Explanation:
Okay, so I need to subtract two numbers in base 9: 58 base 9 minus 18 base 9. Hmm, let me think about how to do this. I remember that when subtracting numbers in different bases, it's similar to subtracting in base 10, but you have to remember the base you're working in. Since both numbers are in base 9, I can perform the subtraction directly in base 9, but I need to be careful with borrowing if necessary. Let me write down the numbers vertically to make it clearer.

So, let me write 58_9 minus 18_9. Let me align them by their place values:

```
  5 8
- 1 8
-------
```

Starting from the rightmost digit, which is the units place. In base 9, the rightmost digit is the 9^0 place, which is the same as base 10. So, the units digit subtraction would be 8 (from 58_9) minus 8 (from 18_9). Let me do that first.

8 - 8 = 0. So, the units digit is 0. Then moving to the next digit, which is the 9^1 place. Here, we have 5 (from 58_9) minus 1 (from 18_9). That would be 5 - 1 = 4. So, putting it together, the result would be 40_9? Wait, that seems too straightforward. Let me check if there's any borrowing involved here.

Wait, in the units place, we had 8 - 8, which is 0, no problem. Then in the next place, 5 - 1 = 4. So, yes, seems like no borrowing is needed. Therefore, the answer is 40_9? Let me verify this by converting both numbers to base 10, performing the subtraction, and then converting back to base 9 to check.

First, converting 58_9 to base 10. The rightmost digit is 8 * 9^0 = 8 * 1 = 8. The next digit is 5 * 9^1 = 5 * 9 = 45. Adding them together: 45 + 8 = 53. So, 58_9 is 53 in base 10.

Now, converting 18_9 to base 10. The rightmost digit is 8 * 9^0 = 8. The next digit is 1 * 9^1 = 9. Adding them: 9 + 8 = 17. So, 18_9 is 17 in base 10.

Subtracting them: 53 - 17 = 36. Now, converting 36 back to base 9. To do this, I need to find how many times 9 goes into 36. 9 * 4 = 36, so that's 4 with no remainder. Therefore, 36 in base 10 is 40_9. So that matches my previous result. Therefore, 58_9 - 18_9 is indeed 40_9.

Wait, but let me try doing the subtraction in base 9 again, just to make sure I didn't skip any steps. Maybe I should write it out more carefully.

So, writing the subtraction:

```
  5 8
- 1 8
-------
```

Starting from the rightmost digit: 8 - 8. Since 8 is equal to 8, we subtract and get 0. No borrowing needed. Then moving to the left: 5 - 1. That's 4. So, the result is 40_9. Yep, same answer. So that seems correct.

But just to be thorough, let me consider if there was a case where borrowing would be necessary. For example, if the top digit was smaller than the bottom digit. Let's say if we had something like 25_9 - 17_9. Then, in the units place, 5 - 7. Since 5 is less than 7, we would need to borrow from the next higher place. But in our original problem, there was no such case.

But let me think again. Maybe I should try another example to check my understanding. Suppose I have 10_9 - 1_9. In base 9, 10_9 is 9 in decimal, and 1_9 is 1 in decimal. So, 9 - 1 = 8, which is 8_9. If I do the subtraction directly:

```
  1 0
-   1
-------
```

Starting from the right: 0 - 1. Can't do that, so borrow from the next digit. The next digit is 1, so we borrow 1 from the 1, which is worth 9 in the units place. So, the 0 becomes 9, and then we subtract 1, getting 8. The next digit was originally 1, but we borrowed 1, so it becomes 0. Therefore, the result is 08_9, which is 8_9. Correct. So, borrowing works here.

But in our original problem, there was no need for borrowing. So, the answer is 40_9. But let me check once more by converting 40_9 back to decimal. 4*9 + 0 = 36, which matches the earlier result. So, that's correct.

Alternatively, maybe I can perform the subtraction in base 9 using another method. Let me think. Let me write down the numbers again:

58_9
-18_9
=?

Another way is to think of 58_9 as 5*9 + 8 and 18_9 as 1*9 + 8. Then subtracting them:

(5*9 + 8) - (1*9 + 8) = (5*9 - 1*9) + (8 - 8) = 4*9 + 0 = 40_9. Yep, same result. So that's straightforward.

Alternatively, if I didn't convert to decimal, but tried to do the subtraction step by step in base 9. Let me try that again.

Starting with the units place: 8 - 8. Since 8 is equal to 8, we subtract and get 0. No problem. Then moving to the 9s place: 5 - 1. That's 4. So, the result is 40_9.

Therefore, all methods lead to the same answer. Therefore, I think the answer is 40_9.

But just to make sure, let me think if there's any chance I misread the original numbers. The problem says 58_9 minus 18_9. So, both numbers are in base 9. The digits are 5 and 8 in the first number, and 1 and 8 in the second. So, yes, that's correct.

Another check: if I add 18_9 to 40_9, I should get 58_9. Let me verify that.

Adding 40_9 and 18_9:

```
  4 0
+ 1 8
-------
```

Starting from the right: 0 + 8 = 8. Then 4 + 1 = 5. So, the result is 58_9. Which matches the original number. Therefore, the subtraction is correct.

Therefore, I can be confident that the answer is 40_9.

But just to recap all steps once again:

1. Align the numbers vertically.
2. Subtract the units digits: 8 - 8 = 0.
3. Subtract the 9s digits: 5 - 1 = 4.
4. Result is 40_9.
5. Verified by converting to decimal, which gave 36, which is 40_9.
6. Verified by adding 40_9 and 18_9 to get back 58_9.

All checks confirm that the answer is correct. Therefore, I think I can safely conclude that the result of 58_9 minus 18_9 is 40_9.

**Final Answer**
The result of $58_9 - 18_9$ is \boxed{40_9}.

To compute $ 58_9 - 18_9 $, we perform the subtraction directly in base 9, following the rules of base arithmetic.

---

### Step 1: Align the Numbers Vertically

We align the numbers by place value:

```
  5 8
- 1 8
-------
```

---

### Step 2: Subtract the Units Place

Start from the rightmost digit (units place):

- $ 8_9 - 8_9 = 0_9 $

No borrowing is needed here.

---

### Step 3: Subtract the 9s Place

Now move to the next digit (9s place):

- $ 5_9 - 1_9 = 4_9 $

Again, no borrowing is required.

---

### Step 4: Combine the Results

Putting the results together:

- Units place: $ 0_9 $
- 9s place: $ 4_9 $

So, the result is:

$$
\boxed{40_9}
$$

---

### Step 5: Verification (Optional)

To ensure correctness, we can convert both numbers to base 10:

- $ 58_9 = 5 \cdot 9 + 8 = 45 + 8 = 53_{10} $
- $ 18_9 = 1 \cdot 9 + 8 = 9 + 8 = 17_{10} $

Subtracting in base 10:

$$
53 - 17 = 36
$$

Now convert $ 36_{10} $ back to base 9:

- $ 36 \div 9 = 4 $ remainder $ 0 $
- So, $ 36_{10} = 40_9 $

This confirms our earlier result.

---

### Final Answer

$$
\boxed{40_9}
$$

[Assistant]
Explanation:
Okay, so I need to subtract two numbers in base 9: 58 base 9 minus 18 base 9. Hmm, let me think about how to do this. I remember that when subtracting numbers in different bases, it's similar to subtracting in base 10, but you have to remember the base you're working in. Since both numbers are in base 9, I can perform the subtraction directly in base 9, but I need to be careful with borrowing if necessary. Let me write down the numbers vertically to make it clearer.

So, let me write 58_9 minus 18_9. Let me align them by their place values:

```
  5 8
- 1 8
-------
```

Starting from the rightmost digit, which is the units place. In base 9, the rightmost digit is the 9^0 place, which is the same as base 10. So, the units digit subtraction would be 8 (from 58_9) minus 8 (from 18_9). Let me do that first.

8 - 8 = 0. So, the units digit is 0. Then moving to the next digit, which is the 9^1 place. Here, we have 5 (from 58_9) minus 1 (from 18_9). That would be 5 - 1 = 4. So, putting it together, the result would be 40_9? Wait, that seems too straightforward. Let me check if there's any borrowing involved here.

Wait, in the units place, we had 8 - 8, which is 0, no problem. Then in the next place, 5 - 1 = 4. So, yes, seems like no borrowing is needed. Therefore, the answer is 40_9? Let me verify this by converting both numbers to base 10, performing the subtraction, and then converting back to base 9 to check.

First, converting 58_9 to base 10. The rightmost digit is 8 * 9^0 = 8 * 1 = 8. The next digit is 5 * 9^1 = 5 * 9 = 45. Adding them together: 45 + 8 = 53. So, 58_9 is 53 in base 10.

Now, converting 18_9 to base 10. The rightmost digit is 8 * 9^0 = 8. The next digit is 1 * 9^1 = 9. Adding them: 9 + 8 = 17. So, 18_9 is 17 in base 10.

Subtracting them: 53 - 17 = 36. Now, converting 36 back to base 9. To do this, I need to find how many times 9 goes into 36. 9 * 4 = 36, so that's 4 with no remainder. Therefore, 36 in base 10 is 40_9. So that matches my previous result. Therefore, 58_9 - 18_9 is indeed 40_9.

Wait, but let me try doing the subtraction in base 9 again, just to make sure I didn't skip any steps. Maybe I should write it out more carefully.

So, writing the subtraction:

```
  5 8
- 1 8
-------
```

Starting from the rightmost digit: 8 - 8. Since 8 is equal to 8, we subtract and get 0. No borrowing needed. Then moving to the left: 5 - 1. That's 4. So, the result is 40_9. Yep, same answer. So that seems correct.

But just to be thorough, let me consider if there was a case where borrowing would be necessary. For example, if the top digit was smaller than the bottom digit. Let's say if we had something like 25_9 - 17_9. Then, in the units place, 5 - 7. Since 5 is less than 7, we would need to borrow from the next higher place. But in our original problem, there was no such case.

But let me think again. Maybe I should try another example to check my understanding. Suppose I have 10_9 - 1_9. In base 9, 10_9 is 9 in decimal, and 1_9 is 1 in decimal. So, 9 - 1 = 8, which is 8_9. If I do the subtraction directly:

```
  1 0
-   1
-------
```

Starting from the right: 0 - 1. Can't do that, so borrow from the next digit. The next digit is 1, so we borrow 1 from the 1, which is worth 9 in the units place. So, the 0 becomes 9, and then we subtract 1, getting 8. The next digit was originally 1, but we borrowed 1, so it becomes 0. Therefore, the result is 08_9, which is 8_9. Correct. So, borrowing works here.

