Explanation:
Okay, so I need to solve the equation (x/2) + (x/3) = 5. Hmm, let me think about how to approach this. I remember from algebra that when you have fractions in an equation, it's often helpful to eliminate the denominators by finding a common denominator or multiplying both sides by the least common multiple (LCM) of the denominators. Let me try that.

First, let me identify the denominators here. They are 2 and 3. The least common multiple of 2 and 3 is 6, right? So if I multiply every term in the equation by 6, that should eliminate the fractions. Let me write that down:

6*(x/2) + 6*(x/3) = 6*5

Now, let me compute each term. For the first term, 6 multiplied by (x/2). Let me think, 6 divided by 2 is 3, so that term becomes 3x. Similarly, 6 multiplied by (x/3) would be 6 divided by 3 is 2, so that term becomes 2x. And on the right side, 6 multiplied by 5 is 30. So now the equation simplifies to:

3x + 2x = 30

Combine like terms. 3x + 2x is 5x. So that gives:

5x = 30

Now, to solve for x, I need to divide both sides by 5. Let me do that:

x = 30 / 5

Which simplifies to:

x = 6

Wait, let me check my work to make sure I didn't make a mistake. Let me substitute x = 6 back into the original equation to verify.

Original equation: (6/2) + (6/3) = ?

Calculating each term: 6 divided by 2 is 3, and 6 divided by 3 is 2. Adding them together: 3 + 2 = 5. Which matches the right side of the equation, which is 5. So that checks out. Therefore, x = 6 is the correct solution.

Alternatively, maybe I can solve it another way to confirm. Let me try combining the fractions first before eliminating denominators. The original equation is (x/2) + (x/3) = 5. To add these two fractions, they need a common denominator, which is 6 again. So let me rewrite each fraction with denominator 6:

(x/2) is equivalent to (3x/6) because 2*3=6 and x*3=3x. Similarly, (x/3) is equivalent to (2x/6) because 3*2=6 and x*2=2x. So adding them together:

3x/6 + 2x/6 = (3x + 2x)/6 = 5x/6

Therefore, the equation becomes:

5x/6 = 5

Now, solving for x, multiply both sides by 6:

5x = 5*6

5x = 30

Divide both sides by 5:

x = 6

Same result. So that's consistent with my earlier method. Therefore, I'm confident that x = 6 is the correct solution.

Another thing I can check is if I made an arithmetic error when multiplying or dividing. Let me go through the steps again quickly. Multiplying both sides by 6: 6*(x/2) is 3x, 6*(x/3) is 2x, and 6*5 is 30. Then 3x + 2x is 5x, so 5x=30. Divide by 5: x=6. Yep, that's correct.

Alternatively, if I didn't think of multiplying by 6, maybe I could factor out x. Let me try that approach. The original equation is (x/2) + (x/3) = 5. Factor out x:

x*(1/2 + 1/3) = 5

Compute the sum inside the parentheses. Again, common denominator is 6. 1/2 is 3/6 and 1/3 is 2/6. Adding them together: 3/6 + 2/6 = 5/6. Therefore:

x*(5/6) = 5

Then, solving for x: multiply both sides by 6/5:

x = 5*(6/5) = 6

Same answer again. So regardless of the method I use, I end up with x=6. Therefore, I think that's the right answer.

Just to make sure, let me visualize this equation. If I think of x as a number, then splitting it into halves and thirds and adding them together gives 5. If x is 6, then half of 6 is 3, third of 6 is 2, adding them gives 5. Perfect. If I tried a different number, like x=5, then half is 2.5 and third is approx 1.666..., adding them gives about 4.166..., which is less than 5. If x=7, half is 3.5, third is approx 2.333..., adding them gives about 5.833..., which is more than 5. So 6 is between 5 and 7, and gives exactly 5. Therefore, this numerical check also confirms that x=6 is correct.

I don't think I made any mistakes here. All methods converge on x=6, and the check with substitution works. So I think that's the solution.

**Final Answer**
The solution is \boxed{6}.

To solve the equation:

$$
\frac{x}{2} + \frac{x}{3} = 5
$$

---

### Step 1: Eliminate Denominators

The denominators in the equation are 2 and 3. The least common multiple (LCM) of these two numbers is **6**. To eliminate the fractions, multiply each term in the equation by 6:

$$
6 \cdot \left( \frac{x}{2} \right) + 6 \cdot \left( \frac{x}{3} \right) = 6 \cdot 5
$$

Simplifying each term:

- $6 \cdot \frac{x}{2} = 3x$
- $6 \cdot \frac{x}{3} = 2x$
- $6 \cdot 5 = 30$

The equation becomes:

$$
3x + 2x = 30
$$

---

### Step 2: Combine Like Terms

$$
3x + 2x = 5x
$$

So the equation simplifies to:

$$
5x = 30
$$

---

### Step 3: Solve for $x$

Divide both sides of the equation by 5:

$$
x = \frac{30}{5} = 6
$$

---

### Step 4: Verify the Solution

Substitute $x = 6$ back into the original equation:

$$
\frac{6}{2} + \frac{6}{3} = 3 + 2 = 5
$$

Since the left-hand side equals the right-hand side, the solution is verified.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation (x/2) + (x/3) = 5. Hmm, let me think about how to approach this. I remember from algebra that when you have fractions in an equation, it's often helpful to eliminate the denominators by finding a common denominator or multiplying both sides by the least common multiple (LCM) of the denominators. Let me try that.

First, let me identify the denominators here. They are 2 and 3. The least common multiple of 2 and 3 is 6, right? So if I multiply every term in the equation by 6, that should eliminate the fractions. Let me write that down:

6*(x/2) + 6*(x/3) = 6*5

Now, let me compute each term. For the first term, 6 multiplied by (x/2). Let me think, 6 divided by 2 is 3, so that term becomes 3x. Similarly, 6 multiplied by (x/3) would be 6 divided by 3 is 2, so that term becomes 2x. And on the right side, 6 multiplied by 5 is 30. So now the equation simplifies to:

3x + 2x = 30

Combine like terms. 3x + 2x is 5x. So that gives:

5x = 30

Now, to solve for x, I need to divide both sides by 5. Let me do that:

x = 30 / 5

Which simplifies to:

x = 6

Wait, let me check my work to make sure I didn't make a mistake. Let me substitute x = 6 back into the original equation to verify.

Original equation: (6/2) + (6/3) = ?

Calculating each term: 6 divided by 2 is 3, and 6 divided by 3 is 2. Adding them together: 3 + 2 = 5. Which matches the right side of the equation, which is 5. So that checks out. Therefore, x = 6 is the correct solution.

Alternatively, maybe I can solve it another way to confirm. Let me try combining the fractions first before eliminating denominators. The original equation is (x/2) + (x/3) = 5. To add these two fractions, they need a common denominator, which is 6 again. So let me rewrite each fraction with denominator 6:

(x/2) is equivalent to (3x/6) because 2*3=6 and x*3=3x. Similarly, (x/3) is equivalent to (2x/6) because 3*2=6 and x*2=2x. So adding them together:

3x/6 + 2x/6 = (3x + 2x)/6 = 5x/6

Therefore, the equation becomes:

5x/6 = 5

Now, solving for x, multiply both sides by 6:

5x = 5*6

5x = 30

Divide both sides by 5:

x = 6

Same result. So that's consistent with my earlier method. Therefore, I'm confident that x = 6 is the correct solution.

Another thing I can check is if I made an arithmetic error when multiplying or dividing. Let me go through the steps again quickly. Multiplying both sides by 6: 6*(x/2) is 3x, 6*(x/3) is 2x, and 6*5 is 30. Then 3x + 2x is 5x, so 5x=30. Divide by 5: x=6. Yep, that's correct.

Alternatively, if I didn't think of multiplying by 6, maybe I could factor out x. Let me try that approach. The original equation is (x/2) + (x/3) = 5. Factor out x:

x*(1/2 + 1/3) = 5

Compute the sum inside the parentheses. Again, common denominator is 6. 1/2 is 3/6 and 1/3 is 2/6. Adding them together: 3/6 + 2/6 = 5/6. Therefore:

x*(5/6) = 5

Then, solving for x: multiply both sides by 6/5:

x = 5*(6/5) = 6

Same answer again. So regardless of the method I use, I end up with x=6. Therefore, I think that's the right answer.

Just to make sure, let me visualize this equation. If I think of x as a number, then splitting it into halves and thirds and adding them together gives 5. If x is 6, then half of 6 is 3, third of 6 is 2, adding them gives 5. Perfect. If I tried a different number, like x=5, then half is 2.5 and third is approx 1.666..., adding them gives about 4.166..., which is less than 5. If x=7, half is 3.5, third is approx 2.333..., adding them gives about 5.833..., which is more than 5. So 6 is between 5 and 7, and gives exactly 5. Therefore, this numerical check also confirms that x=6 is correct.

I don't think I made any mistakes here. All methods converge on x=6, and the check with substitution works. So I think that's the solution.

**Final Answer**
The solution is \boxed{6}.

To solve the equation:

$$
\frac{x}{2} + \frac{x}{3} = 5
$$

---

### Step 1: Eliminate Denominators

The denominators in the equation are 2 and 3. The least common multiple (LCM) of these two numbers is **6**. To eliminate the fractions, multiply each term in the equation by 6:

$$
6 \cdot \left( \frac{x}{2} \right) + 6 \cdot \left( \frac{x}{3} \right) = 6 \cdot 5
$$

Simplifying each term:

- $6 \cdot \frac{x}{2} = 3x$
- $6 \cdot \frac{x}{3} = 2x$
- $6 \cdot 5 = 30$

The equation becomes:

$$
3x + 2x = 30
$$

---

### Step 2: Combine Like Terms

$$
3x + 2x = 5x
$$

So the equation simplifies to:

$$
5x = 30
$$

---

### Step 3: Solve for $x$

Divide both sides of the equation by 5:

$$
x = \frac{30}{5} = 6
$$

---

### Step 4: Verify the Solution

Substitute $x = 6$ back into the original equation:

$$
\frac{6}{2} + \frac{6}{3} = 3 + 2 = 5
$$

Since the left-hand side equals the right-hand side, the solution is verified.

---

### Final Answer

$$
\boxed{6}
$$

Solution: Writing the left-hand side with a common denominator, we have$$\dfrac{x}{2} + \dfrac{x}{3} = \dfrac{3x}{6} + \dfrac{2x}{6} = \dfrac{5x}{6},$$ so our equation is $$\dfrac{5x}{6} = 5.$$Multiplying both sides by $\dfrac{6}{5}$ gives $$x = 5\cdot \dfrac{6}{5} = \boxed{6}.$$


Explanation:
Okay, so I need to figure out Rosa's age based on the given conditions. Let me start by writing down all the information provided:

1. When Rosa's age is divided by 2, 3, 4, or 6, the remainder is 1.
2. When her age is divided by 7, the remainder is 0.
3. She is less than 75 years old.

Alright, so let me break this down. First, let me think about the first condition. If dividing her age by 2, 3, 4, or 6 leaves a remainder of 1 each time, that means her age is 1 more than a multiple of each of these numbers. So mathematically, if her age is $ x $, then:

- $ x \equiv 1 \mod 2 $
- $ x \equiv 1 \mod 3 $
- $ x \equiv 1 \mod 4 $
- $ x \equiv 1 \mod 6 $

And also, $ x \equiv 0 \mod 7 $, and $ x < 75 $.

Hmm, so maybe I can use the Chinese Remainder Theorem here? Or perhaps find a common number that satisfies all those congruences?

But first, let me think about the first set of congruences. If $ x \equiv 1 \mod 2 $, that means x is odd. Similarly, if $ x \equiv 1 \mod 3 $, then x is one more than a multiple of 3. Similarly for 4 and 6.

But maybe there's a way to find the least common multiple (LCM) of 2, 3, 4, 6 and then add 1? Wait, but LCM of those numbers. Let me compute that. The LCM of 2, 3, 4, 6. Let's see:

Prime factors:

- 2 is 2
- 3 is 3
- 4 is 2²
- 6 is 2×3

So LCM would take the highest powers of each prime: 2² × 3 = 4 × 3 = 12. So LCM is 12. Therefore, numbers that are congruent to 1 mod 2, 3, 4, 6 would be numbers of the form 12k + 1, where k is an integer. Is that correct?

Wait, let me check. If x ≡1 mod 12, then x would satisfy x ≡1 mod 2, 3, 4, 6. Because 12 is a multiple of all of those numbers. Therefore, if x is 12k +1, then dividing by 2 would give 6k + 0.5, so the remainder is 1? Wait, actually, let me check with numbers.

Take x = 1. 1 divided by 2 is 0 with remainder 1. Divided by 3, same. Divided by 4, same. Divided by 6, same. So 1 is a solution.

Next one would be 13. Let's check 13 divided by 2: 6*2=12, remainder 1. Divided by 3: 4*3=12, remainder 1. Divided by 4: 3*4=12, remainder 1. Divided by 6: 2*6=12, remainder 1. So yes, 13 also satisfies.

Similarly, 25: divided by 2, remainder 1; divided by 3, 25/3=8*3 +1, remainder 1; divided by 4, 25/4=6*4 +1; divided by 6, 25/6=4*6 +1. So same thing. So numbers of the form 12k +1. So the general solution for the first set of congruences is x ≡1 mod 12. Therefore, x can be 1, 13, 25, 37, 49, 61, 73, etc. But since x <75, then the numbers are 1, 13, 25, 37, 49, 61, 73. Wait, 12*6 +1=73, 12*7 +1=85, which is over 75. So possible ages are 1,13,25,37,49,61,73.

But then we also have the second condition: x is divisible by 7, so x ≡0 mod7. So we have to find which of these numbers (1,13,25,37,49,61,73) is divisible by 7.

