sqrt{3}) \approx (0, 1.732) $
  - $ \frac{1}{r} \in (\frac{1}{3}, \infty) $

---

### Step 2: Compare the Expressions

We examine which expression is **smallest**:

- $ \frac{r}{10} $: Smallest value across the range due to division by 10
- $ r $: Larger than $ r/10 $
- $ 2r $: Much larger
- $ \sqrt{r} $: Larger than $ \frac{r}{10} $
- $ \frac{1}{r} $: Increases as $ r \to 0 $, but always larger than $ \frac{r}{10} $

So, the smallest value is:

$$
\frac{r}{10}
$$

Therefore, **Equation B** has the **smallest base**, and thus gives the **largest $ x $**.

---

### Final Conclusion

All other bases are greater than $ 1 + \frac{r}{10} $, leading to larger $ \ln(\text{base}) $, and thus smaller $ x $. This confirms that Equation B produces the **largest** value of $ x $ for the given constraint $ 0 < r < 3 $.

---

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