But in our original problem, there was no need for borrowing. So, the answer is 40_9. But let me check once more by converting 40_9 back to decimal. 4*9 + 0 = 36, which matches the earlier result. So, that's correct.

Alternatively, maybe I can perform the subtraction in base 9 using another method. Let me think. Let me write down the numbers again:

58_9
-18_9
=?

Another way is to think of 58_9 as 5*9 + 8 and 18_9 as 1*9 + 8. Then subtracting them:

(5*9 + 8) - (1*9 + 8) = (5*9 - 1*9) + (8 - 8) = 4*9 + 0 = 40_9. Yep, same result. So that's straightforward.

Alternatively, if I didn't convert to decimal, but tried to do the subtraction step by step in base 9. Let me try that again.

Starting with the units place: 8 - 8. Since 8 is equal to 8, we subtract and get 0. No problem. Then moving to the 9s place: 5 - 1. That's 4. So, the result is 40_9.

Therefore, all methods lead to the same answer. Therefore, I think the answer is 40_9.

But just to make sure, let me think if there's any chance I misread the original numbers. The problem says 58_9 minus 18_9. So, both numbers are in base 9. The digits are 5 and 8 in the first number, and 1 and 8 in the second. So, yes, that's correct.

Another check: if I add 18_9 to 40_9, I should get 58_9. Let me verify that.

Adding 40_9 and 18_9:

```
  4 0
+ 1 8
-------
```

Starting from the right: 0 + 8 = 8. Then 4 + 1 = 5. So, the result is 58_9. Which matches the original number. Therefore, the subtraction is correct.

Therefore, I can be confident that the answer is 40_9.

But just to recap all steps once again:

1. Align the numbers vertically.
2. Subtract the units digits: 8 - 8 = 0.
3. Subtract the 9s digits: 5 - 1 = 4.
4. Result is 40_9.
5. Verified by converting to decimal, which gave 36, which is 40_9.
6. Verified by adding 40_9 and 18_9 to get back 58_9.

All checks confirm that the answer is correct. Therefore, I think I can safely conclude that the result of 58_9 minus 18_9 is 40_9.

**Final Answer**
The result of $58_9 - 18_9$ is \boxed{40_9}.

To compute $ 58_9 - 18_9 $, we perform the subtraction directly in base 9, following the rules of base arithmetic.

---

### Step 1: Align the Numbers Vertically

We align the numbers by place value:

```
  5 8
- 1 8
-------
```

---

### Step 2: Subtract the Units Place

Start from the rightmost digit (units place):

- $ 8_9 - 8_9 = 0_9 $

No borrowing is needed here.

---

### Step 3: Subtract the 9s Place

Now move to the next digit (9s place):

- $ 5_9 - 1_9 = 4_9 $

Again, no borrowing is required.

---

### Step 4: Combine the Results

Putting the results together:

- Units place: $ 0_9 $
- 9s place: $ 4_9 $

So, the result is:

$$
\boxed{40_9}
$$

---

### Step 5: Verification (Optional)

To ensure correctness, we can convert both numbers to base 10:

- $ 58_9 = 5 \cdot 9 + 8 = 45 + 8 = 53_{10} $
- $ 18_9 = 1 \cdot 9 + 8 = 9 + 8 = 17_{10} $

Subtracting in base 10:

$$
53 - 17 = 36
$$

Now convert $ 36_{10} $ back to base 9:

- $ 36 \div 9 = 4 $ remainder $ 0 $
- So, $ 36_{10} = 40_9 $

This confirms our earlier result.

---

### Final Answer

$$
\boxed{40_9}
$$

Solution: This subtraction is fairly straightforward: we just subtract the respective digits. No borrowing occurs: $$ \begin{array}{c@{}c@{\;}c@{}c} & & 5 & 8_9 \\ &- & 1 & 8_9 \\ \cline{2-4} & & 4 & 0_9 \\ \end{array} $$ Thus, the answer is $\boxed{40_9}.$


Explanation:
Okay, so I need to find the radius of a sphere where the volume and surface area are numerically equal. Hmm, let me think. I remember the formulas for volume and surface area of a sphere. Let me recall them.

The volume of a sphere is (4/3)πr³, right? And the surface area is 4πr². Yeah, that sounds familiar. So the problem says that these two values are numerically equal. That means if I set them equal to each other, I can solve for the radius.

Let me write that down. So, Volume = Surface Area. Therefore:

(4/3)πr³ = 4πr²

Okay, so now I need to solve this equation for r. Let me see. First, maybe I can simplify this equation. Both sides have 4πr², so maybe I can divide both sides by 4πr² to make it simpler. Let me try that.

Dividing both sides by 4πr²:

( (4/3)πr³ ) / (4πr²) ) = (4πr²) / (4πr²)

Simplifying the left side: The 4 cancels out, the π cancels out, and r³ divided by r² is r. So left side becomes (1/3)r. The right side is 1. So:

(1/3)r = 1

Then, solving for r, multiply both sides by 3:

r = 3

Wait, so the radius is 3 units? Let me check that again to make sure I didn't make a mistake.

Let me compute the volume and surface area with r = 3.

Volume: (4/3)π*(3)³ = (4/3)π*27 = 4*9π = 36π

Surface area: 4π*(3)² = 4π*9 = 36π

Yes, both are 36π. So numerically, they are equal. So the radius is indeed 3.

But let me think again if there's another possible solution or if I missed something. For example, when I divided both sides by 4πr², I assumed that r ≠ 0. Well, if r = 0, the sphere would have zero volume and surface area, but that's trivial and not a valid sphere. So r = 3 is the only solution.

Alternatively, maybe I can approach the equation differently. Let me try again:

Starting with (4/3)πr³ = 4πr²

Subtract 4πr² from both sides:

(4/3)πr³ - 4πr² = 0

Factor out 4πr²:

4πr²( (1/3)r - 1 ) = 0

So, setting each factor equal to zero:

4πr² = 0 => r² = 0 => r = 0 (again, trivial solution)

Or (1/3)r - 1 = 0 => (1/3)r = 1 => r = 3

Same result. So that confirms it.

Therefore, the radius is 3 units.

I think that's solid. Let me check if there's any other way someone might approach this problem and maybe get confused. For example, someone might think that since volume is in cubic units and surface area is in square units, they can't be equal? But the problem says they are numerically equal, meaning that if you ignore the units, the numbers are the same. So, for example, if volume was 36π cubic units and surface area was 36π square units, the numbers 36π are equal. So that's what the problem is saying.

So yeah, the answer is 3.

**Final Answer**
The radius of the sphere is \boxed{3} units.

To determine the radius of a sphere where the **volume** and **surface area** are numerically equal, we begin by recalling the standard formulas for these quantities:

- **Volume of a sphere**:
  $$
  V = \frac{4}{3} \pi r^3
  $$

- **Surface area of a sphere**:
  $$
  A = 4 \pi r^2
  $$

We are told that these two values are **numerically equal**, meaning:

$$
\frac{4}{3} \pi r^3 = 4 \pi r^2
$$

---

### Step 1: Simplify the Equation

We can divide both sides of the equation by $ 4 \pi r^2 $, assuming $ r \neq 0 $ (since a sphere with radius 0 is not meaningful in this context):

$$
\frac{\frac{4}{3} \pi r^3}{4 \pi r^2} = \frac{4 \pi r^2}{4 \pi r^2}
$$

Simplifying both sides:

- Left side:
  $$
  \frac{4}{3} \pi r^3 \div 4 \pi r^2 = \frac{1}{3} r
  $$

- Right side:
  $$
  \frac{4 \pi r^2}{4 \pi r^2} = 1
  $$

So the equation becomes:

$$
\frac{1}{3} r = 1
$$

---

### Step 2: Solve for $ r $

Multiply both sides by 3:

$$
r = 3
$$

---

### Step 3: Verify the Solution

Let’s confirm that with $ r = 3 $, the volume and surface area are numerically equal:

- **Volume**:
  $$
  V = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi \cdot 27 = 36 \pi
  $$

- **Surface Area**:
  $$
  A = 4 \pi (3)^2 = 4 \pi \cdot 9 = 36 \pi
  $$

Both are equal to $ 36 \pi $, confirming the solution.

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to find the radius of a sphere where the volume and surface area are numerically equal. Hmm, let me think. I remember the formulas for volume and surface area of a sphere. Let me recall them.

The volume of a sphere is (4/3)πr³, right? And the surface area is 4πr². Yeah, that sounds familiar. So the problem says that these two values are numerically equal. That means if I set them equal to each other, I can solve for the radius.

Let me write that down. So, Volume = Surface Area. Therefore:

(4/3)πr³ = 4πr²

Okay, so now I need to solve this equation for r. Let me see. First, maybe I can simplify this equation. Both sides have 4πr², so maybe I can divide both sides by 4πr² to make it simpler. Let me try that.

Dividing both sides by 4πr²:

( (4/3)πr³ ) / (4πr²) ) = (4πr²) / (4πr²)

Simplifying the left side: The 4 cancels out, the π cancels out, and r³ divided by r² is r. So left side becomes (1/3)r. The right side is 1. So:

(1/3)r = 1

Then, solving for r, multiply both sides by 3:

r = 3

Wait, so the radius is 3 units? Let me check that again to make sure I didn't make a mistake.

Let me compute the volume and surface area with r = 3.

Volume: (4/3)π*(3)³ = (4/3)π*27 = 4*9π = 36π

Surface area: 4π*(3)² = 4π*9 = 36π

Yes, both are 36π. So numerically, they are equal. So the radius is indeed 3.

But let me think again if there's another possible solution or if I missed something. For example, when I divided both sides by 4πr², I assumed that r ≠ 0. Well, if r = 0, the sphere would have zero volume and surface area, but that's trivial and not a valid sphere. So r = 3 is the only solution.

Alternatively, maybe I can approach the equation differently. Let me try again:

Starting with (4/3)πr³ = 4πr²

Subtract 4πr² from both sides:

(4/3)πr³ - 4πr² = 0

Factor out 4πr²:

4πr²( (1/3)r - 1 ) = 0

So, setting each factor equal to zero:

4πr² = 0 => r² = 0 => r = 0 (again, trivial solution)

Or (1/3)r - 1 = 0 => (1/3)r = 1 => r = 3

Same result. So that confirms it.

Therefore, the radius is 3 units.

I think that's solid. Let me check if there's any other way someone might approach this problem and maybe get confused. For example, someone might think that since volume is in cubic units and surface area is in square units, they can't be equal? But the problem says they are numerically equal, meaning that if you ignore the units, the numbers are the same. So, for example, if volume was 36π cubic units and surface area was 36π square units, the numbers 36π are equal. So that's what the problem is saying.