Let me check each one:

- 1: 1 /7 is 0 with remainder 1. Not divisible.
- 13: 13 divided by 7 is 1 with remainder 6. Not divisible.
- 25: 25 /7 is 3*7=21, remainder 4. Not divisible.
- 37: 37 divided by 7: 5*7=35, remainder 2. Not divisible.
- 49: 49 is 7*7. So that's divisible by 7. So 49 is a candidate.
- 61: 61 divided by 7 is 8*7=56, remainder 5. Not divisible.
- 73: 73 divided by 7 is 10*7=70, remainder 3. Not divisible.

So the only number in that list that is divisible by 7 is 49. So is 49 the answer? Let me check again all the conditions.

First, check if 49 divided by 2, 3, 4, 6 leaves remainder 1.

Divided by 2: 49 /2 = 24*2 +1, remainder 1. Good.

Divided by 3: 49 /3 is 16*3 +1, remainder 1. Good.

Divided by 4: 49 /4 is 12*4 +1, remainder 1. Good.

Divided by 6: 49 /6 is 8*6 +1, remainder 1. Good.

Divided by 7: 49 /7 is 7, remainder 0. Perfect. And 49 is less than 75. So yes, seems like 49 is the answer.

But let me just check if there are any other numbers I might have missed. The possible x's were 1,13,25,37,49,61,73. Only 49 is divisible by 7. Therefore, the answer is 49. Wait, but let me think again if the initial assumption that x ≡1 mod 12 is correct.

Because the first set of conditions is x ≡1 mod 2, x≡1 mod3, x≡1 mod4, x≡1 mod6. But is the LCM of 2,3,4,6 equal to 12? Let me check again. LCM of 2,3,4,6. Well, LCM(2,3)=6, LCM(6,4)=12, LCM(12,6)=12. So yes, LCM is 12. Therefore, numbers congruent to 1 mod12 satisfy all those conditions. Therefore, the numbers are 1,13,...,73.

But maybe there's another way to think about it? For example, if x ≡1 mod 4, then x must be 1 mod 4, so also 1 mod 2, since 4 is a multiple of 2. Similarly, x ≡1 mod 6 implies x ≡1 mod 2 and x≡1 mod 3. So, perhaps the conditions x ≡1 mod 2, 3, 4, 6 are equivalent to x ≡1 mod LCM(2,3,4,6)=12. So that is correct.

Alternatively, maybe there's an overlap? For example, if x ≡1 mod 6, then automatically x ≡1 mod 2 and mod 3. So the conditions x≡1 mod 2,3,4,6 are equivalent to x≡1 mod LCM(4,6)=12? Because LCM of 4 and 6 is 12. Since if x ≡1 mod 4 and x≡1 mod 6, then x ≡1 mod LCM(4,6)=12. So that would make sense. So yes, the combined congruence is x ≡1 mod 12.

Therefore, the possible ages are 1,13,25,...,73, and among those, only 49 is divisible by 7. So 49 is the answer.

But just to make sure, let me check if there are other numbers less than 75 that satisfy the first condition (remainder 1 when divided by 2,3,4,6) and divisible by 7. For example, suppose someone thinks that maybe numbers that are 1 more than LCM(2,3,4,6) which is 12, so 13, 25,... but perhaps there's a smaller number?

Wait, but 1 is also a number which satisfies all these conditions, but 1 divided by 7 gives remainder 1, so not 0. Then 13, same thing. 25, same. 37, same. 49 is the first one that is divisible by 7. Then 61, 73. So yes, only 49 is in that list.

Therefore, I think 49 is the answer.

But let me think again. Maybe I made a mistake in assuming that x ≡1 mod 12. Let me check with 49. 49 divided by 12 is 4*12=48, so remainder 1. So yes, 49 ≡1 mod12. So that's correct.

Alternatively, maybe another approach: Let me consider that if x divided by 2 leaves remainder 1, then x is odd. Similarly, divided by 3 leaves remainder 1, so x is 1 more than multiple of 3. Divided by 4 leaves 1, so x is 1 more than multiple of 4. Divided by 6 leaves 1, so x is 1 more than multiple of 6. Therefore, x-1 is divisible by 2,3,4,6. Hence, x-1 is LCM(2,3,4,6)=12, so x=12k+1. Therefore, yes, same conclusion.

Therefore, the same numbers. So 12k+1, and x=12k+1 must be divisible by 7. So 12k +1 ≡0 mod7 => 12k ≡ -1 mod7. Since 12 mod7 is 5. Therefore, 5k ≡ -1 mod7 => 5k ≡6 mod7. So solving for k: Multiply both sides by inverse of 5 mod7. What's inverse of 5 mod7? 5*3=15≡1 mod7, so inverse is 3. Therefore, k≡6*3 mod7 => k≡18 mod7 => 18 divided by7 is 2*7=14, remainder 4. Therefore, k≡4 mod7. Therefore, k=7m +4, where m is integer. Therefore, x=12*(7m +4) +1= 84m +48 +1=84m +49. Therefore, x=84m +49. Since x<75, let's find possible m.

For m=0: x=49.

For m=1: x=84 +49=133, which is more than 75. So only possible value is 49. Therefore, this confirms that the only possible age is 49. Therefore, answer is 49.

Therefore, I think this is solid. So Rosa is 49 years old.

**Final Answer**
\boxed{49}

To determine Rosa’s age, we are given the following conditions:

1. When her age $ x $ is divided by 2, 3, 4, or 6, the remainder is **1**.
2. When her age is divided by **7**, the remainder is **0**.
3. Her age is **less than 75**.

---

### Step 1: Analyze the first condition

We are told that when $ x $ is divided by 2, 3, 4, or 6, the remainder is 1 in all cases. This implies:

$$
x \equiv 1 \pmod{2}, \quad x \equiv 1 \pmod{3}, \quad x \equiv 1 \pmod{4}, \quad x \equiv 1 \pmod{6}
$$

This is equivalent to saying that:

$$
x - 1 \equiv 0 \pmod{2}, \quad x - 1 \equiv 0 \pmod{3}, \quad x - 1 \equiv 0 \pmod{4}, \quad x - 1 \equiv 0 \pmod{6}
$$

This means that $ x - 1 $ is divisible by the least common multiple (LCM) of 2, 3, 4, and 6.

- Prime factorizations:
  - $ 2 = 2 $
  - $ 3 = 3 $
  - $ 4 = 2^2 $
  - $ 6 = 2 \times 3 $

So, LCM(2, 3, 4, 6) = $ 2^2 \times 3 = 12 $

Thus, $ x - 1 = 12k $, for some integer $ k $

$$
\Rightarrow x = 12k + 1
$$

---

### Step 2: Incorporate the second condition

We are told that when $ x $ is divided by 7, the remainder is 0, i.e.,

$$
x \equiv 0 \pmod{7}
$$

From earlier, we have $ x = 12k + 1 $. So:

$$
12k + 1 \equiv 0 \pmod{7} \\
\Rightarrow 12k \equiv -1 \pmod{7}
$$

Simplify $ 12 \mod 7 = 5 $, so:

$$
5k \equiv -1 \pmod{7}
\Rightarrow 5k \equiv 6 \pmod{7}
$$

We now solve for $ k $. To do that, we find the modular inverse of 5 mod 7. Since:

$$
5 \times 3 = 15 \equiv 1 \pmod{7}
$$

Thus, the inverse of 5 mod 7 is 3. Multiply both sides of the congruence by 3:

$$
k \equiv 6 \times 3 = 18 \equiv 4 \pmod{7}
$$

So, $ k = 7m + 4 $ for some integer $ m $

Substitute back into $ x = 12k + 1 $:

$$
x = 12(7m + 4) + 1 = 84m + 48 + 1 = 84m + 49
$$

---

### Step 3: Apply the constraint that $ x < 75 $

We now find values of $ x = 84m + 49 $ less than 75:

- $ m = 0 \Rightarrow x = 49 $
- $ m = 1 \Rightarrow x = 133 $ (too large)

So, the only valid value is **49**

---

### Step 4: Final verification

Let’s verify that 49 meets all the conditions:

- $ 49 \div 2 = 24 \text{ remainder } 1 $ ✅
- $ 49 \div 3 = 16 \text{ remainder } 1 $ ✅
- $ 49 \div 4 = 12 \text{ remainder } 1 $ ✅
- $ 49 \div 6 = 8 \text{ remainder } 1 $ ✅
- $ 49 \div 7 = 7 \text{ remainder } 0 $ ✅
- $ 49 < 75 $ ✅

All conditions are satisfied.

---

### ✅ Final Answer:

$$
\boxed{49}
$$

[Assistant]
Explanation:
Okay, so I need to figure out Rosa's age based on the given conditions. Let me start by writing down all the information provided:

1. When Rosa's age is divided by 2, 3, 4, or 6, the remainder is 1.
2. When her age is divided by 7, the remainder is 0.
3. She is less than 75 years old.

Alright, so let me break this down. First, let me think about the first condition. If dividing her age by 2, 3, 4, or 6 leaves a remainder of 1 each time, that means her age is 1 more than a multiple of each of these numbers. So mathematically, if her age is $ x $, then:

- $ x \equiv 1 \mod 2 $
- $ x \equiv 1 \mod 3 $
- $ x \equiv 1 \mod 4 $
- $ x \equiv 1 \mod 6 $

And also, $ x \equiv 0 \mod 7 $, and $ x < 75 $.

Hmm, so maybe I can use the Chinese Remainder Theorem here? Or perhaps find a common number that satisfies all those congruences?

But first, let me think about the first set of congruences. If $ x \equiv 1 \mod 2 $, that means x is odd. Similarly, if $ x \equiv 1 \mod 3 $, then x is one more than a multiple of 3. Similarly for 4 and 6.

But maybe there's a way to find the least common multiple (LCM) of 2, 3, 4, 6 and then add 1? Wait, but LCM of those numbers. Let me compute that. The LCM of 2, 3, 4, 6. Let's see:

Prime factors:

- 2 is 2
- 3 is 3
- 4 is 2²
- 6 is 2×3

So LCM would take the highest powers of each prime: 2² × 3 = 4 × 3 = 12. So LCM is 12. Therefore, numbers that are congruent to 1 mod 2, 3, 4, 6 would be numbers of the form 12k + 1, where k is an integer. Is that correct?

Wait, let me check. If x ≡1 mod 12, then x would satisfy x ≡1 mod 2, 3, 4, 6. Because 12 is a multiple of all of those numbers. Therefore, if x is 12k +1, then dividing by 2 would give 6k + 0.5, so the remainder is 1? Wait, actually, let me check with numbers.

Take x = 1. 1 divided by 2 is 0 with remainder 1. Divided by 3, same. Divided by 4, same. Divided by 6, same. So 1 is a solution.

Next one would be 13. Let's check 13 divided by 2: 6*2=12, remainder 1. Divided by 3: 4*3=12, remainder 1. Divided by 4: 3*4=12, remainder 1. Divided by 6: 2*6=12, remainder 1. So yes, 13 also satisfies.

Similarly, 25: divided by 2, remainder 1; divided by 3, 25/3=8*3 +1, remainder 1; divided by 4, 25/4=6*4 +1; divided by 6, 25/6=4*6 +1. So same thing. So numbers of the form 12k +1. So the general solution for the first set of congruences is x ≡1 mod 12. Therefore, x can be 1, 13, 25, 37, 49, 61, 73, etc. But since x <75, then the numbers are 1, 13, 25, 37, 49, 61, 73. Wait, 12*6 +1=73, 12*7 +1=85, which is over 75. So possible ages are 1,13,25,37,49,61,73.

But then we also have the second condition: x is divisible by 7, so x ≡0 mod7. So we have to find which of these numbers (1,13,25,37,49,61,73) is divisible by 7.

Let me check each one:

- 1: 1 /7 is 0 with remainder 1. Not divisible.
- 13: 13 divided by 7 is 1 with remainder 6. Not divisible.
- 25: 25 /7 is 3*7=21, remainder 4. Not divisible.
- 37: 37 divided by 7: 5*7=35, remainder 2. Not divisible.
- 49: 49 is 7*7. So that's divisible by 7. So 49 is a candidate.
- 61: 61 divided by 7 is 8*7=56, remainder 5. Not divisible.
- 73: 73 divided by 7 is 10*7=70, remainder 3. Not divisible.

So the only number in that list that is divisible by 7 is 49. So is 49 the answer? Let me check again all the conditions.

First, check if 49 divided by 2, 3, 4, 6 leaves remainder 1.

Divided by 2: 49 /2 = 24*2 +1, remainder 1. Good.

Divided by 3: 49 /3 is 16*3 +1, remainder 1. Good.

Divided by 4: 49 /4 is 12*4 +1, remainder 1. Good.

Divided by 6: 49 /6 is 8*6 +1, remainder 1. Good.

Divided by 7: 49 /7 is 7, remainder 0. Perfect. And 49 is less than 75. So yes, seems like 49 is the answer.

But let me just check if there are any other numbers I might have missed. The possible x's were 1,13,25,37,49,61,73. Only 49 is divisible by 7. Therefore, the answer is 49. Wait, but let me think again if the initial assumption that x ≡1 mod 12 is correct.

Because the first set of conditions is x ≡1 mod 2, x≡1 mod3, x≡1 mod4, x≡1 mod6. But is the LCM of 2,3,4,6 equal to 12? Let me check again. LCM of 2,3,4,6. Well, LCM(2,3)=6, LCM(6,4)=12, LCM(12,6)=12. So yes, LCM is 12. Therefore, numbers congruent to 1 mod12 satisfy all those conditions. Therefore, the numbers are 1,13,...,73.

But maybe there's another way to think about it? For example, if x ≡1 mod 4, then x must be 1 mod 4, so also 1 mod 2, since 4 is a multiple of 2. Similarly, x ≡1 mod 6 implies x ≡1 mod 2 and x≡1 mod 3. So, perhaps the conditions x ≡1 mod 2, 3, 4, 6 are equivalent to x ≡1 mod LCM(2,3,4,6)=12. So that is correct.

Alternatively, maybe there's an overlap? For example, if x ≡1 mod 6, then automatically x ≡1 mod 2 and mod 3. So the conditions x≡1 mod 2,3,4,6 are equivalent to x≡1 mod LCM(4,6)=12? Because LCM of 4 and 6 is 12. Since if x ≡1 mod 4 and x≡1 mod 6, then x ≡1 mod LCM(4,6)=12. So that would make sense. So yes, the combined congruence is x ≡1 mod 12.

Therefore, the possible ages are 1,13,25,...,73, and among those, only 49 is divisible by 7. So 49 is the answer.

But just to make sure, let me check if there are other numbers less than 75 that satisfy the first condition (remainder 1 when divided by 2,3,4,6) and divisible by 7. For example, suppose someone thinks that maybe numbers that are 1 more than LCM(2,3,4,6) which is 12, so 13, 25,... but perhaps there's a smaller number?

Wait, but 1 is also a number which satisfies all these conditions, but 1 divided by 7 gives remainder 1, so not 0. Then 13, same thing. 25, same. 37, same. 49 is the first one that is divisible by 7. Then 61, 73. So yes, only 49 is in that list.

Therefore, I think 49 is the answer.

But let me think again. Maybe I made a mistake in assuming that x ≡1 mod 12. Let me check with 49. 49 divided by 12 is 4*12=48, so remainder 1. So yes, 49 ≡1 mod12. So that's correct.

Alternatively, maybe another approach: Let me consider that if x divided by 2 leaves remainder 1, then x is odd. Similarly, divided by 3 leaves remainder 1, so x is 1 more than multiple of 3. Divided by 4 leaves 1, so x is 1 more than multiple of 4. Divided by 6 leaves 1, so x is 1 more than multiple of 6. Therefore, x-1 is divisible by 2,3,4,6. Hence, x-1 is LCM(2,3,4,6)=12, so x=12k+1. Therefore, yes, same conclusion.

Therefore, the same numbers. So 12k+1, and x=12k+1 must be divisible by 7. So 12k +1 ≡0 mod7 => 12k ≡ -1 mod7. Since 12 mod7 is 5. Therefore, 5k ≡ -1 mod7 => 5k ≡6 mod7. So solving for k: Multiply both sides by inverse of 5 mod7. What's inverse of 5 mod7? 5*3=15≡1 mod7, so inverse is 3. Therefore, k≡6*3 mod7 => k≡18 mod7 => 18 divided by7 is 2*7=14, remainder 4. Therefore, k≡4 mod7. Therefore, k=7m +4, where m is integer. Therefore, x=12*(7m +4) +1= 84m +48 +1=84m +49. Therefore, x=84m +49. Since x<75, let's find possible m.

For m=0: x=49.

For m=1: x=84 +49=133, which is more than 75. So only possible value is 49. Therefore, this confirms that the only possible age is 49. Therefore, answer is 49.

Therefore, I think this is solid. So Rosa is 49 years old.

**Final Answer**
\boxed{49}

To determine Rosa’s age, we are given the following conditions:

1. When her age $ x $ is divided by 2, 3, 4, or 6, the remainder is **1**.
2. When her age is divided by **7**, the remainder is **0**.
3. Her age is **less than 75**.

---

### Step 1: Analyze the first condition

We are told that when $ x $ is divided by 2, 3, 4, or 6, the remainder is 1 in all cases. This implies:

$$
x \equiv 1 \pmod{2}, \quad x \equiv 1 \pmod{3}, \quad x \equiv 1 \pmod{4}, \quad x \equiv 1 \pmod{6}
$$

This is equivalent to saying that:

$$
x - 1 \equiv 0 \pmod{2}, \quad x - 1 \equiv 0 \pmod{3}, \quad x - 1 \equiv 0 \pmod{4}, \quad x - 1 \equiv 0 \pmod{6}
$$

This means that $ x - 1 $ is divisible by the least common multiple (LCM) of 2, 3, 4, and 6.

- Prime factorizations:
  - $ 2 = 2 $
  - $ 3 = 3 $
  - $ 4 = 2^2 $
  - $ 6 = 2 \times 3 $

So, LCM(2, 3, 4, 6) = $ 2^2 \times 3 = 12 $

Thus, $ x - 1 = 12k $, for some integer $ k $

$$
\Rightarrow x = 12k + 1
$$

---

### Step 2: Incorporate the second condition

We are told that when $ x $ is divided by 7, the remainder is 0, i.e.,

$$
x \equiv 0 \pmod{7}
$$

From earlier, we have $ x = 12k + 1 $. So:

$$
12k + 1 \equiv 0 \pmod{7} \\
\Rightarrow 12k \equiv -1 \pmod{7}
$$

Simplify $ 12 \mod 7 = 5 $, so:

$$
5k \equiv -1 \pmod{7}
\Rightarrow 5k \equiv 6 \pmod{7}
$$

We now solve for $ k $. To do that, we find the modular inverse of 5 mod 7. Since:

$$
5 \times 3 = 15 \equiv 1 \pmod{7}
$$

Thus, the inverse of 5 mod 7 is 3. Multiply both sides of the congruence by 3:

$$
k \equiv 6 \times 3 = 18 \equiv 4 \pmod{7}
$$

So, $ k = 7m + 4 $ for some integer $ m $

Substitute back into $ x = 12k + 1 $:

$$
x = 12(7m + 4) + 1 = 84m + 48 + 1 = 84m + 49
$$

---

### Step 3: Apply the constraint that $ x < 75 $

We now find values of $ x = 84m + 49 $ less than 75:

- $ m = 0 \Rightarrow x = 49 $
- $ m = 1 \Rightarrow x = 133 $ (too large)

So, the only valid value is **49**

---

### Step 4: Final verification

Let’s verify that 49 meets all the conditions:

- $ 49 \div 2 = 24 \text{ remainder } 1 $ ✅
- $ 49 \div 3 = 16 \text{ remainder } 1 $ ✅
- $ 49 \div 4 = 12 \text{ remainder } 1 $ ✅
- $ 49 \div 6 = 8 \text{ remainder } 1 $ ✅
- $ 49 \div 7 = 7 \text{ remainder } 0 $ ✅
- $ 49 < 75 $ ✅

All conditions are satisfied.

---

### ✅ Final Answer:

$$
\boxed{49}
$$

Solution:
Since her age divided by 7 results in a remainder of 0, her age must be a multiple of 7. If her age is $n$, we notice that $n-1$ must be a multiple of 2, 3, 4, and 6. The least common multiple of those numbers is 12, so $n-1$ must be a multiple of 12. The multiples of 12 less than 75 are 12, 24, 36, 48, and 60. Adding 1 results in 13, 25, 37, 49, and 61, where 49 is the only multiple of 7. So Rosa is $\boxed{49}$ years old.

OR

We look for a multiple of 7 that is not divisible by 2, 3, 4, or 6. First we list all odd multiples of 7 less than 75, which are 7, 21, 35, 49, and 63. Since 21 and 63 are multiples of 3, we're left with 7, 35, and 49 as possibilities. Only $\boxed{49}$ has a remainder of 1 when divided by 2, 3, 4, or 6.


Explanation:
Okay, so I need to find the value of a + b + c, where a, b, and c are the roots of the equation given: 1/x + 5x² = 6x - 24. Hmm, let me think about how to approach this.

First, I remember that when dealing with equations involving roots, Vieta's formulas are usually helpful. Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. However, the given equation isn't a polynomial equation yet. It has a term with 1/x, which complicates things. So my first step should be to convert this equation into a polynomial equation.

Let me write down the equation again:

1/x + 5x² = 6x - 24.

To eliminate the fraction, I can multiply both sides of the equation by x. But I need to be careful because multiplying by x might introduce an extraneous solution if x = 0 is a solution, but in this case, since there's a 1/x term, x cannot be zero. So x ≠ 0, and multiplying both sides by x should be safe.

Multiplying both sides by x:

x*(1/x) + x*(5x²) = x*(6x) - x*(24)

Simplifying each term:

1 + 5x³ = 6x² - 24x

Now, let me bring all terms to one side to set the equation equal to zero:

5x³ - 6x² + 24x + 1 = 0

Wait, let me check that again. Starting from:

1 + 5x³ = 6x² - 24x

Subtracting 6x² and adding 24x to both sides:

5x³ - 6x² + 24x + 1 = 0

Yes, that seems right. Let me verify each term:

Left side after multiplying by x: 1 + 5x³

Right side after multiplying by x: 6x² - 24x

Bringing everything to the left: 1 + 5x³ - 6x² + 24x = 0, which rearranged is 5x³ - 6x² + 24x + 1 = 0. Okay, that seems correct.

So now we have a cubic equation: 5x³ - 6x² + 24x + 1 = 0. The roots of this equation are a, b, c. Wait, but hold on. The original equation had 1/x, so when we multiplied by x, we assumed x ≠ 0, but the original equation also doesn't have x=0 as a solution, since plugging x=0 would make 1/x undefined. So the roots of the original equation are the same as the roots of this cubic equation. Therefore, a, b, c are the roots of 5x³ - 6x² + 24x + 1 = 0.

Now, Vieta's formula for a cubic equation. Let me recall that for a general cubic equation ax³ + bx² + cx + d = 0, the sum of the roots is -b/a, the sum of the product of roots two at a time is c/a, and the product of the roots is -d/a.

Therefore, for our equation 5x³ - 6x² + 24x + 1 = 0, the coefficients are:

a_cubic = 5,

b_cubic = -6,

c_cubic = 24,

d_cubic = 1.

Therefore, the sum of the roots a + b + c = -b_cubic / a_cubic = -(-6)/5 = 6/5. Wait, so is that the answer? 6/5? But let me make sure I didn't mix up the coefficients.

Wait, Vieta's formula says sum of roots is -b/a. In standard form, the cubic is ax³ + bx² + cx + d = 0. So yes, here the cubic is 5x³ + (-6)x² + 24x + 1 = 0. Therefore, the sum of roots is -(-6)/5 = 6/5.

But hold on, this seems too straightforward. Let me check my steps again. Because sometimes when you manipulate equations, especially when multiplying both sides by x, you can sometimes introduce extraneous roots or lose roots, but in this case, since x ≠ 0, and the original equation is 1/x + 5x² = 6x -24, if x is a root of the original equation, then multiplying by x would give a root of the cubic equation. Conversely, any root of the cubic equation (other than x=0, which isn't a root here) would satisfy the original equation. So the cubic equation's roots are exactly the roots of the original equation. Therefore, a, b, c are roots of the cubic, so Vieta applies here. Therefore, sum a + b + c is 6/5.

But let me check if I transformed the equation correctly. Let me go back step by step.

Original equation: 1/x + 5x² = 6x -24.

Multiply both sides by x: 1 + 5x³ = 6x² -24x.

Then bringing all terms to left side: 5x³ -6x² +24x +1=0. Yes, that's correct. So the cubic equation is correct. Then Vieta's formula says sum of roots is 6/5. Therefore, answer is 6/5. But let me check with an example. Let me suppose that maybe there's a mistake here.

Alternatively, maybe I need to check if the cubic equation is correct. Let me take a simple value. Let me plug in x = -1 into original equation and see if it satisfies.

Original equation: 1/x +5x² = 6x -24.

Left side: 1/(-1) + 5*(-1)^2 = -1 + 5(1) = -1 +5 = 4.

Right side: 6*(-1) -24 = -6 -24 = -30. 4 ≠ -30, so x=-1 is not a root. Now, let's check if x=-1 is a root of the cubic equation.

Cubic equation: 5x³ -6x² +24x +1. Plugging x=-1:

5*(-1)^3 -6*(-1)^2 +24*(-1) +1 = -5 -6 -24 +1 = -34. Not zero. So that's correct, because x=-1 is not a root.

How about x=0? Well, original equation is undefined. Cubic equation at x=0 is 0 -0 +0 +1=1≠0. So that's good.

How about maybe another value? Let me try to find a root of the cubic equation. Let me check x= -1/5? Let me try that. But maybe this is too tedious. Alternatively, perhaps I can check my Vieta's formula calculation again.

Alternatively, perhaps the user made a mistake in the sign? Let me think.

Original cubic equation: 5x³ -6x² +24x +1 = 0. So coefficients are:

Leading coefficient: 5.

Next coefficient: -6.

Then +24.

Then +1.

So Vieta's formula says sum of roots is -(-6)/5 = 6/5. Correct. So if the cubic is written as 5x³ + (-6)x² + 24x + 1 = 0, then yes. Therefore, sum of roots is 6/5. Therefore, answer is 6/5, which is 1.2. But let me think again if this is correct.

Alternatively, maybe I made a mistake when moving terms. Let me check again:

Starting from:

1/x + 5x² = 6x -24

Multiply both sides by x:

1 + 5x³ = 6x² -24x

Then subtracting 6x² and adding 24x:

5x³ -6x² +24x +1=0. Yes, that is correct. So the cubic is correct. Then Vieta's formula applies, so sum a + b + c is 6/5. Therefore, answer is 6/5. So maybe this is correct?

But I have a doubt because the original equation is 1/x + 5x² = 6x -24. Let me think about degrees. The original equation is not a polynomial equation, but after multiplying by x, it becomes a cubic. The roots of that cubic are the solutions to the original equation. So unless there are some restrictions, but since x ≠ 0, which is already considered.

Alternatively, maybe the problem is expecting something else? Let me check if the problem says "a, b, c are the roots of the equation", so maybe there are three roots? But a cubic equation can have up to three roots. However, maybe the original equation has three roots? Let me think.