So yeah, the answer is 3.

**Final Answer**
The radius of the sphere is \boxed{3} units.

To determine the radius of a sphere where the **volume** and **surface area** are numerically equal, we begin by recalling the standard formulas for these quantities:

- **Volume of a sphere**:
  $$
  V = \frac{4}{3} \pi r^3
  $$

- **Surface area of a sphere**:
  $$
  A = 4 \pi r^2
  $$

We are told that these two values are **numerically equal**, meaning:

$$
\frac{4}{3} \pi r^3 = 4 \pi r^2
$$

---

### Step 1: Simplify the Equation

We can divide both sides of the equation by $ 4 \pi r^2 $, assuming $ r \neq 0 $ (since a sphere with radius 0 is not meaningful in this context):

$$
\frac{\frac{4}{3} \pi r^3}{4 \pi r^2} = \frac{4 \pi r^2}{4 \pi r^2}
$$

Simplifying both sides:

- Left side:
  $$
  \frac{4}{3} \pi r^3 \div 4 \pi r^2 = \frac{1}{3} r
  $$

- Right side:
  $$
  \frac{4 \pi r^2}{4 \pi r^2} = 1
  $$

So the equation becomes:

$$
\frac{1}{3} r = 1
$$

---

### Step 2: Solve for $ r $

Multiply both sides by 3:

$$
r = 3
$$

---

### Step 3: Verify the Solution

Let’s confirm that with $ r = 3 $, the volume and surface area are numerically equal:

- **Volume**:
  $$
  V = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi \cdot 27 = 36 \pi
  $$

- **Surface Area**:
  $$
  A = 4 \pi (3)^2 = 4 \pi \cdot 9 = 36 \pi
  $$

Both are equal to $ 36 \pi $, confirming the solution.

---

### Final Answer

$$
\boxed{3}
$$

Solution:
The volume of a sphere is $\frac{4}{3}\pi r^3$ and the surface area is $4\pi r^2$, so
\[\frac{4}{3} \pi r^3 = 4 \pi r^2.\]We can divide both sides by $4 \pi r^2$, to get
\[\frac{1}{3} r = 1.\]Therefore, $r = \boxed{3}.$


Explanation:
Okay, so I need to simplify this expression: $\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}$. Hmm, let me think. First, I remember that secant is the reciprocal of cosine, right? So $\sec x = \frac{1}{\cos x}$. Maybe I can rewrite the first term using that. Let me try that.

So substituting $\sec x$ with $\frac{1}{\cos x}$, the first term becomes $\frac{1}{\cos x \cdot \sin x}$. Wait, is that right? Let me check. The original term is $\frac{\sec x}{\sin x}$, so replacing sec x gives $\frac{1/\cos x}{\sin x}$, which is the same as $\frac{1}{\cos x \cdot \sin x}$. Yeah, that seems correct.

So now the expression is $\frac{1}{\cos x \sin x} - \frac{\sin x}{\cos x}$. Hmm, okay. Now I need to combine these two terms. To do that, they need a common denominator. The first term has a denominator of $\cos x \sin x$, and the second term has a denominator of $\cos x$. So if I multiply the numerator and denominator of the second term by $\sin x$, that should give me the same denominator as the first term. Let me try that.

So the second term becomes $\frac{\sin x \cdot \sin x}{\cos x \cdot \sin x}$, which is $\frac{\sin^2 x}{\cos x \sin x}$. Therefore, the entire expression is $\frac{1 - \sin^2 x}{\cos x \sin x}$. Wait, because I subtract the second term, so it's $\frac{1}{\cos x \sin x} - \frac{\sin^2 x}{\cos x \sin x} = \frac{1 - \sin^2 x}{\cos x \sin x}$.

Now, I remember that $1 - \sin^2 x$ is equal to $\cos^2 x$. That's one of the Pythagorean identities. So substituting that in, the numerator becomes $\cos^2 x$. Therefore, the expression simplifies to $\frac{\cos^2 x}{\cos x \sin x}$.

Now, let me simplify that fraction. The numerator is $\cos^2 x$ and the denominator is $\cos x \sin x$. I can cancel out one $\cos x$ from the numerator and denominator. That leaves me with $\frac{\cos x}{\sin x}$.

Wait, $\frac{\cos x}{\sin x}$ is $\cot x$, right? Because cotangent is cosine over sine. So does that mean the entire expression simplifies to $\cot x$? Let me check my steps again to make sure I didn't make a mistake.

Starting with $\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}$:

1. Convert $\sec x$ to $1/\cos x$: $\frac{1}{\cos x \sin x} - \frac{\sin x}{\cos x}$.
2. Find common denominator: Multiply second term by $\sin x/\sin x$: $\frac{1 - \sin^2 x}{\cos x \sin x}$.
3. Use identity $1 - \sin^2 x = \cos^2 x$: $\frac{\cos^2 x}{\cos x \sin x}$.
4. Simplify: $\frac{\cos x}{\sin x} = \cot x$.

Seems correct. Let me verify with a specific angle to make sure. Let's take $x = 45^\circ$ (or $\pi/4$ radians). Let me compute the original expression and the simplified version.

Original expression:

$\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}$.

First, compute each part:

$\sec(\pi/4) = 1/\cos(\pi/4) = \sqrt{2}$. Then divide by $\sin(\pi/4) = \sqrt{2}/2$. So $\frac{\sqrt{2}}{\sqrt{2}/2} = 2$.

Then the second term: $\sin(\pi/4)/\cos(\pi/4) = (\sqrt{2}/2)/(\sqrt{2}/2) = 1$.

So original expression is $2 - 1 = 1$.

Now, $\cot(\pi/4) = \cos(\pi/4)/\sin(\pi/4) = 1$. So that matches. Good.

Let me try another angle, maybe $x = 60^\circ$ ($\pi/3$):

Original expression:

$\sec(\pi/3) = 1/\cos(\pi/3) = 1/(1/2) = 2$. Divide by $\sin(\pi/3) = \sqrt{3}/2$. So first term is $2 / (\sqrt{3}/2) = 4/\sqrt{3}$.

Second term: $\sin(\pi/3)/\cos(\pi/3) = (\sqrt{3}/2)/(1/2) = \sqrt{3}$.

So original expression is $4/\sqrt{3} - \sqrt{3}$. Let me compute that:

Convert to common denominator: $4/\sqrt{3} - \sqrt{3} = 4/\sqrt{3} - 3/\sqrt{3} = 1/\sqrt{3}$.

Now, $\cot(\pi/3) = \cos(\pi/3)/\sin(\pi/3) = (1/2)/(\sqrt{3}/2) = 1/\sqrt{3}$. Which matches. So that works too.

Therefore, my simplification seems correct. So the answer is $\cot x$. Therefore, the simplified form is $\cot x$.

But let me check if there's another way to approach this problem, just to confirm. Maybe using different identities?

Alternatively, starting with the original expression:

$\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}$.

Another thought: Maybe express everything in terms of sine and cosine from the start. Let me try that.

Original expression:

$\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x} = \frac{1/\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{1}{\sin x \cos x} - \frac{\sin x}{\cos x}$, which is the same as before. So same steps.

Alternatively, factor out $\frac{1}{\cos x}$? Let me see:

$\frac{1}{\cos x} \left( \frac{1}{\sin x} - \sin x \right)$. Let me check that:

Yes, because $\frac{1}{\cos x} \times \frac{1}{\sin x} = \frac{1}{\sin x \cos x}$ and $\frac{1}{\cos x} \times (-\sin x) = -\frac{\sin x}{\cos x}$. So that works.

So then inside the parentheses: $\frac{1}{\sin x} - \sin x$. Let me combine those terms. Common denominator would be $\sin x$:

$\frac{1 - \sin^2 x}{\sin x}$. Which again is $\frac{\cos^2 x}{\sin x}$. Therefore, the entire expression is $\frac{1}{\cos x} \times \frac{\cos^2 x}{\sin x} = \frac{\cos x}{\sin x} = \cot x$. Same result. So that's another way to look at it.

Therefore, regardless of the method, I end up with $\cot x$. So I think that's the correct simplified form.

Wait, but let me check if there's a possibility of another identity or if there's a different way to write it. For example, sometimes problems might expect a different form, but in this case, $\cot x$ is pretty straightforward. Let me check if the original problem had any restrictions or if there's a different form required. The problem just says "simplify," so $\cot x$ is likely the answer.

Alternatively, maybe express it in terms of tangent? Since $\cot x = 1/\tan x$, but I don't think that's simpler. So probably $\cot x$ is the simplest form.

Alternatively, maybe write it as $\frac{\cos x}{\sin x}$, but that's equivalent. Since $\cot x$ is a standard trigonometric function, it's better to write it as $\cot x$.

Therefore, I think the simplified form is $\cot x$.

**Final Answer**
\boxed{\cot x}

To simplify the expression:

$$
\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}
$$

we begin by recalling the definition of the secant function:

$$
\sec x = \frac{1}{\cos x}
$$

Substituting this into the first term:

$$
\frac{\sec x}{\sin x} = \frac{1/\cos x}{\sin x} = \frac{1}{\cos x \sin x}
$$

So the expression becomes:

$$
\frac{1}{\cos x \sin x} - \frac{\sin x}{\cos x}
$$

---

### Step 1: Find a common denominator

The first term has a denominator of $ \cos x \sin x $, and the second term has a denominator of $ \cos x $. To combine them, we rewrite the second term with the same denominator:

$$
\frac{\sin x}{\cos x} = \frac{\sin^2 x}{\cos x \sin x}
$$

Now the expression becomes:

$$
\frac{1}{\cos x \sin x} - \frac{\sin^2 x}{\cos x \sin x} = \frac{1 - \sin^2 x}{\cos x \sin x}
$$

---

### Step 2: Use a Pythagorean identity

Recall the identity:

$$
1 - \sin^2 x = \cos^2 x
$$

Substituting this in:

$$
\frac{1 - \sin^2 x}{\cos x \sin x} = \frac{\cos^2 x}{\cos x \sin x}
$$

---

### Step 3: Simplify the fraction

Cancel one $ \cos x $ from the numerator and denominator:

$$
\frac{\cos^2 x}{\cos x \sin x} = \frac{\cos x}{\sin x}
$$

---

### Step 4: Recognize the result

The expression $ \frac{\cos x}{\sin x} $ is the definition of the cotangent function:

$$
\frac{\cos x}{\sin x} = \cot x
$$

---

### Final Answer

$$
\boxed{\cot x}
$$

[Assistant]
Explanation:
Okay, so I need to simplify this expression: $\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}$. Hmm, let me think. First, I remember that secant is the reciprocal of cosine, right? So $\sec x = \frac{1}{\cos x}$. Maybe I can rewrite the first term using that. Let me try that.