Original equation: 1/x +5x² = 6x -24. Let me rearrange it:

5x² -6x +24 +1/x =0. So 5x² -6x +24 +1/x =0. Which is equivalent to the cubic equation. Therefore, there are three roots. Therefore, Vieta applies. Therefore, answer is 6/5. So I think this is correct. But maybe the problem is presented in a different way?

Alternatively, maybe I need to check the problem again. The problem says: Compute a + b + c, given that a, b, c are the roots of 1/x +5x² =6x -24. So assuming that the equation is equivalent to the cubic, so the answer is 6/5.

But maybe I made a mistake in the sign? Let me check Vieta again. For a cubic equation ax³ + bx² + cx + d = 0, sum of roots is -b/a. So in our case, the cubic is 5x³ -6x² +24x +1 = 0. Therefore, a_cubic = 5, b_cubic = -6, so sum is -b_cubic /a_cubic = -(-6)/5 = 6/5. Correct.

Alternatively, maybe the problem expects integer roots or something else? Let me check if the cubic factors. Maybe there is a rational root. Using Rational Root Theorem, possible rational roots are ±1, ±1/5.

Testing x= -1: 5(-1)^3 -6(-1)^2 +24(-1) +1 = -5 -6 -24 +1 = -34 ≠0

x=1: 5 -6 +24 +1= 24 ≠0

x=1/5: 5*(1/5)^3 -6*(1/5)^2 +24*(1/5) +1 = 5*(1/125) -6*(1/25) +24/5 +1 = 1/25 -6/25 + 24/5 +1 = (-5/25) + 24/5 +1 = (-1/5) + 24/5 + 5/5 = ( -1 +24 +5)/5 = 28/5 ≠0

x=-1/5: 5*(-1/5)^3 -6*(-1/5)^2 +24*(-1/5) +1 = 5*(-1/125) -6*(1/25) -24/5 +1 = -1/25 -6/25 -24/5 +1 = (-7/25) - 24/5 +1 = (-7/25 -120/25) +25/25 = (-127/25) +25/25 = -102/25 ≠0

Therefore, there are no rational roots. Therefore, the cubic doesn't factor nicely, but that doesn't matter because Vieta's formula still applies regardless of whether roots are rational or not. Therefore, the answer is 6/5.

Alternatively, maybe there's a mistake in the problem transcription? Let me check again. The original equation is 1/x +5x² =6x -24. Yes. Then multiplying by x gives 1 +5x³=6x² -24x. Moving terms: 5x³ -6x² +24x +1=0. Correct. Then Vieta gives sum of roots 6/5. Therefore, the answer is 6/5. Therefore, I think that's correct.

But maybe I need to check once more with another approach? Let me think. Suppose I let a, b, c be roots of the cubic equation. Then the cubic can be written as 5(x - a)(x - b)(x - c) = 0. Expanding this:

First, (x - a)(x - b)(x - c) = x³ - (a + b + c)x² + (ab + ac + bc)x - abc. Therefore, multiplying by 5:

5x³ -5(a + b + c)x² +5(ab + ac + bc)x -5abc = 0.

Comparing this with the original cubic equation 5x³ -6x² +24x +1=0. Therefore, we can equate coefficients:

- Coefficient of x³: 5 = 5, okay.

- Coefficient of x²: -5(a + b + c) = -6. Therefore, solving for a + b + c: -5(a + b + c) = -6 ⇒ a + b + c = (-6)/(-5) = 6/5. Same result. Therefore, this confirms that a + b + c = 6/5. Therefore, the answer is 6/5.

Therefore, I think I can be confident that the answer is 6/5. Therefore, the problem is solved.

**Final Answer**
\boxed{\dfrac{6}{5}}

To find the value of $ a + b + c $, where $ a, b, c $ are the roots of the equation

$$
\frac{1}{x} + 5x^2 = 6x - 24,
$$

we begin by transforming the equation into a standard polynomial form.

---

### Step 1: Eliminate the Fraction

Multiply both sides of the equation by $ x $ (noting that $ x \ne 0 $ since $ \frac{1}{x} $ is undefined at $ x = 0 $):

$$
x \cdot \left( \frac{1}{x} + 5x^2 \right) = x \cdot (6x - 24)
$$

Simplifying both sides:

$$
1 + 5x^3 = 6x^2 - 24x
$$

---

### Step 2: Rearrange into Standard Polynomial Form

Bring all terms to one side of the equation:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

This is now a cubic polynomial equation.

---

### Step 3: Apply Vieta’s Formulas

A general cubic equation of the form

$$
ax^3 + bx^2 + cx + d = 0
$$

has the following properties regarding its roots $ a, b, c $:

- Sum of roots: $ a + b + c = -\frac{b}{a} $
- Sum of product of roots taken two at a time: $ ab + bc + ac = \frac{c}{a} $
- Product of roots: $ abc = -\frac{d}{a} $

In our case, the cubic is:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

Here:

- $ a = 5 $
- $ b = -6 $
- $ c = 24 $
- $ d = 1 $

Thus, the sum of the roots is:

$$
a + b + c = -\frac{-6}{5} = \frac{6}{5}
$$

---

### Final Answer:

$$
\boxed{\dfrac{6}{5}}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of a + b + c, where a, b, and c are the roots of the equation given: 1/x + 5x² = 6x - 24. Hmm, let me think about how to approach this.

First, I remember that when dealing with equations involving roots, Vieta's formulas are usually helpful. Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. However, the given equation isn't a polynomial equation yet. It has a term with 1/x, which complicates things. So my first step should be to convert this equation into a polynomial equation.

Let me write down the equation again:

1/x + 5x² = 6x - 24.

To eliminate the fraction, I can multiply both sides of the equation by x. But I need to be careful because multiplying by x might introduce an extraneous solution if x = 0 is a solution, but in this case, since there's a 1/x term, x cannot be zero. So x ≠ 0, and multiplying both sides by x should be safe.

Multiplying both sides by x:

x*(1/x) + x*(5x²) = x*(6x) - x*(24)

Simplifying each term:

1 + 5x³ = 6x² - 24x

Now, let me bring all terms to one side to set the equation equal to zero:

5x³ - 6x² + 24x + 1 = 0

Wait, let me check that again. Starting from:

1 + 5x³ = 6x² - 24x

Subtracting 6x² and adding 24x to both sides:

5x³ - 6x² + 24x + 1 = 0

Yes, that seems right. Let me verify each term:

Left side after multiplying by x: 1 + 5x³

Right side after multiplying by x: 6x² - 24x

Bringing everything to the left: 1 + 5x³ - 6x² + 24x = 0, which rearranged is 5x³ - 6x² + 24x + 1 = 0. Okay, that seems correct.

So now we have a cubic equation: 5x³ - 6x² + 24x + 1 = 0. The roots of this equation are a, b, c. Wait, but hold on. The original equation had 1/x, so when we multiplied by x, we assumed x ≠ 0, but the original equation also doesn't have x=0 as a solution, since plugging x=0 would make 1/x undefined. So the roots of the original equation are the same as the roots of this cubic equation. Therefore, a, b, c are the roots of 5x³ - 6x² + 24x + 1 = 0.

Now, Vieta's formula for a cubic equation. Let me recall that for a general cubic equation ax³ + bx² + cx + d = 0, the sum of the roots is -b/a, the sum of the product of roots two at a time is c/a, and the product of the roots is -d/a.

Therefore, for our equation 5x³ - 6x² + 24x + 1 = 0, the coefficients are:

a_cubic = 5,

b_cubic = -6,

c_cubic = 24,

d_cubic = 1.

Therefore, the sum of the roots a + b + c = -b_cubic / a_cubic = -(-6)/5 = 6/5. Wait, so is that the answer? 6/5? But let me make sure I didn't mix up the coefficients.

Wait, Vieta's formula says sum of roots is -b/a. In standard form, the cubic is ax³ + bx² + cx + d = 0. So yes, here the cubic is 5x³ + (-6)x² + 24x + 1 = 0. Therefore, the sum of roots is -(-6)/5 = 6/5.

But hold on, this seems too straightforward. Let me check my steps again. Because sometimes when you manipulate equations, especially when multiplying both sides by x, you can sometimes introduce extraneous roots or lose roots, but in this case, since x ≠ 0, and the original equation is 1/x + 5x² = 6x -24, if x is a root of the original equation, then multiplying by x would give a root of the cubic equation. Conversely, any root of the cubic equation (other than x=0, which isn't a root here) would satisfy the original equation. So the cubic equation's roots are exactly the roots of the original equation. Therefore, a, b, c are roots of the cubic, so Vieta applies here. Therefore, sum a + b + c is 6/5.

But let me check if I transformed the equation correctly. Let me go back step by step.

Original equation: 1/x + 5x² = 6x -24.

Multiply both sides by x: 1 + 5x³ = 6x² -24x.

Then bringing all terms to left side: 5x³ -6x² +24x +1=0. Yes, that's correct. So the cubic equation is correct. Then Vieta's formula says sum of roots is 6/5. Therefore, answer is 6/5. But let me check with an example. Let me suppose that maybe there's a mistake here.

Alternatively, maybe I need to check if the cubic equation is correct. Let me take a simple value. Let me plug in x = -1 into original equation and see if it satisfies.

Original equation: 1/x +5x² = 6x -24.

Left side: 1/(-1) + 5*(-1)^2 = -1 + 5(1) = -1 +5 = 4.

Right side: 6*(-1) -24 = -6 -24 = -30. 4 ≠ -30, so x=-1 is not a root. Now, let's check if x=-1 is a root of the cubic equation.

Cubic equation: 5x³ -6x² +24x +1. Plugging x=-1:

5*(-1)^3 -6*(-1)^2 +24*(-1) +1 = -5 -6 -24 +1 = -34. Not zero. So that's correct, because x=-1 is not a root.

How about x=0? Well, original equation is undefined. Cubic equation at x=0 is 0 -0 +0 +1=1≠0. So that's good.

How about maybe another value? Let me try to find a root of the cubic equation. Let me check x= -1/5? Let me try that. But maybe this is too tedious. Alternatively, perhaps I can check my Vieta's formula calculation again.

Alternatively, perhaps the user made a mistake in the sign? Let me think.

Original cubic equation: 5x³ -6x² +24x +1 = 0. So coefficients are:

Leading coefficient: 5.

Next coefficient: -6.

Then +24.

Then +1.

So Vieta's formula says sum of roots is -(-6)/5 = 6/5. Correct. So if the cubic is written as 5x³ + (-6)x² + 24x + 1 = 0, then yes. Therefore, sum of roots is 6/5. Therefore, answer is 6/5, which is 1.2. But let me think again if this is correct.

Alternatively, maybe I made a mistake when moving terms. Let me check again:

Starting from:

1/x + 5x² = 6x -24

Multiply both sides by x:

1 + 5x³ = 6x² -24x

Then subtracting 6x² and adding 24x:

5x³ -6x² +24x +1=0. Yes, that is correct. So the cubic is correct. Then Vieta's formula applies, so sum a + b + c is 6/5. Therefore, answer is 6/5. So maybe this is correct?

But I have a doubt because the original equation is 1/x + 5x² = 6x -24. Let me think about degrees. The original equation is not a polynomial equation, but after multiplying by x, it becomes a cubic. The roots of that cubic are the solutions to the original equation. So unless there are some restrictions, but since x ≠ 0, which is already considered.

Alternatively, maybe the problem is expecting something else? Let me check if the problem says "a, b, c are the roots of the equation", so maybe there are three roots? But a cubic equation can have up to three roots. However, maybe the original equation has three roots? Let me think.

Original equation: 1/x +5x² = 6x -24. Let me rearrange it:

5x² -6x +24 +1/x =0. So 5x² -6x +24 +1/x =0. Which is equivalent to the cubic equation. Therefore, there are three roots. Therefore, Vieta applies. Therefore, answer is 6/5. So I think this is correct. But maybe the problem is presented in a different way?

Alternatively, maybe I need to check the problem again. The problem says: Compute a + b + c, given that a, b, c are the roots of 1/x +5x² =6x -24. So assuming that the equation is equivalent to the cubic, so the answer is 6/5.

But maybe I made a mistake in the sign? Let me check Vieta again. For a cubic equation ax³ + bx² + cx + d = 0, sum of roots is -b/a. So in our case, the cubic is 5x³ -6x² +24x +1 = 0. Therefore, a_cubic = 5, b_cubic = -6, so sum is -b_cubic /a_cubic = -(-6)/5 = 6/5. Correct.

Alternatively, maybe the problem expects integer roots or something else? Let me check if the cubic factors. Maybe there is a rational root. Using Rational Root Theorem, possible rational roots are ±1, ±1/5.

Testing x= -1: 5(-1)^3 -6(-1)^2 +24(-1) +1 = -5 -6 -24 +1 = -34 ≠0

x=1: 5 -6 +24 +1= 24 ≠0

x=1/5: 5*(1/5)^3 -6*(1/5)^2 +24*(1/5) +1 = 5*(1/125) -6*(1/25) +24/5 +1 = 1/25 -6/25 + 24/5 +1 = (-5/25) + 24/5 +1 = (-1/5) + 24/5 + 5/5 = ( -1 +24 +5)/5 = 28/5 ≠0

x=-1/5: 5*(-1/5)^3 -6*(-1/5)^2 +24*(-1/5) +1 = 5*(-1/125) -6*(1/25) -24/5 +1 = -1/25 -6/25 -24/5 +1 = (-7/25) - 24/5 +1 = (-7/25 -120/25) +25/25 = (-127/25) +25/25 = -102/25 ≠0

Therefore, there are no rational roots. Therefore, the cubic doesn't factor nicely, but that doesn't matter because Vieta's formula still applies regardless of whether roots are rational or not. Therefore, the answer is 6/5.

Alternatively, maybe there's a mistake in the problem transcription? Let me check again. The original equation is 1/x +5x² =6x -24. Yes. Then multiplying by x gives 1 +5x³=6x² -24x. Moving terms: 5x³ -6x² +24x +1=0. Correct. Then Vieta gives sum of roots 6/5. Therefore, the answer is 6/5. Therefore, I think that's correct.

But maybe I need to check once more with another approach? Let me think. Suppose I let a, b, c be roots of the cubic equation. Then the cubic can be written as 5(x - a)(x - b)(x - c) = 0. Expanding this:

First, (x - a)(x - b)(x - c) = x³ - (a + b + c)x² + (ab + ac + bc)x - abc. Therefore, multiplying by 5:

5x³ -5(a + b + c)x² +5(ab + ac + bc)x -5abc = 0.

Comparing this with the original cubic equation 5x³ -6x² +24x +1=0. Therefore, we can equate coefficients:

- Coefficient of x³: 5 = 5, okay.

- Coefficient of x²: -5(a + b + c) = -6. Therefore, solving for a + b + c: -5(a + b + c) = -6 ⇒ a + b + c = (-6)/(-5) = 6/5. Same result. Therefore, this confirms that a + b + c = 6/5. Therefore, the answer is 6/5.

Therefore, I think I can be confident that the answer is 6/5. Therefore, the problem is solved.

**Final Answer**
\boxed{\dfrac{6}{5}}

To find the value of $ a + b + c $, where $ a, b, c $ are the roots of the equation

$$
\frac{1}{x} + 5x^2 = 6x - 24,
$$

we begin by transforming the equation into a standard polynomial form.

---

### Step 1: Eliminate the Fraction

Multiply both sides of the equation by $ x $ (noting that $ x \ne 0 $ since $ \frac{1}{x} $ is undefined at $ x = 0 $):

$$
x \cdot \left( \frac{1}{x} + 5x^2 \right) = x \cdot (6x - 24)
$$

Simplifying both sides:

$$
1 + 5x^3 = 6x^2 - 24x
$$

---

### Step 2: Rearrange into Standard Polynomial Form

Bring all terms to one side of the equation:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

This is now a cubic polynomial equation.

---

### Step 3: Apply Vieta’s Formulas

A general cubic equation of the form

$$
ax^3 + bx^2 + cx + d = 0
$$

has the following properties regarding its roots $ a, b, c $:

- Sum of roots: $ a + b + c = -\frac{b}{a} $
- Sum of product of roots taken two at a time: $ ab + bc + ac = \frac{c}{a} $
- Product of roots: $ abc = -\frac{d}{a} $

In our case, the cubic is:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

Here:

- $ a = 5 $
- $ b = -6 $
- $ c = 24 $
- $ d = 1 $

Thus, the sum of the roots is:

$$
a + b + c = -\frac{-6}{5} = \frac{6}{5}
$$

---

### Final Answer:

$$
\boxed{\dfrac{6}{5}}
$$

Solution: We want to apply Vieta's formulas, but the given equation is not a polynomial equation, because of the $\frac1x$ term. To turn this equation into an equivalent polynomial equation, we multiply both sides by $x$ and rearrange: \[\begin{aligned} 1+5x^3 &= 6x^2 - 24x \\ 5x^3 - 6x^2 + 24x + 1 &= 0 .\end{aligned}\]Now we can use Vieta: the sum of the roots is $a+b+c=\boxed{\frac65}.$


Explanation:
Okay, so I need to find two consecutive positive even integers whose product is 288, and then figure out which one is the greater of the two. Let me start by recalling what consecutive even integers are. They are even numbers that follow each other in order, like 2 and 4, or 6 and 8, right? So each consecutive even integer is 2 more than the previous one.

Let me think about how to represent them algebraically. If I let the first even integer be x, then the next consecutive even integer would be x + 2. That makes sense because they're two apart. So their product is x times (x + 2), and that equals 288.

So, translating that into an equation: x(x + 2) = 288. Let me write that down:

x(x + 2) = 288

Now I need to solve for x. To do that, I can expand the left side and then rearrange the equation into a standard quadratic form. Let me do that step by step.

First, expanding x(x + 2):

x * x + x * 2 = x² + 2x

So the equation becomes:

x² + 2x = 288

Now, I need to bring all terms to one side to form a quadratic equation. Let me subtract 288 from both sides:

x² + 2x - 288 = 0

Alright, so now I have a quadratic equation: x² + 2x - 288 = 0. To solve this quadratic equation, I can try factoring, completing the square, or using the quadratic formula. Let me try factoring first because if it factors nicely, that might be quicker.

Looking for two numbers that multiply to -288 and add up to 2. Hmm, factors of 288... Let me think. Since the product is negative, one of the numbers has to be positive and the other negative. Also, their sum is positive 2. So the positive number must be larger in magnitude than the negative one.

Let me list some factor pairs of 288:

1 and 288

2 and 144

3 and 96

4 and 72

6 and 48

8 and 36

9 and 32

12 and 24

16 and 18

Now, since one is positive and the other is negative, and their sum is 2, let me think. Let me try combinations where the positive number is larger. For example, if I take 18 and -16: 18 * (-16) = -288, and 18 + (-16) = 2. Yes! That works.

So the equation can be factored as:

(x + 18)(x - 16) = 0

Wait, let me check that. If I expand (x + 18)(x - 16), I get x² -16x + 18x - 288 = x² + 2x - 288. Perfect, that's exactly our equation.

So the solutions are x + 18 = 0 or x - 16 = 0. Solving these gives x = -18 or x = 16.

But the problem specifies that the integers are positive. So x = -18 is negative, which we can disregard. Therefore, the first integer is 16, and the next consecutive even integer is 16 + 2 = 18.

Let me check if their product is indeed 288. 16 * 18. Let me calculate that. 16*10=160, 16*8=128, so 160+128=288. Perfect, that works.

So the two integers are 16 and 18, and the greater one is 18. Therefore, the answer should be 18.

Wait, but let me just make sure I didn't make any mistakes in my factoring. Sometimes when factoring quadratics, it's easy to mix up the signs. Let me verify again. The quadratic equation was x² + 2x - 288 = 0. Factored into (x + 18)(x - 16) = 0. Multiplying those gives x² -16x +18x -288 = x² + 2x - 288. Yep, that's correct.

Alternatively, maybe I can use the quadratic formula to double-check. The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a). For the equation x² + 2x - 288 = 0, a=1, b=2, c=-288.

Calculating discriminant: b² - 4ac = (2)^2 - 4*1*(-288) = 4 + 1152 = 1156.

Square root of 1156 is 34, since 34^2 = 1156.

So x = [-2 ± 34]/2.

First solution: (-2 + 34)/2 = 32/2 = 16.

Second solution: (-2 - 34)/2 = -36/2 = -18.

Same results as before, so that confirms that x=16 is the positive solution.

Therefore, the two integers are 16 and 18, and the greater one is 18.

Just to ensure there's no alternative interpretation, let me check if there's another pair of consecutive even integers that multiply to 288. For example, suppose someone thought consecutive even integers could be 14 and 16? Let me check 14*16=224, which is less than 288. How about 18*20? That's 360, which is too big. So 16*18 is in between. So that seems to be the correct pair.

Alternatively, maybe negative numbers? But the problem says positive integers, so negatives are out.

Another way to approach it could be to take the square root of 288 and see what's around there. The square root of 288 is approximately 16.97. So the two numbers should be around 17, but since they are even integers, the closest even numbers would be 16 and 18. Which matches our previous result.

So all methods lead to the conclusion that the greater integer is 18. Therefore, I think that's the correct answer.

**Final Answer**
The greater of the two integers is \boxed{18}.

To solve the problem, we start by identifying the nature of the two integers. The question specifies that they are **two consecutive positive even integers**, meaning they follow one after the other in the sequence of even numbers, such as 16 and 18, or 4 and 6.

---

### Step 1: Represent the Integers Algebraically

Let the first even integer be:

$$
x
$$

Then the next consecutive even integer is:

$$
x + 2
$$

Their product is given to be:

$$
x(x + 2) = 288
$$

---

### Step 2: Expand and Rearrange into a Quadratic Equation

Expanding the left-hand side:

$$
x^2 + 2x = 288
$$

Now, subtract 288 from both sides to set the equation to zero:

$$
x^2 + 2x - 288 = 0
$$

---

### Step 3: Solve the Quadratic Equation

We attempt to factor the quadratic:

$$
x^2 + 2x - 288 = 0
$$

Looking for two numbers that multiply to -288 and add to 2. These numbers are **18** and **-16**:

$$
(x + 18)(x - 16) = 0
$$

Setting each factor equal to zero:

$$
x + 18 = 0 \Rightarrow x = -18 \quad (\text{not valid, as we need positive integers}) \\
x - 16 = 0 \Rightarrow x = 16
$$

So, the first integer is 16, and the second is:

$$
x + 2 = 16 + 2 = 18
$$

---

### Step 4: Verify the Solution

Check the product:

$$
16 \times 18 = 288
$$

This confirms our solution is correct.

---

### Final Answer

The greater of the two integers is:

$$
\boxed{18}
$$

[Assistant]
Explanation:
Okay, so I need to find two consecutive positive even integers whose product is 288, and then figure out which one is the greater of the two. Let me start by recalling what consecutive even integers are. They are even numbers that follow each other in order, like 2 and 4, or 6 and 8, right? So each consecutive even integer is 2 more than the previous one.

Let me think about how to represent them algebraically. If I let the first even integer be x, then the next consecutive even integer would be x + 2. That makes sense because they're two apart. So their product is x times (x + 2), and that equals 288.

So, translating that into an equation: x(x + 2) = 288. Let me write that down:

x(x + 2) = 288

Now I need to solve for x. To do that, I can expand the left side and then rearrange the equation into a standard quadratic form. Let me do that step by step.

First, expanding x(x + 2):

x * x + x * 2 = x² + 2x

So the equation becomes:

x² + 2x = 288

Now, I need to bring all terms to one side to form a quadratic equation. Let me subtract 288 from both sides:

x² + 2x - 288 = 0

Alright, so now I have a quadratic equation: x² + 2x - 288 = 0. To solve this quadratic equation, I can try factoring, completing the square, or using the quadratic formula. Let me try factoring first because if it factors nicely, that might be quicker.

Looking for two numbers that multiply to -288 and add up to 2. Hmm, factors of 288... Let me think. Since the product is negative, one of the numbers has to be positive and the other negative. Also, their sum is positive 2. So the positive number must be larger in magnitude than the negative one.

Let me list some factor pairs of 288:

1 and 288

2 and 144

3 and 96

4 and 72

6 and 48

8 and 36

9 and 32

12 and 24

16 and 18

Now, since one is positive and the other is negative, and their sum is 2, let me think. Let me try combinations where the positive number is larger. For example, if I take 18 and -16: 18 * (-16) = -288, and 18 + (-16) = 2. Yes! That works.

So the equation can be factored as:

(x + 18)(x - 16) = 0

Wait, let me check that. If I expand (x + 18)(x - 16), I get x² -16x + 18x - 288 = x² + 2x - 288. Perfect, that's exactly our equation.

So the solutions are x + 18 = 0 or x - 16 = 0. Solving these gives x = -18 or x = 16.

But the problem specifies that the integers are positive. So x = -18 is negative, which we can disregard. Therefore, the first integer is 16, and the next consecutive even integer is 16 + 2 = 18.

Let me check if their product is indeed 288. 16 * 18. Let me calculate that. 16*10=160, 16*8=128, so 160+128=288. Perfect, that works.

So the two integers are 16 and 18, and the greater one is 18. Therefore, the answer should be 18.

Wait, but let me just make sure I didn't make any mistakes in my factoring. Sometimes when factoring quadratics, it's easy to mix up the signs. Let me verify again. The quadratic equation was x² + 2x - 288 = 0. Factored into (x + 18)(x - 16) = 0. Multiplying those gives x² -16x +18x -288 = x² + 2x - 288. Yep, that's correct.

Alternatively, maybe I can use the quadratic formula to double-check. The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a). For the equation x² + 2x - 288 = 0, a=1, b=2, c=-288.

Calculating discriminant: b² - 4ac = (2)^2 - 4*1*(-288) = 4 + 1152 = 1156.

Square root of 1156 is 34, since 34^2 = 1156.

So x = [-2 ± 34]/2.

First solution: (-2 + 34)/2 = 32/2 = 16.

Second solution: (-2 - 34)/2 = -36/2 = -18.

Same results as before, so that confirms that x=16 is the positive solution.

Therefore, the two integers are 16 and 18, and the greater one is 18.

Just to ensure there's no alternative interpretation, let me check if there's another pair of consecutive even integers that multiply to 288. For example, suppose someone thought consecutive even integers could be 14 and 16? Let me check 14*16=224, which is less than 288. How about 18*20? That's 360, which is too big. So 16*18 is in between. So that seems to be the correct pair.

Alternatively, maybe negative numbers? But the problem says positive integers, so negatives are out.

Another way to approach it could be to take the square root of 288 and see what's around there. The square root of 288 is approximately 16.97. So the two numbers should be around 17, but since they are even integers, the closest even numbers would be 16 and 18. Which matches our previous result.

So all methods lead to the conclusion that the greater integer is 18. Therefore, I think that's the correct answer.

**Final Answer**
The greater of the two integers is \boxed{18}.

To solve the problem, we start by identifying the nature of the two integers. The question specifies that they are **two consecutive positive even integers**, meaning they follow one after the other in the sequence of even numbers, such as 16 and 18, or 4 and 6.