So substituting $\sec x$ with $\frac{1}{\cos x}$, the first term becomes $\frac{1}{\cos x \cdot \sin x}$. Wait, is that right? Let me check. The original term is $\frac{\sec x}{\sin x}$, so replacing sec x gives $\frac{1/\cos x}{\sin x}$, which is the same as $\frac{1}{\cos x \cdot \sin x}$. Yeah, that seems correct.

So now the expression is $\frac{1}{\cos x \sin x} - \frac{\sin x}{\cos x}$. Hmm, okay. Now I need to combine these two terms. To do that, they need a common denominator. The first term has a denominator of $\cos x \sin x$, and the second term has a denominator of $\cos x$. So if I multiply the numerator and denominator of the second term by $\sin x$, that should give me the same denominator as the first term. Let me try that.

So the second term becomes $\frac{\sin x \cdot \sin x}{\cos x \cdot \sin x}$, which is $\frac{\sin^2 x}{\cos x \sin x}$. Therefore, the entire expression is $\frac{1 - \sin^2 x}{\cos x \sin x}$. Wait, because I subtract the second term, so it's $\frac{1}{\cos x \sin x} - \frac{\sin^2 x}{\cos x \sin x} = \frac{1 - \sin^2 x}{\cos x \sin x}$.

Now, I remember that $1 - \sin^2 x$ is equal to $\cos^2 x$. That's one of the Pythagorean identities. So substituting that in, the numerator becomes $\cos^2 x$. Therefore, the expression simplifies to $\frac{\cos^2 x}{\cos x \sin x}$.

Now, let me simplify that fraction. The numerator is $\cos^2 x$ and the denominator is $\cos x \sin x$. I can cancel out one $\cos x$ from the numerator and denominator. That leaves me with $\frac{\cos x}{\sin x}$.

Wait, $\frac{\cos x}{\sin x}$ is $\cot x$, right? Because cotangent is cosine over sine. So does that mean the entire expression simplifies to $\cot x$? Let me check my steps again to make sure I didn't make a mistake.

Starting with $\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}$:

1. Convert $\sec x$ to $1/\cos x$: $\frac{1}{\cos x \sin x} - \frac{\sin x}{\cos x}$.
2. Find common denominator: Multiply second term by $\sin x/\sin x$: $\frac{1 - \sin^2 x}{\cos x \sin x}$.
3. Use identity $1 - \sin^2 x = \cos^2 x$: $\frac{\cos^2 x}{\cos x \sin x}$.
4. Simplify: $\frac{\cos x}{\sin x} = \cot x$.

Seems correct. Let me verify with a specific angle to make sure. Let's take $x = 45^\circ$ (or $\pi/4$ radians). Let me compute the original expression and the simplified version.

Original expression:

$\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}$.

First, compute each part:

$\sec(\pi/4) = 1/\cos(\pi/4) = \sqrt{2}$. Then divide by $\sin(\pi/4) = \sqrt{2}/2$. So $\frac{\sqrt{2}}{\sqrt{2}/2} = 2$.

Then the second term: $\sin(\pi/4)/\cos(\pi/4) = (\sqrt{2}/2)/(\sqrt{2}/2) = 1$.

So original expression is $2 - 1 = 1$.

Now, $\cot(\pi/4) = \cos(\pi/4)/\sin(\pi/4) = 1$. So that matches. Good.

Let me try another angle, maybe $x = 60^\circ$ ($\pi/3$):

Original expression:

$\sec(\pi/3) = 1/\cos(\pi/3) = 1/(1/2) = 2$. Divide by $\sin(\pi/3) = \sqrt{3}/2$. So first term is $2 / (\sqrt{3}/2) = 4/\sqrt{3}$.

Second term: $\sin(\pi/3)/\cos(\pi/3) = (\sqrt{3}/2)/(1/2) = \sqrt{3}$.

So original expression is $4/\sqrt{3} - \sqrt{3}$. Let me compute that:

Convert to common denominator: $4/\sqrt{3} - \sqrt{3} = 4/\sqrt{3} - 3/\sqrt{3} = 1/\sqrt{3}$.

Now, $\cot(\pi/3) = \cos(\pi/3)/\sin(\pi/3) = (1/2)/(\sqrt{3}/2) = 1/\sqrt{3}$. Which matches. So that works too.

Therefore, my simplification seems correct. So the answer is $\cot x$. Therefore, the simplified form is $\cot x$.

But let me check if there's another way to approach this problem, just to confirm. Maybe using different identities?

Alternatively, starting with the original expression:

$\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}$.

Another thought: Maybe express everything in terms of sine and cosine from the start. Let me try that.

Original expression:

$\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x} = \frac{1/\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{1}{\sin x \cos x} - \frac{\sin x}{\cos x}$, which is the same as before. So same steps.

Alternatively, factor out $\frac{1}{\cos x}$? Let me see:

$\frac{1}{\cos x} \left( \frac{1}{\sin x} - \sin x \right)$. Let me check that:

Yes, because $\frac{1}{\cos x} \times \frac{1}{\sin x} = \frac{1}{\sin x \cos x}$ and $\frac{1}{\cos x} \times (-\sin x) = -\frac{\sin x}{\cos x}$. So that works.

So then inside the parentheses: $\frac{1}{\sin x} - \sin x$. Let me combine those terms. Common denominator would be $\sin x$:

$\frac{1 - \sin^2 x}{\sin x}$. Which again is $\frac{\cos^2 x}{\sin x}$. Therefore, the entire expression is $\frac{1}{\cos x} \times \frac{\cos^2 x}{\sin x} = \frac{\cos x}{\sin x} = \cot x$. Same result. So that's another way to look at it.

Therefore, regardless of the method, I end up with $\cot x$. So I think that's the correct simplified form.

Wait, but let me check if there's a possibility of another identity or if there's a different way to write it. For example, sometimes problems might expect a different form, but in this case, $\cot x$ is pretty straightforward. Let me check if the original problem had any restrictions or if there's a different form required. The problem just says "simplify," so $\cot x$ is likely the answer.

Alternatively, maybe express it in terms of tangent? Since $\cot x = 1/\tan x$, but I don't think that's simpler. So probably $\cot x$ is the simplest form.

Alternatively, maybe write it as $\frac{\cos x}{\sin x}$, but that's equivalent. Since $\cot x$ is a standard trigonometric function, it's better to write it as $\cot x$.

Therefore, I think the simplified form is $\cot x$.

**Final Answer**
\boxed{\cot x}

To simplify the expression:

$$
\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}
$$

we begin by recalling the definition of the secant function:

$$
\sec x = \frac{1}{\cos x}
$$

Substituting this into the first term:

$$
\frac{\sec x}{\sin x} = \frac{1/\cos x}{\sin x} = \frac{1}{\cos x \sin x}
$$

So the expression becomes:

$$
\frac{1}{\cos x \sin x} - \frac{\sin x}{\cos x}
$$

---

### Step 1: Find a common denominator

The first term has a denominator of $ \cos x \sin x $, and the second term has a denominator of $ \cos x $. To combine them, we rewrite the second term with the same denominator:

$$
\frac{\sin x}{\cos x} = \frac{\sin^2 x}{\cos x \sin x}
$$

Now the expression becomes:

$$
\frac{1}{\cos x \sin x} - \frac{\sin^2 x}{\cos x \sin x} = \frac{1 - \sin^2 x}{\cos x \sin x}
$$

---

### Step 2: Use a Pythagorean identity

Recall the identity:

$$
1 - \sin^2 x = \cos^2 x
$$

Substituting this in:

$$
\frac{1 - \sin^2 x}{\cos x \sin x} = \frac{\cos^2 x}{\cos x \sin x}
$$

---

### Step 3: Simplify the fraction

Cancel one $ \cos x $ from the numerator and denominator:

$$
\frac{\cos^2 x}{\cos x \sin x} = \frac{\cos x}{\sin x}
$$

---

### Step 4: Recognize the result

The expression $ \frac{\cos x}{\sin x} $ is the definition of the cotangent function:

$$
\frac{\cos x}{\sin x} = \cot x
$$

---

### Final Answer

$$
\boxed{\cot x}
$$

Solution:
We can write
\begin{align*}
\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x} &= \frac{1}{\cos x \sin x} - \frac{\sin x}{\cos x} \\
&= \frac{1 - \sin^2 x}{\cos x \sin x} \\
&= \frac{\cos^2 x}{\cos x \sin x} \\
&= \frac{\cos x}{\sin x} \\
&= \boxed{\cot x}.
\end{align*}


Explanation:
Okay, so I need to find the domain of the function log(x²) and express it as x < a or x > b, then find a + b. Hmm, let me think. First, I remember that the domain of a logarithmic function log(f(x)) is all real numbers x for which f(x) > 0. So, in this case, the function is log(x²), which means the argument of the logarithm is x squared. Therefore, I need to find all x such that x² > 0.

Wait, but hold on. Let me make sure. The logarithm is only defined when its argument is positive. So, log(x²) is defined when x² > 0. Now, x squared is always non-negative, right? Because squaring any real number gives a non-negative result. But x² is equal to zero when x is zero, and positive otherwise. Therefore, x² > 0 when x ≠ 0. So, the domain of log(x²) should be all real numbers except zero.

But the problem says the domain is x < a or x > b. That suggests that there are two intervals: one to the left of a and one to the right of b. If the domain is all real numbers except zero, then that would mean the domain is (-∞, 0) ∪ (0, ∞). Comparing that to x < a or x > b, it seems like a is 0 and b is 0? But then a + b would be 0 + 0 = 0. But that seems too straightforward. Maybe I'm missing something here?

Wait, let me check again. Maybe there's a different interpretation. Is the function log(x²) or is it (log x)²? The original problem says log x². Depending on the order of operations, log x² could be interpreted as log(x²) or (log x)². But in mathematics, typically, log x² is interpreted as log(x²), not (log x)². However, maybe the problem is written in a way that could be ambiguous? Let me check the original problem again.

The problem says: "the domain of the function log x² is x < a or x > b". So, if it's log(x²), then as I thought before, the domain is x ≠ 0. But if it's (log x)², then the domain would be all x > 0 because log x is defined only for x > 0, and squaring it would still be defined. But then the domain would be x > 0, which is not in the form x < a or x > b. So that can't be. Therefore, the original function must be log(x²).