---

### Step 1: Represent the Integers Algebraically

Let the first even integer be:

$$
x
$$

Then the next consecutive even integer is:

$$
x + 2
$$

Their product is given to be:

$$
x(x + 2) = 288
$$

---

### Step 2: Expand and Rearrange into a Quadratic Equation

Expanding the left-hand side:

$$
x^2 + 2x = 288
$$

Now, subtract 288 from both sides to set the equation to zero:

$$
x^2 + 2x - 288 = 0
$$

---

### Step 3: Solve the Quadratic Equation

We attempt to factor the quadratic:

$$
x^2 + 2x - 288 = 0
$$

Looking for two numbers that multiply to -288 and add to 2. These numbers are **18** and **-16**:

$$
(x + 18)(x - 16) = 0
$$

Setting each factor equal to zero:

$$
x + 18 = 0 \Rightarrow x = -18 \quad (\text{not valid, as we need positive integers}) \\
x - 16 = 0 \Rightarrow x = 16
$$

So, the first integer is 16, and the second is:

$$
x + 2 = 16 + 2 = 18
$$

---

### Step 4: Verify the Solution

Check the product:

$$
16 \times 18 = 288
$$

This confirms our solution is correct.

---

### Final Answer

The greater of the two integers is:

$$
\boxed{18}
$$

Solution: First we find the prime factorization of 288 to be $2^5\cdot 3^2$, and we have to split these factors among two consecutive even integers. The 3 must be with at least one 2 for the integer to be even, meaning one of the factors must be a multiple of $6.$ After some playing around, we find that when one factor is 18, that leaves us with $2^4=16$. So, our two integers are 16 and 18, with the greater integer being $\boxed{18}$.


Explanation:
Okay, so I need to figure out what log base 2 of 64 is. Hmm, let me remember. Logarithms are the inverse of exponents, right? So log base 2 of 64 means I need to find the exponent that I have to raise 2 to, to get 64.

Let me write that down. So, log₂(64) = x means that 2^x = 64. Now, I need to figure out what x is.

Hmm, 2 to the power of what gives me 64? Let me think about powers of 2.

2^1 is 2.

2^2 is 4.

2^3 is 8.

2^4 is 16.

2^5 is 32.

2^6 is 64. Oh, there we go! So 2^6 equals 64. That means x is 6. Therefore, log base 2 of 64 should be 6.

Wait, let me double-check. Maybe I can break down 64 into factors of 2. 64 divided by 2 is 32. Divided by 2 again is 16, then 8, 4, 2, and 1. So that's 6 divisions by 2, which means 2^6 is 64. Yep, that confirms it.

Alternatively, I can think about prime factorization. 64 is 2 multiplied by itself 6 times, so 2^6. Therefore, the logarithm base 2 of 2^6 is just 6. Because log base a of a^b is b.

Is there another way to verify this? Maybe using logarithm properties? Let's see. If I take log₂(64), and I know that 64 is 8 squared, so 8^2. And 8 is 2^3. Therefore, 64 is (2^3)^2 = 2^(3*2) = 2^6. So again, log₂(2^6) = 6.

Or maybe using change of base formula? Although that might be overkill here. The change of base formula says log_b(a) = log_c(a)/log_c(b). So if I use base 10, log₂(64) = log₁₀(64)/log₁₀(2). Let me compute that.

First, log₁₀(64). I know that log₁₀(10) is 1, log₁₀(100) is 2. 64 is between 10^1 and 10^2. But I need the exact value? Wait, maybe approximate? Well, if I use a calculator, log₁₀(64) is approximately 1.806, and log₁₀(2) is approximately 0.3010. Dividing 1.806 by 0.3010 gives roughly 6. So that's about 6. So that also confirms it.

But since I don't have a calculator here, maybe I can remember that 2^6 = 64, so log₂(64) is definitely 6.

Another thought: if I know that 2^6 = 64, then taking log base 2 of both sides gives log₂(2^6) = log₂(64). And by logarithm identity, log_b(b^x) = x, so that simplifies to 6 = log₂(64). Therefore, log₂(64) is 6.

So all these different methods point to the same answer. I don't see any contradictions here.

Wait, just to make sure I didn't mix up any numbers. Let me list powers of 2 again:

2^0 = 1

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

Yep, so 2^6 is indeed 64. So the exponent is 6, so the logarithm is 6.

I think that's solid. I can't think of any other way this could go wrong. Maybe if I confused the base or the argument? Like, if I thought log base 64 of 2, but no, the question is log base 2 of 64.

So, no, I think I'm confident that the answer is 6.

**Final Answer**
The value of $\log_2 64$ is \boxed{6}.

To evaluate $ \log_2 64 $, we begin by understanding the meaning of a logarithm. The expression $ \log_b a $ asks the question: *"To what power must the base $ b $ be raised to obtain $ a $?*"

---

### Step 1: Convert the Logarithmic Expression to an Exponential Form

We are asked to evaluate:
$$
\log_2 64
$$
This is equivalent to asking:
$$
2^x = 64
$$
Our task is to find the value of $ x $ that satisfies this equation.

---

### Step 2: Use Powers of 2 to Find the Exponent

We compute successive powers of 2:
$$
\begin{align*}
2^1 &= 2 \\
2^2 &= 4 \\
2^3 &= 8 \\
2^4 &= 16 \\
2^5 &= 32 \\
2^6 &= 64 \\
\end{align*}
$$

From this, we see that:
$$
2^6 = 64
$$

Therefore, the exponent $ x $ that satisfies $ 2^x = 64 $ is 6.

---

### Step 3: Apply Logarithmic Identity

Using the fundamental logarithmic identity:
$$
\log_b (b^x) = x
$$
we have:
$$
\log_2 (2^6) = 6
$$

So:
$$
\log_2 64 = 6
$$

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what log base 2 of 64 is. Hmm, let me remember. Logarithms are the inverse of exponents, right? So log base 2 of 64 means I need to find the exponent that I have to raise 2 to, to get 64.

Let me write that down. So, log₂(64) = x means that 2^x = 64. Now, I need to figure out what x is.

Hmm, 2 to the power of what gives me 64? Let me think about powers of 2.

2^1 is 2.

2^2 is 4.

2^3 is 8.

2^4 is 16.

2^5 is 32.

2^6 is 64. Oh, there we go! So 2^6 equals 64. That means x is 6. Therefore, log base 2 of 64 should be 6.

Wait, let me double-check. Maybe I can break down 64 into factors of 2. 64 divided by 2 is 32. Divided by 2 again is 16, then 8, 4, 2, and 1. So that's 6 divisions by 2, which means 2^6 is 64. Yep, that confirms it.

Alternatively, I can think about prime factorization. 64 is 2 multiplied by itself 6 times, so 2^6. Therefore, the logarithm base 2 of 2^6 is just 6. Because log base a of a^b is b.

Is there another way to verify this? Maybe using logarithm properties? Let's see. If I take log₂(64), and I know that 64 is 8 squared, so 8^2. And 8 is 2^3. Therefore, 64 is (2^3)^2 = 2^(3*2) = 2^6. So again, log₂(2^6) = 6.

Or maybe using change of base formula? Although that might be overkill here. The change of base formula says log_b(a) = log_c(a)/log_c(b). So if I use base 10, log₂(64) = log₁₀(64)/log₁₀(2). Let me compute that.

First, log₁₀(64). I know that log₁₀(10) is 1, log₁₀(100) is 2. 64 is between 10^1 and 10^2. But I need the exact value? Wait, maybe approximate? Well, if I use a calculator, log₁₀(64) is approximately 1.806, and log₁₀(2) is approximately 0.3010. Dividing 1.806 by 0.3010 gives roughly 6. So that's about 6. So that also confirms it.

But since I don't have a calculator here, maybe I can remember that 2^6 = 64, so log₂(64) is definitely 6.

Another thought: if I know that 2^6 = 64, then taking log base 2 of both sides gives log₂(2^6) = log₂(64). And by logarithm identity, log_b(b^x) = x, so that simplifies to 6 = log₂(64). Therefore, log₂(64) is 6.

So all these different methods point to the same answer. I don't see any contradictions here.

Wait, just to make sure I didn't mix up any numbers. Let me list powers of 2 again:

2^0 = 1

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

Yep, so 2^6 is indeed 64. So the exponent is 6, so the logarithm is 6.

I think that's solid. I can't think of any other way this could go wrong. Maybe if I confused the base or the argument? Like, if I thought log base 64 of 2, but no, the question is log base 2 of 64.

So, no, I think I'm confident that the answer is 6.

**Final Answer**
The value of $\log_2 64$ is \boxed{6}.

To evaluate $ \log_2 64 $, we begin by understanding the meaning of a logarithm. The expression $ \log_b a $ asks the question: *"To what power must the base $ b $ be raised to obtain $ a $?*"

---

### Step 1: Convert the Logarithmic Expression to an Exponential Form

We are asked to evaluate:
$$
\log_2 64
$$
This is equivalent to asking:
$$
2^x = 64
$$
Our task is to find the value of $ x $ that satisfies this equation.

---

### Step 2: Use Powers of 2 to Find the Exponent

We compute successive powers of 2:
$$
\begin{align*}
2^1 &= 2 \\
2^2 &= 4 \\
2^3 &= 8 \\
2^4 &= 16 \\
2^5 &= 32 \\
2^6 &= 64 \\
\end{align*}
$$

From this, we see that:
$$
2^6 = 64
$$

Therefore, the exponent $ x $ that satisfies $ 2^x = 64 $ is 6.

---

### Step 3: Apply Logarithmic Identity

Using the fundamental logarithmic identity:
$$
\log_b (b^x) = x
$$
we have:
$$
\log_2 (2^6) = 6
$$

So:
$$
\log_2 64 = 6
$$

---

### Final Answer

$$
\boxed{6}
$$

Solution: We have $2^6=64$, so $\log_2 64 = \boxed{6}$.


Explanation:
Okay, so I need to simplify the expression tan 100° + 4 sin 100°. Hmm, let me think about how to approach this. First, maybe I should recall some trigonometric identities that could help simplify this. Let me jot down what I know:

- Tangent is sine over cosine, so tan θ = sin θ / cos θ.
- Maybe I can write everything in terms of sine and cosine? Let me try that.

So, tan 100° + 4 sin 100° = (sin 100° / cos 100°) + 4 sin 100°. Hmm, that seems a bit messy. Maybe I can factor out sin 100°?

Let me see: sin 100° [1 / cos 100° + 4]. But I don't know if that helps. Maybe I need another approach. Let me think about the angles involved. 100° is in the second quadrant. I know that tan(180° - θ) = -tan θ, so tan 100° = tan(180° - 80°) = -tan 80°. Similarly, sin 100° = sin(180° - 80°) = sin 80°. So maybe rewriting the original expression in terms of 80°?

Let me do that:

tan 100° + 4 sin 100° = -tan 80° + 4 sin 80°.

Hmm, that might be helpful. Let me write that down:

- tan 80° + 4 sin 80°. Now, tan 80° is sin 80° / cos 80°, so substituting that in:

- (sin 80° / cos 80°) + 4 sin 80°. Let me factor out sin 80°:

sin 80° [ -1 / cos 80° + 4 ].

Hmm, still not sure. Let me compute the numerical values to see if there's a pattern or if it simplifies to something nice. Wait, maybe I should check if there's an identity that can combine these terms. Alternatively, perhaps using complex numbers or some other trick?

Alternatively, maybe express everything in terms of sine or cosine? Let me see:

Let me consider writing the expression as:

- tan 80° + 4 sin 80° = 4 sin 80° - tan 80°.

Maybe converting tan 80° to sin 80° / cos 80°:

= 4 sin 80° - sin 80° / cos 80°.

Factor sin 80°:

= sin 80° [4 - 1 / cos 80°].

Hmm, perhaps not helpful. Let me compute the value numerically to see if it's a standard angle or something.

Let me calculate each term:

First, compute tan 100°:

100° is in the second quadrant. Let me convert it to decimal or check approximate value.

But maybe I can use calculator-like approximations. Let me recall that tan 80° is approximately 5.671, so tan 100° is -tan 80° ≈ -5.671.

Then sin 100° is sin 80° ≈ 0.9848. Therefore 4 * sin 100° ≈ 4 * 0.9848 ≈ 3.9392.

So adding them together: -5.671 + 3.9392 ≈ -1.7318. Wait, that's approximately -√3 ≈ -1.732. That seems close. Is the exact value -√3? Let me check.

Wait, √3 is approximately 1.732, so -√3 is approximately -1.732. So my approximate calculation gives -1.7318, which is very close. Therefore, maybe the exact value is -√3?

But let me verify this more carefully.

If tan 100° + 4 sin 100° = -√3, is that correct?

Let me try to prove that.

Let me start with tan 100° + 4 sin 100°. Let me write tan as sin/cos:

sin 100° / cos 100° + 4 sin 100° = sin 100° (1 / cos 100° + 4).

But 1 / cos 100° is sec 100°, which is negative because cos 100° is negative (since 100° is in the second quadrant). But maybe I can express this in terms of 80°, since 100° = 180° - 80°, so cos 100° = -cos 80°, and sin 100° = sin 80°, so:

sin 80° [ 1 / (-cos 80°) + 4 ] = sin 80° [ 4 - 1 / cos 80° ].

Hmm, so that's sin 80° times (4 - sec 80°). If this equals -√3, then maybe there's a relationship here.

Alternatively, maybe I can write this as:

Let me consider the expression 4 sin 80° - tan 80°, which is equal to tan 100° + 4 sin 100°.

Let me compute tan 80°:

tan 80° is approximately 5.671, so 4 sin 80° is approximately 3.939, so the difference is approximately -1.732, which is -√3. Therefore, if I can show algebraically that 4 sin 80° - tan 80° = -√3, that would be great.

Let me try to do that. Let me write 4 sin 80° - tan 80°.

Express tan 80° as sin 80° / cos 80°:

4 sin 80° - sin 80° / cos 80° = sin 80° (4 - 1 / cos 80°).

So, sin 80° * [ (4 cos 80° - 1) / cos 80° ].

Therefore, the entire expression is [ sin 80° (4 cos 80° - 1) ] / cos 80°.

So if this is equal to -√3, then:

[ sin 80° (4 cos 80° - 1) ] / cos 80° = -√3.

Let me compute the numerator:

sin 80° (4 cos 80° - 1).

Let me compute 4 cos 80° - 1. Let me use exact values or relations. Hmm. Let me recall that cos 80° is sin 10°, but not sure if that helps. Alternatively, maybe I can relate angles to known values. Alternatively, let me think of 80° as 60° + 20°, but not sure.

Alternatively, perhaps I can compute sin 80° (4 cos 80° - 1) and see if it relates to √3 cos 80°.

Wait, if I suppose that [ sin 80° (4 cos 80° - 1) ] / cos 80° = -√3, then:

Multiply both sides by cos 80°:

sin 80° (4 cos 80° - 1) = -√3 cos 80°.

Therefore:

sin 80° * 4 cos 80° - sin 80° = -√3 cos 80°.

Let me compute sin 80° * 4 cos 80°:

That is 4 sin 80° cos 80°. Recall that sin 2θ = 2 sin θ cos θ, so sin 160° = 2 sin 80° cos 80°, so 4 sin 80° cos 80° = 2 sin 160°.

Therefore, the left-hand side becomes:

2 sin 160° - sin 80° = -√3 cos 80°.

But sin 160° is sin (180° - 20°) = sin 20°, so:

2 sin 20° - sin 80° = -√3 cos 80°.

Let me compute the left side:

2 sin 20° - sin 80°.

Let me note that sin 80° = sin (60° + 20°) = sin 60° cos 20° + cos 60° sin 20° = (√3/2) cos 20° + (1/2) sin 20°.

But maybe that's not helpful. Alternatively, let me compute 2 sin 20° - sin 80°.

Alternatively, use sine subtraction formulas or other identities.

Alternatively, write sin 80° as sin(60° + 20°). Wait, sin 80° = sin(60° + 20°). Wait, 60 + 20 is 80. Let me expand that:

sin(60° + 20°) = sin 60° cos 20° + cos 60° sin 20°, which is same as before.

But maybe I can write 2 sin 20° - sin(60° + 20°) = ?

Let me compute 2 sin 20° - [ sin 60° cos 20° + cos 60° sin 20° ].

Expanding this:

2 sin 20° - sin 60° cos 20° - cos 60° sin 20°.

Let me group terms with sin 20°:

[2 - cos 60°] sin 20° - sin 60° cos 20°.

Since cos 60° = 0.5, so:

[2 - 0.5] sin 20° - sin 60° cos 20° = 1.5 sin 20° - (√3/2) cos 20°.

Hmm, 1.5 is 3/2. Let me write that:

(3/2) sin 20° - (√3/2) cos 20°.

This looks similar to the formula for a sine of some angle. Remember that a sin θ + b cos θ = C sin(θ + φ) or something like that. Wait, but here it's (3/2) sin 20° - (√3/2) cos 20°. Let me write this as:

Let me think of this as A sin θ + B cos θ, where A = 3/2, B = -√3/2. The magnitude would be sqrt(A² + B²) = sqrt( (9/4) + (3/4) ) = sqrt(12/4) = sqrt(3) ≈ 1.732.

So the expression is sqrt(3) times ( (3/2)/sqrt(3) sin θ + (-√3/2)/sqrt(3) cos θ ) ?

Wait, let me recall that A sin θ + B cos θ = R sin(θ + φ), where R = sqrt(A² + B²). Let me compute:

Alternatively, maybe write it as:

Let me factor out sqrt(3):

sqrt(3) [ ( (3/2)/sqrt(3) ) sin 20° - ( (√3/2)/sqrt(3) ) cos 20° ]

Simplify the coefficients:

(3/2)/sqrt(3) = (sqrt(3)/2) because (3)/sqrt(3) is sqrt(3). Similarly, (√3/2)/sqrt(3) is 1/2. Therefore:

sqrt(3) [ (sqrt(3)/2 sin 20° - 1/2 cos 20° ) ].

Now, note that sqrt(3)/2 is cos 30°, and 1/2 is sin 30°, but with a negative sign. Wait:

Wait, cos(30°) = sqrt(3)/2, sin(30°) = 1/2. Therefore:

sqrt(3) [ cos 30° sin 20° - sin 30° cos 20° ].

Which is sqrt(3) [ sin 20° cos 30° - cos 20° sin 30° ].

That's sqrt(3) sin(20° - 30°) = sqrt(3) sin(-10°) = -sqrt(3) sin 10°.

So the entire expression 2 sin 20° - sin 80° equals -sqrt(3) sin 10°, according to this.

Wait, so going back, we had earlier:

Left-hand side was 2 sin 20° - sin 80° = -sqrt(3) sin 10°, and we supposed this equals -sqrt(3) cos 80°.

Wait, so we have:

- sqrt(3) sin 10° = - sqrt(3) cos 80° ?

Since sin 10° = cos(80°) because sin θ = cos(90° - θ). Yes! Because 90° - 80° = 10°, so sin 10° = cos 80°. Therefore, -sqrt(3) sin 10° = -sqrt(3) cos 80°, which matches the right-hand side.

Therefore, the equation holds true! So this means that:

[ sin 80° (4 cos 80° - 1) ] / cos 80° = -√3.

Therefore, tan 100° + 4 sin 100° = -√3.

Therefore, the simplified form is -√3.

But let me check once more to make sure I didn't make any errors in the steps.

Starting from tan 100° + 4 sin 100°, converted to -tan 80° + 4 sin 80°. Then expressed tan 80° as sin 80° / cos 80°, leading to -sin 80° / cos 80° + 4 sin 80°. Then factoring sin 80°, and proceeding through the steps, leading to the expression involving 2 sin 20° - sin 80°, which after some identity manipulations became -sqrt(3) sin 10°, which equals -sqrt(3) cos 80°, so equality holds. Therefore, the entire expression equals -√3. Therefore, the answer is -√3.

Alternatively, to check with approximate values:

Compute tan(100°):

First, convert 100° to radians for calculator input? Well, maybe I can remember or approximate:

tan(90°) is undefined, approaching infinity. tan(80°) is about 5.671, tan(90°) is infinity, tan(100°) is tan(180° - 80°) = -tan 80° ≈ -5.671.

Then 4 sin 100° is 4 sin(80°). sin(80°) is approx 0.9848, so 4 * 0.9848 ≈ 3.939.

Adding them: -5.671 + 3.939 ≈ -1.732, which is approximately -√3 (since √3 ≈ 1.732). Therefore, the approximate calculation matches the exact value. Therefore, I can be confident that the exact value is indeed -√3.

Hence, the simplified form of tan 100° + 4 sin 100° is -√3.

**Final Answer**
\boxed{-\sqrt{3}}

To simplify the expression:

$$
\tan 100^\circ + 4 \sin 100^\circ
$$

we begin by leveraging trigonometric identities and angle relationships.

---

### Step 1: Use angle identities

We observe that:

$$
\tan(180^\circ - \theta) = -\tan(\theta)
$$
$$
\sin(180^\circ - \theta) = \sin(\theta)
$$

So, applying these to the given angles:

$$
\tan 100^\circ = \tan(180^\circ - 80^\circ) = -\tan 80^\circ
$$
$$
\sin 100^\circ = \sin(180^\circ - 80^\circ) = \sin 80^\circ
$$

Substituting into the original expression:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\tan 80^\circ + 4 \sin 80^\circ
$$

---

### Step 2: Express everything in terms of sine and cosine

Recall that:

$$
\tan 80^\circ = \frac{\sin 80^\circ}{\cos 80^\circ}
$$

So:

$$
-\tan 80^\circ + 4 \sin 80^\circ = -\frac{\sin 80^\circ}{\cos 80^\circ} + 4 \sin 80^\circ
$$

Factor out $\sin 80^\circ$:

$$
= \sin 80^\circ \left( 4 - \frac{1}{\cos 80^\circ} \right)
$$

Now simplify the expression inside the parentheses:

$$
4 - \frac{1}{\cos 80^\circ} = \frac{4 \cos 80^\circ - 1}{\cos 80^\circ}
$$

Thus, the entire expression becomes:

$$
\sin 80^\circ \cdot \frac{4 \cos 80^\circ - 1}{\cos 80^\circ}
$$

So:

$$
\frac{\sin 80^\circ (4 \cos 80^\circ - 1)}{\cos 80^\circ}
$$

---

### Step 3: Simplify using trigonometric identities

Compute:

$$
4 \sin 80^\circ \cos 80^\circ = 2 \cdot 2 \sin 80^\circ \cos 80^\circ = 2 \sin 160^\circ
$$

Because:

$$
\sin(2\theta) = 2 \sin \theta \cos \theta \Rightarrow \sin 160^\circ = 2 \sin 80^\circ \cos 80^\circ
$$

Then:

$$
4 \sin 80^\circ \cos 80^\circ - \sin 80^\circ = 2 \sin 160^\circ - \sin 80^\circ
$$

But $\sin 160^\circ = \sin(20^\circ)$, so:

$$
2 \sin 20^\circ - \sin 80^\circ
$$

Now apply the identity:

$$
\sin(60^\circ + 20^\circ) = \sin 80^\circ = \sin 60^\circ \cos 20^\circ + \cos 60^\circ \sin 20^\circ
$$

Then:

$$
2 \sin 20^\circ - \sin 80^\circ = 2 \sin 20^\circ - \left( \frac{\sqrt{3}}{2} \cos 20^\circ + \frac{1}{2} \sin 20^\circ \right)
$$

Simplify:

$$
= 2 \sin 20^\circ - \frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ
= \left( 2 - \frac{1}{2} \right) \sin 20^\circ - \frac{\sqrt{3}}{2} \cos 20^\circ
= \frac{3}{2} \sin 20^\circ - \frac{\sqrt{3}}{2} \cos 20^\circ
$$

This can be rewritten as:

$$
= \sqrt{3} \left( \frac{\sqrt{3}}{2} \sin 20^\circ - \frac{1}{2} \cos 20^\circ \right)
$$

Recognize the identity:

$$
\sin(\theta - 30^\circ) = \sin \theta \cos 30^\circ - \cos \theta \sin 30^\circ
$$

So the expression becomes:

$$
\sqrt{3} \sin(-10^\circ) = -\sqrt{3} \sin 10^\circ
$$

Since $\sin 10^\circ = \cos 80^\circ$, we have:

$$
-\sqrt{3} \cos 80^\circ
$$

Therefore:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\sqrt{3}
$$

---

### Final Answer

$$
\boxed{-\sqrt{3}}
$$

[Assistant]
Explanation:
Okay, so I need to simplify the expression tan 100° + 4 sin 100°. Hmm, let me think about how to approach this. First, maybe I should recall some trigonometric identities that could help simplify this. Let me jot down what I know:

- Tangent is sine over cosine, so tan θ = sin θ / cos θ.
- Maybe I can write everything in terms of sine and cosine? Let me try that.

So, tan 100° + 4 sin 100° = (sin 100° / cos 100°) + 4 sin 100°. Hmm, that seems a bit messy. Maybe I can factor out sin 100°?

Let me see: sin 100° [1 / cos 100° + 4]. But I don't know if that helps. Maybe I need another approach. Let me think about the angles involved. 100° is in the second quadrant. I know that tan(180° - θ) = -tan θ, so tan 100° = tan(180° - 80°) = -tan 80°. Similarly, sin 100° = sin(180° - 80°) = sin 80°. So maybe rewriting the original expression in terms of 80°?

Let me do that:

tan 100° + 4 sin 100° = -tan 80° + 4 sin 80°.

Hmm, that might be helpful. Let me write that down:

- tan 80° + 4 sin 80°. Now, tan 80° is sin 80° / cos 80°, so substituting that in:

- (sin 80° / cos 80°) + 4 sin 80°. Let me factor out sin 80°:

sin 80° [ -1 / cos 80° + 4 ].

Hmm, still not sure. Let me compute the numerical values to see if there's a pattern or if it simplifies to something nice. Wait, maybe I should check if there's an identity that can combine these terms. Alternatively, perhaps using complex numbers or some other trick?

Alternatively, maybe express everything in terms of sine or cosine? Let me see:

Let me consider writing the expression as:

- tan 80° + 4 sin 80° = 4 sin 80° - tan 80°.

Maybe converting tan 80° to sin 80° / cos 80°:

= 4 sin 80° - sin 80° / cos 80°.

Factor sin 80°:

= sin 80° [4 - 1 / cos 80°].

Hmm, perhaps not helpful. Let me compute the value numerically to see if it's a standard angle or something.

Let me calculate each term:

First, compute tan 100°:

100° is in the second quadrant. Let me convert it to decimal or check approximate value.

But maybe I can use calculator-like approximations. Let me recall that tan 80° is approximately 5.671, so tan 100° is -tan 80° ≈ -5.671.

Then sin 100° is sin 80° ≈ 0.9848. Therefore 4 * sin 100° ≈ 4 * 0.9848 ≈ 3.9392.

So adding them together: -5.671 + 3.9392 ≈ -1.7318. Wait, that's approximately -√3 ≈ -1.732. That seems close. Is the exact value -√3? Let me check.

Wait, √3 is approximately 1.732, so -√3 is approximately -1.732. So my approximate calculation gives -1.7318, which is very close. Therefore, maybe the exact value is -√3?

But let me verify this more carefully.

If tan 100° + 4 sin 100° = -√3, is that correct?

Let me try to prove that.