But then, as I said, the domain is x ≠ 0, which is x < 0 or x > 0. So, comparing to x < a or x > b, that would mean a = 0 and b = 0. Therefore, a + b = 0 + 0 = 0. But maybe the problem is expecting different values? Maybe there's a mistake in my reasoning?

Wait, let me think again. Maybe the problem is written as log(x²) but with some different base? Wait, no, the base of the logarithm doesn't affect the domain, as long as it's a positive base not equal to 1. The domain is determined by the argument being positive. So regardless of the base, log(x²) is defined when x² > 0, which is x ≠ 0.

Alternatively, maybe the problem is written in a different notation where log x² is meant to be log base 10 or natural log, but that shouldn't affect the domain.

Wait, but maybe the problem is in a context where the logarithm is only defined for positive real numbers, and maybe there's some other consideration? For example, if we consider complex numbers, but no, the domain is specified as x < a or x > b, which are real numbers. So, the problem is definitely in the context of real numbers.

Therefore, I think my original conclusion is correct. The domain is all real numbers except zero, so x < 0 or x > 0. Therefore, a is 0 and b is 0, so a + b is 0. But maybe the problem is expecting different values? Let me check again.

Wait, maybe there's a misunderstanding in the problem statement. If the function was log(x)², which is (log x)², then the domain is x > 0, but that's not in the form x < a or x > b. Alternatively, if the function was log(x^2) with some other constraints?

Alternatively, maybe the problem is written as log(x^2) but with a different interpretation. For example, if the function is log(x)^2, which is [log(x)]², then the domain is x > 0. But again, that's not matching the given form.

Alternatively, maybe the problem is written in a way that the logarithm is defined only for positive arguments, but if we consider x² inside the log, then x² must be positive. Which again, is x ≠ 0.

Wait, but maybe the problem is from a source where the logarithm is defined with a different domain? For example, sometimes in some contexts, people might consider log(x²) as log(x) + log(x) if x is positive, but that's not really changing the domain.

Alternatively, maybe the problem is in a different language or notation? For example, in some countries, log might refer to base 10, but that still doesn't affect the domain.

Alternatively, maybe the problem is referring to the domain of the function log(x²) as x < a or x > b, but there's a typo or something?

Wait, let me check if there's another way to interpret log(x²). For example, if the function is log(x²) where log is a complex logarithm? But then the domain would be all complex numbers except zero, but the problem is talking about x < a or x > b, which are real numbers.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in a restricted way? For example, if the function is defined piecewise or something?

Alternatively, maybe the problem is written incorrectly? For example, if it was log(x)², but that's not the case.

Alternatively, maybe the problem is referring to the domain of the inverse function? Wait, no, the question is about the domain of log(x²).

Wait, let me check once again. If the function is log(x²), then the domain is x² > 0, which is x ≠ 0. Therefore, the domain is (-∞, 0) ∪ (0, ∞). Therefore, if we write this as x < a or x > b, then a is 0 and b is 0. Therefore, a + b = 0.

But maybe the problem is expecting a different answer? Maybe there's a mistake in my reasoning? Let me check with an example. Let me take x = 1, which is in the domain. log(1²) = log(1) = 0, which is defined. x = -1, log((-1)^2) = log(1) = 0, which is defined. x = 0, log(0²) = log(0), which is undefined. So yes, the domain is all real numbers except 0.

Therefore, the answer should be a + b = 0 + 0 = 0.

But maybe the problem is written in a different way? For example, if the function was log(x^2) with some restrictions? Wait, maybe the original problem was log(x^2) where x is in some specific context?

Alternatively, maybe the problem is from a textbook or a course where they consider the domain of log(x²) as x < -a or x > a? But that would be if the argument was x^2 - a or something else.

Wait, let me think again. Suppose the problem was log(x^2 - a), then the domain would be x^2 - a > 0, which is x < -sqrt(a) or x > sqrt(a). Then, in that case, a + b would be -sqrt(a) + sqrt(a) = 0? Wait, no. If the domain is x < -sqrt(a) or x > sqrt(a), then comparing to x < a or x > b, then a would be -sqrt(a) and b would be sqrt(a). But that seems conflicting.

But in our problem, the function is log(x²), not log(x² - something). Therefore, unless there's a typo in the problem, I think my original conclusion is correct.

Alternatively, maybe the problem is written as log(x)^2, which is [log(x)]², which would have domain x > 0, but that's not in the form x < a or x > b.

Alternatively, maybe the problem is written as log(x^2) with a different base? For example, if the base is negative? But logarithm with negative base is not defined in real numbers.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in a different way? For example, if we consider the function log(x²) as a real function, then yes, it's defined for x ≠ 0.

Alternatively, maybe the problem is in a context where x is a complex number? But then the domain would be all complex numbers except zero, which is not expressible as x < a or x > b.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of some other constraints? For example, if the function is part of a larger problem where x is restricted? But the problem doesn't mention that.

Alternatively, maybe the problem is written incorrectly, and the function is supposed to be log(x)², but that's not the case.

Alternatively, maybe the problem is referring to the domain of the function log(x²) where x is a real number, but considering that log(x²) can be written as 2 log|x|, so maybe there's some consideration with absolute value? But even so, the domain is still x ≠ 0.

Wait, but if we write log(x²) as 2 log|x|, then the domain is still x ≠ 0. So, same result.

Alternatively, maybe the problem is expecting me to consider that log(x²) is defined for x ≠ 0, but the question says "the domain of the function log x² is x < a or x > b", so maybe they are considering that the domain is x < a or x > b, which would mean that a and b are the same?

But if a and b are different, then maybe there's a different interpretation. Wait, but if the domain is x < a or x > b, then for the domain to be all real numbers except zero, we need that a = 0 and b = 0. Because if a and b are different, then the domain would be x < a or x > b, which would be two separate intervals. For example, if a = -1 and b = 1, then the domain would be x < -1 or x > 1, which is not the case here.

But in our case, the domain is x < 0 or x > 0, so a = 0 and b = 0. Therefore, a + b = 0.

But maybe the problem is expecting a different answer? Maybe I need to check if there's a different way to interpret the original function.

Wait, let me check if the function is log(x^2) or log(x)^2. If it's log(x)^2, then the domain is x > 0, which is not in the form x < a or x > b. If it's log(x^2), then domain is x ≠ 0.

Alternatively, maybe the problem is written in a way that log x² is meant to be log base 2 or something else? But again, the base doesn't affect the domain.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in a restricted way, such as in a specific interval? But the problem doesn't mention that.

Alternatively, maybe the problem is from a source where the answer is different? For example, if the function was log(x^2 - 1), then the domain would be x < -1 or x > 1, so a = -1 and b = 1, so a + b = 0. Wait, that's interesting. If the function was log(x² - 1), then the domain is x² - 1 > 0 => x² > 1 => x < -1 or x > 1, so a = -1 and b = 1, so a + b = 0.

But the problem says log x², not log(x² - 1). However, maybe there was a typo in the problem? If that's the case, then the answer would be 0. But if the problem is correct as stated, then the answer is 0.

Alternatively, maybe the problem is correct, and the answer is 0.

But maybe I need to check once again. Let me think. If the domain is x < a or x > b, then the domain is all real numbers except between a and b. But in our case, the domain is all real numbers except 0. So, if we want to write that as x < a or x > b, then the interval between a and b must be [a, b], and the domain is everything outside of that. Therefore, if the domain is x ≠ 0, then the interval that's excluded is [0, 0], which is just the single point 0. Therefore, to write the domain as x < a or x > b, we need that a = 0 and b = 0. Because if a = 0 and b = 0, then x < 0 or x > 0, which is the domain.

Alternatively, if we take a = 0 and b = 0, then the excluded interval is [0, 0], which is just the point 0. Therefore, the domain is all real numbers except 0.

Therefore, the answer is a + b = 0 + 0 = 0.

But maybe the problem is expecting a different answer? For example, if the function was log(x^2) with some other consideration?

Alternatively, maybe the problem is in a context where the logarithm is defined only for positive arguments, and the function log(x²) is considered as log(x) + log(x) for x > 0, but then again, that's not changing the domain.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of the inverse function? But no, the question is about the domain of the function itself.

Alternatively, maybe the problem is from a calculus textbook where they consider the domain of the derivative or something else? But no, the question is straightforward.

Alternatively, maybe the problem is written incorrectly, and the function is supposed to be log(x^2 - 4), which would have domain x < -2 or x > 2, so a + b = -2 + 2 = 0. But again, that's assuming a typo.

But given the problem as stated, I think the answer is 0.

Wait, but maybe the problem is written in a way that log x² is meant to be log base 10 of x squared, but with some restrictions? For example, if the problem is in a context where x is a positive real number? But then the domain would be x ≠ 0, but x is already positive?

No, if x is restricted to positive real numbers, then x² is positive, so log(x²) is defined for all x > 0. But the problem says the domain is x < a or x > b, which would not make sense if x is restricted to positive numbers.

Alternatively, maybe the problem is in a context where x is a complex number? But again, the answer would not be in the form x < a or x > b.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of the real numbers, but considering that log(x²) is an even function? But that doesn't affect the domain.

Alternatively, maybe the problem is expecting me to consider that log(x²) is defined for x ≠ 0, but the question says "the domain of the function log x² is x < a or x > b", so maybe they are considering that the domain is x < a or x > b, which is equivalent to x ≠ a or x ≠ b? No, that doesn't make sense.

Alternatively, maybe the problem is written incorrectly, and the function is log(x)², which is [log(x)]², which is defined for x > 0, but again, that's not in the form x < a or x > b.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of the range? No, the question is about the domain.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in a different way, such as considering the function's output? But no, the domain is about the input.

At this point, I think I have to conclude that the answer is 0. However, since the problem says "x < a or x > b", and if a and b are different, then maybe there's a different interpretation. Wait, but if the domain is x < a or x > b, then the domain is all real numbers except between a and b. So, if the domain is x ≠ 0, then the excluded interval is [0, 0], which is just the point 0. Therefore, to write the domain as x < a or x > b, we need that a = 0 and b = 0. Therefore, a + b = 0.

Alternatively, if the problem had said the domain is x < a or x > b, and the domain is all real numbers except between a and b, then if the domain is x ≠ 0, then the excluded interval is [0, 0], so a and b must both be 0.

Therefore, I think the answer is 0.

But maybe the problem is expecting me to think that log(x²) is defined for x > 0? But that's not true. For example, if x is negative, x² is positive, so log(x²) is defined.