Let me start with tan 100° + 4 sin 100°. Let me write tan as sin/cos:

sin 100° / cos 100° + 4 sin 100° = sin 100° (1 / cos 100° + 4).

But 1 / cos 100° is sec 100°, which is negative because cos 100° is negative (since 100° is in the second quadrant). But maybe I can express this in terms of 80°, since 100° = 180° - 80°, so cos 100° = -cos 80°, and sin 100° = sin 80°, so:

sin 80° [ 1 / (-cos 80°) + 4 ] = sin 80° [ 4 - 1 / cos 80° ].

Hmm, so that's sin 80° times (4 - sec 80°). If this equals -√3, then maybe there's a relationship here.

Alternatively, maybe I can write this as:

Let me consider the expression 4 sin 80° - tan 80°, which is equal to tan 100° + 4 sin 100°.

Let me compute tan 80°:

tan 80° is approximately 5.671, so 4 sin 80° is approximately 3.939, so the difference is approximately -1.732, which is -√3. Therefore, if I can show algebraically that 4 sin 80° - tan 80° = -√3, that would be great.

Let me try to do that. Let me write 4 sin 80° - tan 80°.

Express tan 80° as sin 80° / cos 80°:

4 sin 80° - sin 80° / cos 80° = sin 80° (4 - 1 / cos 80°).

So, sin 80° * [ (4 cos 80° - 1) / cos 80° ].

Therefore, the entire expression is [ sin 80° (4 cos 80° - 1) ] / cos 80°.

So if this is equal to -√3, then:

[ sin 80° (4 cos 80° - 1) ] / cos 80° = -√3.

Let me compute the numerator:

sin 80° (4 cos 80° - 1).

Let me compute 4 cos 80° - 1. Let me use exact values or relations. Hmm. Let me recall that cos 80° is sin 10°, but not sure if that helps. Alternatively, maybe I can relate angles to known values. Alternatively, let me think of 80° as 60° + 20°, but not sure.

Alternatively, perhaps I can compute sin 80° (4 cos 80° - 1) and see if it relates to √3 cos 80°.

Wait, if I suppose that [ sin 80° (4 cos 80° - 1) ] / cos 80° = -√3, then:

Multiply both sides by cos 80°:

sin 80° (4 cos 80° - 1) = -√3 cos 80°.

Therefore:

sin 80° * 4 cos 80° - sin 80° = -√3 cos 80°.

Let me compute sin 80° * 4 cos 80°:

That is 4 sin 80° cos 80°. Recall that sin 2θ = 2 sin θ cos θ, so sin 160° = 2 sin 80° cos 80°, so 4 sin 80° cos 80° = 2 sin 160°.

Therefore, the left-hand side becomes:

2 sin 160° - sin 80° = -√3 cos 80°.

But sin 160° is sin (180° - 20°) = sin 20°, so:

2 sin 20° - sin 80° = -√3 cos 80°.

Let me compute the left side:

2 sin 20° - sin 80°.

Let me note that sin 80° = sin (60° + 20°) = sin 60° cos 20° + cos 60° sin 20° = (√3/2) cos 20° + (1/2) sin 20°.

But maybe that's not helpful. Alternatively, let me compute 2 sin 20° - sin 80°.

Alternatively, use sine subtraction formulas or other identities.

Alternatively, write sin 80° as sin(60° + 20°). Wait, sin 80° = sin(60° + 20°). Wait, 60 + 20 is 80. Let me expand that:

sin(60° + 20°) = sin 60° cos 20° + cos 60° sin 20°, which is same as before.

But maybe I can write 2 sin 20° - sin(60° + 20°) = ?

Let me compute 2 sin 20° - [ sin 60° cos 20° + cos 60° sin 20° ].

Expanding this:

2 sin 20° - sin 60° cos 20° - cos 60° sin 20°.

Let me group terms with sin 20°:

[2 - cos 60°] sin 20° - sin 60° cos 20°.

Since cos 60° = 0.5, so:

[2 - 0.5] sin 20° - sin 60° cos 20° = 1.5 sin 20° - (√3/2) cos 20°.

Hmm, 1.5 is 3/2. Let me write that:

(3/2) sin 20° - (√3/2) cos 20°.

This looks similar to the formula for a sine of some angle. Remember that a sin θ + b cos θ = C sin(θ + φ) or something like that. Wait, but here it's (3/2) sin 20° - (√3/2) cos 20°. Let me write this as:

Let me think of this as A sin θ + B cos θ, where A = 3/2, B = -√3/2. The magnitude would be sqrt(A² + B²) = sqrt( (9/4) + (3/4) ) = sqrt(12/4) = sqrt(3) ≈ 1.732.

So the expression is sqrt(3) times ( (3/2)/sqrt(3) sin θ + (-√3/2)/sqrt(3) cos θ ) ?

Wait, let me recall that A sin θ + B cos θ = R sin(θ + φ), where R = sqrt(A² + B²). Let me compute:

Alternatively, maybe write it as:

Let me factor out sqrt(3):

sqrt(3) [ ( (3/2)/sqrt(3) ) sin 20° - ( (√3/2)/sqrt(3) ) cos 20° ]

Simplify the coefficients:

(3/2)/sqrt(3) = (sqrt(3)/2) because (3)/sqrt(3) is sqrt(3). Similarly, (√3/2)/sqrt(3) is 1/2. Therefore:

sqrt(3) [ (sqrt(3)/2 sin 20° - 1/2 cos 20° ) ].

Now, note that sqrt(3)/2 is cos 30°, and 1/2 is sin 30°, but with a negative sign. Wait:

Wait, cos(30°) = sqrt(3)/2, sin(30°) = 1/2. Therefore:

sqrt(3) [ cos 30° sin 20° - sin 30° cos 20° ].

Which is sqrt(3) [ sin 20° cos 30° - cos 20° sin 30° ].

That's sqrt(3) sin(20° - 30°) = sqrt(3) sin(-10°) = -sqrt(3) sin 10°.

So the entire expression 2 sin 20° - sin 80° equals -sqrt(3) sin 10°, according to this.

Wait, so going back, we had earlier:

Left-hand side was 2 sin 20° - sin 80° = -sqrt(3) sin 10°, and we supposed this equals -sqrt(3) cos 80°.

Wait, so we have:

- sqrt(3) sin 10° = - sqrt(3) cos 80° ?

Since sin 10° = cos(80°) because sin θ = cos(90° - θ). Yes! Because 90° - 80° = 10°, so sin 10° = cos 80°. Therefore, -sqrt(3) sin 10° = -sqrt(3) cos 80°, which matches the right-hand side.

Therefore, the equation holds true! So this means that:

[ sin 80° (4 cos 80° - 1) ] / cos 80° = -√3.

Therefore, tan 100° + 4 sin 100° = -√3.

Therefore, the simplified form is -√3.

But let me check once more to make sure I didn't make any errors in the steps.

Starting from tan 100° + 4 sin 100°, converted to -tan 80° + 4 sin 80°. Then expressed tan 80° as sin 80° / cos 80°, leading to -sin 80° / cos 80° + 4 sin 80°. Then factoring sin 80°, and proceeding through the steps, leading to the expression involving 2 sin 20° - sin 80°, which after some identity manipulations became -sqrt(3) sin 10°, which equals -sqrt(3) cos 80°, so equality holds. Therefore, the entire expression equals -√3. Therefore, the answer is -√3.

Alternatively, to check with approximate values:

Compute tan(100°):

First, convert 100° to radians for calculator input? Well, maybe I can remember or approximate:

tan(90°) is undefined, approaching infinity. tan(80°) is about 5.671, tan(90°) is infinity, tan(100°) is tan(180° - 80°) = -tan 80° ≈ -5.671.

Then 4 sin 100° is 4 sin(80°). sin(80°) is approx 0.9848, so 4 * 0.9848 ≈ 3.939.

Adding them: -5.671 + 3.939 ≈ -1.732, which is approximately -√3 (since √3 ≈ 1.732). Therefore, the approximate calculation matches the exact value. Therefore, I can be confident that the exact value is indeed -√3.

Hence, the simplified form of tan 100° + 4 sin 100° is -√3.

**Final Answer**
\boxed{-\sqrt{3}}

To simplify the expression:

$$
\tan 100^\circ + 4 \sin 100^\circ
$$

we begin by leveraging trigonometric identities and angle relationships.

---

### Step 1: Use angle identities

We observe that:

$$
\tan(180^\circ - \theta) = -\tan(\theta)
$$
$$
\sin(180^\circ - \theta) = \sin(\theta)
$$

So, applying these to the given angles:

$$
\tan 100^\circ = \tan(180^\circ - 80^\circ) = -\tan 80^\circ
$$
$$
\sin 100^\circ = \sin(180^\circ - 80^\circ) = \sin 80^\circ
$$

Substituting into the original expression:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\tan 80^\circ + 4 \sin 80^\circ
$$

---

### Step 2: Express everything in terms of sine and cosine

Recall that:

$$
\tan 80^\circ = \frac{\sin 80^\circ}{\cos 80^\circ}
$$

So:

$$
-\tan 80^\circ + 4 \sin 80^\circ = -\frac{\sin 80^\circ}{\cos 80^\circ} + 4 \sin 80^\circ
$$

Factor out $\sin 80^\circ$:

$$
= \sin 80^\circ \left( 4 - \frac{1}{\cos 80^\circ} \right)
$$

Now simplify the expression inside the parentheses:

$$
4 - \frac{1}{\cos 80^\circ} = \frac{4 \cos 80^\circ - 1}{\cos 80^\circ}
$$

Thus, the entire expression becomes:

$$
\sin 80^\circ \cdot \frac{4 \cos 80^\circ - 1}{\cos 80^\circ}
$$

So:

$$
\frac{\sin 80^\circ (4 \cos 80^\circ - 1)}{\cos 80^\circ}
$$

---

### Step 3: Simplify using trigonometric identities

Compute:

$$
4 \sin 80^\circ \cos 80^\circ = 2 \cdot 2 \sin 80^\circ \cos 80^\circ = 2 \sin 160^\circ
$$

Because:

$$
\sin(2\theta) = 2 \sin \theta \cos \theta \Rightarrow \sin 160^\circ = 2 \sin 80^\circ \cos 80^\circ
$$

Then:

$$
4 \sin 80^\circ \cos 80^\circ - \sin 80^\circ = 2 \sin 160^\circ - \sin 80^\circ
$$

But $\sin 160^\circ = \sin(20^\circ)$, so:

$$
2 \sin 20^\circ - \sin 80^\circ
$$

Now apply the identity:

$$
\sin(60^\circ + 20^\circ) = \sin 80^\circ = \sin 60^\circ \cos 20^\circ + \cos 60^\circ \sin 20^\circ
$$

Then:

$$
2 \sin 20^\circ - \sin 80^\circ = 2 \sin 20^\circ - \left( \frac{\sqrt{3}}{2} \cos 20^\circ + \frac{1}{2} \sin 20^\circ \right)
$$

Simplify:

$$
= 2 \sin 20^\circ - \frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ
= \left( 2 - \frac{1}{2} \right) \sin 20^\circ - \frac{\sqrt{3}}{2} \cos 20^\circ
= \frac{3}{2} \sin 20^\circ - \frac{\sqrt{3}}{2} \cos 20^\circ
$$

This can be rewritten as:

$$
= \sqrt{3} \left( \frac{\sqrt{3}}{2} \sin 20^\circ - \frac{1}{2} \cos 20^\circ \right)
$$

Recognize the identity:

$$
\sin(\theta - 30^\circ) = \sin \theta \cos 30^\circ - \cos \theta \sin 30^\circ
$$

So the expression becomes:

$$
\sqrt{3} \sin(-10^\circ) = -\sqrt{3} \sin 10^\circ
$$

Since $\sin 10^\circ = \cos 80^\circ$, we have:

$$
-\sqrt{3} \cos 80^\circ
$$

Therefore:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\sqrt{3}
$$

---

### Final Answer

$$
\boxed{-\sqrt{3}}
$$

Solution:
We have that
\begin{align*}
\tan 100^\circ + 4 \sin 100^\circ &= \frac{\sin 100^\circ}{\cos 100^\circ} + 4 \sin 100^\circ \\
&= \frac{\sin 80^\circ}{-\cos 80^\circ} + 4 \sin 80^\circ \\
&= -\frac{\cos 10^\circ}{\sin 10^\circ} + 4 \cos 10^\circ \\
&= \frac{4 \cos 10^\circ \sin 10^\circ - \cos 10^\circ}{\sin 10^\circ}.
\end{align*}By double angle formula,
\begin{align*}
\frac{4 \cos 10^\circ \sin 10^\circ - \cos 10^\circ}{\sin 10^\circ} &= \frac{2 \sin 20^\circ - \cos 10^\circ}{\sin 10^\circ} \\
&= \frac{\sin 20^\circ + \sin 20^\circ - \sin 80^\circ}{\sin 10^\circ}.
\end{align*}By sum-to-product,
\[\sin 20^\circ - \sin 80^\circ = 2 \cos 50^\circ \sin (-30^\circ) = -\cos 50^\circ,\]so
\begin{align*}
\frac{\sin 20^\circ + \sin 20^\circ - \sin 80^\circ}{\sin 10^\circ} &= \frac{\sin 20^\circ - \cos 50^\circ}{\sin 10^\circ} \\
&= \frac{\sin 20^\circ - \sin 40^\circ}{\sin 10^\circ}.
\end{align*}By sum-to-product,
\[\sin 20^\circ - \sin 40^\circ = 2 \cos 30^\circ \sin (-10^\circ) = -\sqrt{3} \sin 10^\circ,\]so $\frac{\sin 20^\circ - \sin 40^\circ}{\sin 10^\circ} = \boxed{-\sqrt{3}}.$