Wait, let me check with specific numbers. Let me take x = -1. Then log((-1)^2) = log(1) = 0, which is defined. Similarly, x = 2, log(4) is defined. x = 0, log(0) is undefined. So yes, the domain is all real numbers except 0.

Therefore, the answer is a + b = 0 + 0 = 0.

But maybe the problem is from a source where the answer is different? For example, if the problem was log(x^2) with a different base? But no, the base doesn't affect the domain.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of the principal branch of the complex logarithm? But again, that's not expressible as x < a or x > b.

Alternatively, maybe the problem is in a context where x is a positive real number, and the function is log(x²), so the domain is x > 0, but then the problem says x < a or x > b, which would not fit.

Alternatively, maybe the problem is written incorrectly, and the function is log(x^2) with x being a complex number, but again, that's not helpful.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of the inverse function? For example, if we consider the inverse of log(x²), but that's not relevant here.

At this point, I think I have to go with my original conclusion. The domain of log(x²) is all real numbers except 0, which can be written as x < 0 or x > 0. Therefore, a = 0 and b = 0, so a + b = 0.

But just to make sure, let me check if there's any other possible interpretation. For example, if the function is log(x²) and the problem is considering the domain where the function is real and defined, which is x ≠ 0. Therefore, the answer is 0.

Alternatively, maybe the problem is expecting me to consider that log(x²) is equal to 2 log|x|, and then the domain is x ≠ 0, so again, same result.

Therefore, I think the answer is 0.

**Final Answer**
\boxed{0}

To determine the domain of the function $ \log(x^2) $, we must consider the conditions under which the logarithmic function is defined.

---

### Step 1: Understand the Domain of a Logarithmic Function

The logarithmic function $ \log(f(x)) $ is defined **only when the argument $ f(x) > 0 $**. In this case, the function is $ \log(x^2) $, so we require:

$$
x^2 > 0
$$

---

### Step 2: Solve the Inequality

We solve the inequality:

$$
x^2 > 0
$$

This is true for all real numbers **except** $ x = 0 $, because $ x^2 = 0 $ when $ x = 0 $, and the logarithm of zero is undefined.

So, the domain of $ \log(x^2) $ is:

$$
x \in (-\infty, 0) \cup (0, \infty)
$$

---

### Step 3: Express the Domain in the Given Form

The problem states that the domain is of the form:

$$
x < a \quad \text{or} \quad x > b
$$

Comparing this with the domain we found:

$$
x < 0 \quad \text{or} \quad x > 0
$$

We can match this to the form $ x < a $ or $ x > b $ by setting:

$$
a = 0, \quad b = 0
$$

---

### Step 4: Compute $ a + b $

$$
a + b = 0 + 0 = \boxed{0}
$$

---

### Final Answer

$$
\boxed{0}
$$

[Assistant]
Explanation:
Okay, so I need to find the domain of the function log(x²) and express it as x < a or x > b, then find a + b. Hmm, let me think. First, I remember that the domain of a logarithmic function log(f(x)) is all real numbers x for which f(x) > 0. So, in this case, the function is log(x²), which means the argument of the logarithm is x squared. Therefore, I need to find all x such that x² > 0.

Wait, but hold on. Let me make sure. The logarithm is only defined when its argument is positive. So, log(x²) is defined when x² > 0. Now, x squared is always non-negative, right? Because squaring any real number gives a non-negative result. But x² is equal to zero when x is zero, and positive otherwise. Therefore, x² > 0 when x ≠ 0. So, the domain of log(x²) should be all real numbers except zero.

But the problem says the domain is x < a or x > b. That suggests that there are two intervals: one to the left of a and one to the right of b. If the domain is all real numbers except zero, then that would mean the domain is (-∞, 0) ∪ (0, ∞). Comparing that to x < a or x > b, it seems like a is 0 and b is 0? But then a + b would be 0 + 0 = 0. But that seems too straightforward. Maybe I'm missing something here?

Wait, let me check again. Maybe there's a different interpretation. Is the function log(x²) or is it (log x)²? The original problem says log x². Depending on the order of operations, log x² could be interpreted as log(x²) or (log x)². But in mathematics, typically, log x² is interpreted as log(x²), not (log x)². However, maybe the problem is written in a way that could be ambiguous? Let me check the original problem again.

The problem says: "the domain of the function log x² is x < a or x > b". So, if it's log(x²), then as I thought before, the domain is x ≠ 0. But if it's (log x)², then the domain would be all x > 0 because log x is defined only for x > 0, and squaring it would still be defined. But then the domain would be x > 0, which is not in the form x < a or x > b. So that can't be. Therefore, the original function must be log(x²).

But then, as I said, the domain is x ≠ 0, which is x < 0 or x > 0. So, comparing to x < a or x > b, that would mean a = 0 and b = 0. Therefore, a + b = 0 + 0 = 0. But maybe the problem is expecting different values? Maybe there's a mistake in my reasoning?

Wait, let me think again. Maybe the problem is written as log(x²) but with some different base? Wait, no, the base of the logarithm doesn't affect the domain, as long as it's a positive base not equal to 1. The domain is determined by the argument being positive. So regardless of the base, log(x²) is defined when x² > 0, which is x ≠ 0.

Alternatively, maybe the problem is written in a different notation where log x² is meant to be log base 10 or natural log, but that shouldn't affect the domain.

Wait, but maybe the problem is in a context where the logarithm is only defined for positive real numbers, and maybe there's some other consideration? For example, if we consider complex numbers, but no, the domain is specified as x < a or x > b, which are real numbers. So, the problem is definitely in the context of real numbers.

Therefore, I think my original conclusion is correct. The domain is all real numbers except zero, so x < 0 or x > 0. Therefore, a is 0 and b is 0, so a + b is 0. But maybe the problem is expecting different values? Let me check again.

Wait, maybe there's a misunderstanding in the problem statement. If the function was log(x)², which is (log x)², then the domain is x > 0, but that's not in the form x < a or x > b. Alternatively, if the function was log(x^2) with some other constraints?

Alternatively, maybe the problem is written as log(x^2) but with a different interpretation. For example, if the function is log(x)^2, which is [log(x)]², then the domain is x > 0. But again, that's not matching the given form.

Alternatively, maybe the problem is written in a way that the logarithm is defined only for positive arguments, but if we consider x² inside the log, then x² must be positive. Which again, is x ≠ 0.

Wait, but maybe the problem is from a source where the logarithm is defined with a different domain? For example, sometimes in some contexts, people might consider log(x²) as log(x) + log(x) if x is positive, but that's not really changing the domain.

Alternatively, maybe the problem is in a different language or notation? For example, in some countries, log might refer to base 10, but that still doesn't affect the domain.

Alternatively, maybe the problem is referring to the domain of the function log(x²) as x < a or x > b, but there's a typo or something?

Wait, let me check if there's another way to interpret log(x²). For example, if the function is log(x²) where log is a complex logarithm? But then the domain would be all complex numbers except zero, but the problem is talking about x < a or x > b, which are real numbers.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in a restricted way? For example, if the function is defined piecewise or something?

Alternatively, maybe the problem is written incorrectly? For example, if it was log(x)², but that's not the case.

Alternatively, maybe the problem is referring to the domain of the inverse function? Wait, no, the question is about the domain of log(x²).

Wait, let me check once again. If the function is log(x²), then the domain is x² > 0, which is x ≠ 0. Therefore, the domain is (-∞, 0) ∪ (0, ∞). Therefore, if we write this as x < a or x > b, then a is 0 and b is 0. Therefore, a + b = 0.

But maybe the problem is expecting a different answer? Maybe there's a mistake in my reasoning? Let me check with an example. Let me take x = 1, which is in the domain. log(1²) = log(1) = 0, which is defined. x = -1, log((-1)^2) = log(1) = 0, which is defined. x = 0, log(0²) = log(0), which is undefined. So yes, the domain is all real numbers except 0.

Therefore, the answer should be a + b = 0 + 0 = 0.

But maybe the problem is written in a different way? For example, if the function was log(x^2) with some restrictions? Wait, maybe the original problem was log(x^2) where x is in some specific context?

Alternatively, maybe the problem is from a textbook or a course where they consider the domain of log(x²) as x < -a or x > a? But that would be if the argument was x^2 - a or something else.

Wait, let me think again. Suppose the problem was log(x^2 - a), then the domain would be x^2 - a > 0, which is x < -sqrt(a) or x > sqrt(a). Then, in that case, a + b would be -sqrt(a) + sqrt(a) = 0? Wait, no. If the domain is x < -sqrt(a) or x > sqrt(a), then comparing to x < a or x > b, then a would be -sqrt(a) and b would be sqrt(a). But that seems conflicting.

But in our problem, the function is log(x²), not log(x² - something). Therefore, unless there's a typo in the problem, I think my original conclusion is correct.

Alternatively, maybe the problem is written as log(x)^2, which is [log(x)]², which would have domain x > 0, but that's not in the form x < a or x > b.

Alternatively, maybe the problem is written as log(x^2) with a different base? For example, if the base is negative? But logarithm with negative base is not defined in real numbers.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in a different way? For example, if we consider the function log(x²) as a real function, then yes, it's defined for x ≠ 0.

Alternatively, maybe the problem is in a context where x is a complex number? But then the domain would be all complex numbers except zero, which is not expressible as x < a or x > b.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of some other constraints? For example, if the function is part of a larger problem where x is restricted? But the problem doesn't mention that.

Alternatively, maybe the problem is written incorrectly, and the function is supposed to be log(x)², but that's not the case.

Alternatively, maybe the problem is referring to the domain of the function log(x²) where x is a real number, but considering that log(x²) can be written as 2 log|x|, so maybe there's some consideration with absolute value? But even so, the domain is still x ≠ 0.

Wait, but if we write log(x²) as 2 log|x|, then the domain is still x ≠ 0. So, same result.

Alternatively, maybe the problem is expecting me to consider that log(x²) is defined for x ≠ 0, but the question says "the domain of the function log x² is x < a or x > b", so maybe they are considering that the domain is x < a or x > b, which would mean that a and b are the same?

But if a and b are different, then maybe there's a different interpretation. Wait, but if the domain is x < a or x > b, then for the domain to be all real numbers except zero, we need that a = 0 and b = 0. Because if a and b are different, then the domain would be x < a or x > b, which would be two separate intervals. For example, if a = -1 and b = 1, then the domain would be x < -1 or x > 1, which is not the case here.

But in our case, the domain is x < 0 or x > 0, so a = 0 and b = 0. Therefore, a + b = 0.

But maybe the problem is expecting a different answer? Maybe I need to check if there's a different way to interpret the original function.

Wait, let me check if the function is log(x^2) or log(x)^2. If it's log(x)^2, then the domain is x > 0, which is not in the form x < a or x > b. If it's log(x^2), then domain is x ≠ 0.

Alternatively, maybe the problem is written in a way that log x² is meant to be log base 2 or something else? But again, the base doesn't affect the domain.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in a restricted way, such as in a specific interval? But the problem doesn't mention that.

Alternatively, maybe the problem is from a source where the answer is different? For example, if the function was log(x^2 - 1), then the domain would be x < -1 or x > 1, so a = -1 and b = 1, so a + b = 0. Wait, that's interesting. If the function was log(x² - 1), then the domain is x² - 1 > 0 => x² > 1 => x < -1 or x > 1, so a = -1 and b = 1, so a + b = 0.

But the problem says log x², not log(x² - 1). However, maybe there was a typo in the problem? If that's the case, then the answer would be 0. But if the problem is correct as stated, then the answer is 0.

Alternatively, maybe the problem is correct, and the answer is 0.

But maybe I need to check once again. Let me think. If the domain is x < a or x > b, then the domain is all real numbers except between a and b. But in our case, the domain is all real numbers except 0. So, if we want to write that as x < a or x > b, then the interval between a and b must be [a, b], and the domain is everything outside of that. Therefore, if the domain is x ≠ 0, then the interval that's excluded is [0, 0], which is just the single point 0. Therefore, to write the domain as x < a or x > b, we need that a = 0 and b = 0. Because if a = 0 and b = 0, then x < 0 or x > 0, which is the domain.

Alternatively, if we take a = 0 and b = 0, then the excluded interval is [0, 0], which is just the point 0. Therefore, the domain is all real numbers except 0.

Therefore, the answer is a + b = 0 + 0 = 0.

But maybe the problem is expecting a different answer? For example, if the function was log(x^2) with some other consideration?

Alternatively, maybe the problem is in a context where the logarithm is defined only for positive arguments, and the function log(x²) is considered as log(x) + log(x) for x > 0, but then again, that's not changing the domain.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of the inverse function? But no, the question is about the domain of the function itself.

Alternatively, maybe the problem is from a calculus textbook where they consider the domain of the derivative or something else? But no, the question is straightforward.

Alternatively, maybe the problem is written incorrectly, and the function is supposed to be log(x^2 - 4), which would have domain x < -2 or x > 2, so a + b = -2 + 2 = 0. But again, that's assuming a typo.

But given the problem as stated, I think the answer is 0.

Wait, but maybe the problem is written in a way that log x² is meant to be log base 10 of x squared, but with some restrictions? For example, if the problem is in a context where x is a positive real number? But then the domain would be x ≠ 0, but x is already positive?

No, if x is restricted to positive real numbers, then x² is positive, so log(x²) is defined for all x > 0. But the problem says the domain is x < a or x > b, which would not make sense if x is restricted to positive numbers.

Alternatively, maybe the problem is in a context where x is a complex number? But again, the answer would not be in the form x < a or x > b.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of the real numbers, but considering that log(x²) is an even function? But that doesn't affect the domain.

Alternatively, maybe the problem is expecting me to consider that log(x²) is defined for x ≠ 0, but the question says "the domain of the function log x² is x < a or x > b", so maybe they are considering that the domain is x < a or x > b, which is equivalent to x ≠ a or x ≠ b? No, that doesn't make sense.

Alternatively, maybe the problem is written incorrectly, and the function is log(x)², which is [log(x)]², which is defined for x > 0, but again, that's not in the form x < a or x > b.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of the range? No, the question is about the domain.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in a different way, such as considering the function's output? But no, the domain is about the input.

At this point, I think I have to conclude that the answer is 0. However, since the problem says "x < a or x > b", and if a and b are different, then maybe there's a different interpretation. Wait, but if the domain is x < a or x > b, then the domain is all real numbers except between a and b. So, if the domain is x ≠ 0, then the excluded interval is [0, 0], which is just the point 0. Therefore, to write the domain as x < a or x > b, we need that a = 0 and b = 0. Therefore, a + b = 0.

Alternatively, if the problem had said the domain is x < a or x > b, and the domain is all real numbers except between a and b, then if the domain is x ≠ 0, then the excluded interval is [0, 0], so a and b must both be 0.

Therefore, I think the answer is 0.

But maybe the problem is expecting me to think that log(x²) is defined for x > 0? But that's not true. For example, if x is negative, x² is positive, so log(x²) is defined.

Wait, let me check with specific numbers. Let me take x = -1. Then log((-1)^2) = log(1) = 0, which is defined. Similarly, x = 2, log(4) is defined. x = 0, log(0) is undefined. So yes, the domain is all real numbers except 0.

Therefore, the answer is a + b = 0 + 0 = 0.

But maybe the problem is from a source where the answer is different? For example, if the problem was log(x^2) with a different base? But no, the base doesn't affect the domain.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of the principal branch of the complex logarithm? But again, that's not expressible as x < a or x > b.

Alternatively, maybe the problem is in a context where x is a positive real number, and the function is log(x²), so the domain is x > 0, but then the problem says x < a or x > b, which would not fit.

Alternatively, maybe the problem is written incorrectly, and the function is log(x^2) with x being a complex number, but again, that's not helpful.

Alternatively, maybe the problem is referring to the domain of the function log(x²) in terms of the inverse function? For example, if we consider the inverse of log(x²), but that's not relevant here.

At this point, I think I have to go with my original conclusion. The domain of log(x²) is all real numbers except 0, which can be written as x < 0 or x > 0. Therefore, a = 0 and b = 0, so a + b = 0.

But just to make sure, let me check if there's any other possible interpretation. For example, if the function is log(x²) and the problem is considering the domain where the function is real and defined, which is x ≠ 0. Therefore, the answer is 0.

Alternatively, maybe the problem is expecting me to consider that log(x²) is equal to 2 log|x|, and then the domain is x ≠ 0, so again, same result.

Therefore, I think the answer is 0.

**Final Answer**
\boxed{0}

To determine the domain of the function $ \log(x^2) $, we must consider the conditions under which the logarithmic function is defined.

---

### Step 1: Understand the Domain of a Logarithmic Function

The logarithmic function $ \log(f(x)) $ is defined **only when the argument $ f(x) > 0 $**. In this case, the function is $ \log(x^2) $, so we require:

$$
x^2 > 0
$$

---

### Step 2: Solve the Inequality

We solve the inequality:

$$
x^2 > 0
$$

This is true for all real numbers **except** $ x = 0 $, because $ x^2 = 0 $ when $ x = 0 $, and the logarithm of zero is undefined.

So, the domain of $ \log(x^2) $ is:

$$
x \in (-\infty, 0) \cup (0, \infty)
$$

---

### Step 3: Express the Domain in the Given Form

The problem states that the domain is of the form:

$$
x < a \quad \text{or} \quad x > b
$$

Comparing this with the domain we found:

$$
x < 0 \quad \text{or} \quad x > 0
$$

We can match this to the form $ x < a $ or $ x > b $ by setting:

$$
a = 0, \quad b = 0
$$

---

### Step 4: Compute $ a + b $

$$
a + b = 0 + 0 = \boxed{0}
$$

---

### Final Answer

$$
\boxed{0}
$$

Solution: In order for $\log x^2$ to be defined, we must have $x^2 > 0$. This true for all $x$, except for $x = 0$. It follows that the domain of this function is $x < 0$ or $x > 0$. Therefore, our answer is $0 + 0 = \boxed{0}$.


Explanation:
Okay, so I need to solve the equation $3^{2x} + 19 = 10^x$. Let me think about how to approach this. Hmm, first, maybe I can simplify the equation a bit. Let me see... The term $3^{2x}$ can be rewritten as $(3^2)^x$, which is $9^x$. So the equation becomes $9^x + 19 = 10^x$. That seems a bit simpler.

Now, I have an equation with two exponential terms: $9^x$ and $10^x$. I remember that equations with different bases can sometimes be tricky. Maybe I can take logarithms? But I'm not sure if that would work directly here. Let me think. If I take the natural logarithm of both sides, I would get $\ln(9^x + 19) = \ln(10^x)$. But that doesn't seem helpful because the logarithm of a sum isn't easily simplified. Maybe there's another way.

Alternatively, maybe I can rearrange the equation to get all terms on one side. Let me try that. Subtract $9^x$ from both sides: $19 = 10^x - 9^x$. Hmm, so $10^x - 9^x = 19$. Not sure if that helps yet.

I wonder if there's an integer solution. Let me test some integer values for x. Let's try x=1: $10^1 - 9^1 = 10 - 9 = 1$, which is not 19. How about x=2: $10^2 - 9^2 = 100 - 81 = 19$. Oh! That works! So x=2 is a solution. Wait, so is that the only solution? Let me check x=3: $10^3 - 9^3 = 1000 - 729 = 271$, which is way bigger than 19. What about x=0: $10^0 - 9^0 = 1 - 1 = 0$, not 19. How about x=1.5? Let me calculate that.

First, 10^1.5 is sqrt(10^3) = sqrt(1000) ≈ 31.62. 9^1.5 is sqrt(9^3) = sqrt(729) = 27. So 31.62 - 27 ≈ 4.62, which is less than 19. So between x=1.5 and x=2, the value goes from ~4.62 to 19. Wait, but at x=2, it's exactly 19. What about x=1. Let me check again. At x=1, it's 1. So maybe there's another solution? Wait, but when x approaches infinity, which one grows faster? 10^x grows faster than 9^x, so 10^x - 9^x tends to infinity. But for negative x? Let me check x=-1: 10^{-1} - 9^{-1} = 1/10 - 1/9 ≈ -0.011, which is negative. So maybe there's another solution for some negative x? But the original equation was 3^{2x} + 19 = 10^x. Let me check for x negative. Let's say x=-1: 3^{-2} + 19 = 1/9 + 19 ≈ 19.11, and 10^{-1} = 0.1. So 19.11 ≈ 0.1? No, that's not equal. So maybe there's only one solution at x=2?

But let me think again. Let me consider the function f(x) = 10^x - 9^x - 19. We found that f(2) = 0. Let me check the behavior of f(x). For x approaching negative infinity, 10^x and 9^x both approach 0, so f(x) approaches -19. At x=0, f(0) = 1 - 1 -19 = -19. At x=1, f(1)=10 -9 -19= -18. At x=2, f(2)=100 -81 -19=0. For x>2, since 10^x grows faster than 9^x, f(x) will be positive and increasing. So maybe x=2 is the only solution?

But let me check between x=1 and x=2. At x=1, f(x)=-18, at x=2, f(x)=0. So it's increasing? Let me check the derivative to see if the function is increasing. The derivative of f(x) is f’(x)= ln(10)*10^x - ln(9)*9^x. Since ln(10) > ln(9) and 10^x > 9^x for x>0, so f’(x) is positive for x>0. Therefore, f(x) is increasing for x>0. Therefore, since f(2)=0 and it's increasing, there are no solutions for x>2. For x between 0 and 2, since f(0)=-19 and f(2)=0, and it's increasing, there's only one solution at x=2. For x<0, f(x) is always negative, as we saw. Therefore, x=2 is the only solution.

But let me check if there's another solution. Suppose someone says maybe x= something else? Let me try x= log base 10 of something? Wait, but maybe not. Let me think again.

Alternatively, maybe I can write the equation as 10^x - 9^x = 19. Let me consider substituting variables. Let me set y = x. Wait, that's not helpful. Maybe set t = something. Let me think. Let me take natural logs? But again, not sure.

Alternatively, maybe write 10^x = 9^x + 19. Let me divide both sides by 9^x: (10/9)^x = 1 + 19/(9^x). Let me denote t = (10/9)^x. Then, since 19/(9^x) = 19/(9^x) = 19*(1/9)^x = 19*( (1/9)^x ). But since t = (10/9)^x, then (1/9)^x = (1/10)^x * 10^x / 9^x? Wait, maybe not helpful. Let me think again.

Let me note that (10/9)^x = t. Then, 10^x = t * 9^x. Therefore, substituting into the equation 10^x = 9^x + 19, we get t*9^x = 9^x + 19. Then, divide both sides by 9^x: t = 1 + 19/(9^x). But t = (10/9)^x. Therefore, (10/9)^x = 1 + 19/(9^x). Hmm, not sure if that helps. Maybe take logarithms?

Alternatively, let me set u = 9^x. Then, since 10^x = (10^x). But 10^x can be written as (10/9)^x * 9^x = t * u. Wait, but that's similar to before. So, if u = 9^x, then 10^x = (10/9)^x * u. But then, the equation is u + 19 = (10/9)^x * u. Which is u + 19 = t * u, where t = (10/9)^x. But t is also equal to (10/9)^x. However, since u = 9^x, then taking natural logs: ln u = x ln 9, so x = (ln u)/ln 9. Therefore, t = (10/9)^x = e^{x ln(10/9)} = e^{(ln u / ln 9) * ln(10/9)}. This seems complicated. Maybe not helpful.

Alternatively, maybe think of the equation as 10^x - 9^x = 19. Let me consider that for x=2, it works. Since the function is strictly increasing for x>0, as we saw, and since for x approaching infinity, 10^x - 9^x approaches infinity, and at x=2 it's exactly 19, then x=2 is the only solution.

But maybe there's another solution? Let me check for x between 1 and 2. Let me try x=1.5 again. As before, 10^1.5 ≈ 31.62, 9^1.5 = 27, so difference is ~4.62. At x=1.8: Let me compute 10^1.8 and 9^1.8.

First, 10^1.8 = 10^(1 + 0.8) = 10 * 10^0.8. 10^0.8: Let me recall that ln(10^0.8) = 0.8 ln 10 ≈ 0.8 * 2.3026 ≈ 1.842. So e^1.842 ≈ 6.3 (since e^1.6 is about 4.95, e^1.8 is about 6.05, e^1.842 is a bit more). Let me check with calculator steps. Alternatively, approximate 10^0.8. Since 10^0.8 = e^{0.8 ln10} ≈ e^{0.8*2.302585} ≈ e^{1.842068} ≈ 6.3. So 10^1.8 ≈ 10 * 6.3 = 63.

Similarly, 9^1.8 = e^{1.8 ln9} ≈ e^{1.8 * 2.1972} ≈ e^{3.95496} ≈ 52. So 63 - 52 ≈ 11. Which is still less than 19.

At x=1.9: 10^1.9 = 10^1 * 10^0.9 ≈ 10 * 10^0.9. 10^0.9 ≈ e^{0.9 * 2.3026} ≈ e^{2.0723} ≈ 7.94. So 10^1.9 ≈ 10 * 7.94 ≈ 79.4.

9^1.9 = e^{1.9 * ln9} ≈ e^{1.9 * 2.1972} ≈ e^{4.1747} ≈ 64.8. So difference is 79.4 - 64.8 ≈ 14.6. Still less than 19.

At x=1.95: Let me compute 10^1.95 and 9^1.95.

First, 10^1.95 = 10^(2 - 0.05) = 10^2 / 10^0.05 ≈ 100 / 1.122 ≈ 89.1.

9^1.95 = 9^(2 - 0.05) = 81 / 9^0.05. Let me compute 9^0.05. ln(9^0.05) = 0.05 * ln9 ≈ 0.05 * 2.1972 ≈ 0.10986. So e^0.10986 ≈ 1.116. Therefore, 9^1.95 ≈ 81 / 1.116 ≈ 72.6. Therefore, difference is 89.1 - 72.6 ≈ 16.5. Closer to 19, but still not there.

At x=1.98: Let me try.

10^1.98 = 10^(2 - 0.02) = 100 / 10^0.02. 10^0.02 ≈ e^{0.02 * ln10} ≈ e^{0.04605} ≈ 1.047. Therefore, 10^1.98 ≈ 100 / 1.047 ≈ 95.5.

9^1.98 = 9^(2 - 0.02) = 81 / 9^0.02. 9^0.02 = e^{0.02 * ln9} ≈ e^{0.04394} ≈ 1.045. Therefore, 9^1.98 ≈ 81 / 1.045 ≈ 77.5. So difference is 95.5 - 77.5 ≈ 18.

Almost 19. So at x=1.98, difference is ~18. Let me try x=1.99.

10^1.99 = 100 / 10^0.01 ≈ 100 / 1.023 ≈ 97.7.

9^1.99 = 81 / 9^0.01 ≈ 81 / 1.021 ≈ 79.3. Difference ≈ 97.7 - 79.3 ≈ 18.4.

Wait, but earlier at x=1.98, difference was ~18, and at x=2, it's 19. So maybe between x=1.98 and x=2, the difference increases from 18 to 19. So maybe there's another solution? Wait, but earlier I thought that the function is increasing. Wait, but according to the derivative, f’(x) = ln(10)*10^x - ln(9)*9^x. For x>0, since ln(10) > ln(9) and 10^x > 9^x, so f’(x) is positive. Therefore, the function is strictly increasing for x>0. Therefore, since at x=2, f(x)=0, and it's increasing, there can't be another solution. But according to my approximate calculations, at x=1.98, f(x)=~18, and at x=2, f(x)=19. Wait, but that would mean that f(x) increases from 18 to 19 as x goes from 1.98 to 2. But that would mean that the function crosses 19 at x=2. But if f(x) is increasing, then there's only one solution at x=2.

Wait, but earlier when I tried x=1.98, I thought the difference was ~18, but maybe my approximations were rough. Let me check more accurately.

Alternatively, maybe I can use linear approximation. Let me consider that between x=1.98 and x=2, the function increases from 18 to 19. So if I assume that the function is approximately linear in that interval, then to reach 19, which is 1 unit above 18, and since the derivative at x=2 is f’(2) = ln(10)*10^2 - ln(9)*9^2 ≈ 2.3026*100 - 2.1972*81 ≈ 230.26 - 177.97 ≈ 52.29. So the slope is about 52.29. Therefore, to increase by 1, the change in x would be approximately 1 / 52.29 ≈ 0.019. So x ≈ 1.98 + 0.019 ≈ 1.999. Which is very close to 2. Therefore, the solution is exactly at x=2.

Therefore, my initial thought that x=2 is the only solution seems correct.

But let me check if there's another solution for x < 0. Suppose x is negative. Let me try x=-1: 10^{-1} - 9^{-1} = 1/10 - 1/9 ≈ -0.011, which is negative. So f(x) = -19 + something? Wait, original equation was 10^x - 9^x = 19. For x negative, 10^x and 9^x are both positive, but since 10^x < 9^x when x is negative (because 10 < 9 when raised to negative exponents), so 10^x - 9^x is negative. Therefore, f(x) = negative number, which can't be 19. Therefore, no solution for x < 0.

Therefore, the only solution is x=2.

But let me check if there's another solution for x>2. For example, suppose x=3, as before, 10^3 - 9^3 = 1000 - 729 = 271, which is greater than 19. Since the function is increasing, it will keep increasing, so no other solutions.

Therefore, the only solution is x=2.

But let me check if there's a way to solve this algebraically, instead of just testing values. Maybe taking logarithms? Let me try again.

Original equation: 9^x + 19 = 10^x. Let me rearrange it as 10^x - 9^x = 19.

Let me consider taking logarithms, but since it's a difference, it's not straightforward. However, if I assume that x is an integer, which we saw x=2 works, but maybe there's another way.

Alternatively, maybe write 10^x = 9^x + 19. Let me divide both sides by 9^x: (10/9)^x = 1 + 19/(9^x). Let me denote y = (10/9)^x. Then, since 9^x = (9^x), but also note that (10/9)^x = y, so 9^x = 10^x / y. Wait, maybe not helpful.

Alternatively, take natural logs of both sides of the equation 10^x = 9^x + 19. But again, log(9^x + 19) = x log 10. Not helpful.

Alternatively, maybe set t = x, and think of it as a transcendental equation, which might not have an algebraic solution. Therefore, maybe the only way is to find that x=2 is the solution by testing.

Alternatively, maybe consider that 10^x - 9^x = 19. Let me think about the difference between 10^x and 9^x. For x=2, it's 19. For x=1, it's 1. For x=0, it's 0. For x= -1, it's negative. So since the function is strictly increasing, there's only one solution.

Therefore, the answer is x=2.

But just to make sure, let me plug x=2 back into the original equation.

Left side: 3^{2*2} + 19 = 3^4 + 19 = 81 + 19 = 100.

Right side: 10^2 = 100.

Yes, they are equal. Therefore, x=2 is indeed the solution.

I think that's it. So the solution is x=2.

**Final Answer**
The solution is \boxed{2}.

To solve the equation:

$$
3^{2x} + 19 = 10^x
$$

---

### Step 1: Simplify the Equation

We begin by simplifying the left-hand side using exponent rules:

$$
3^{2x} = (3^2)^x = 9^x
$$

So the equation becomes:

$$
9^x + 19 = 10^x
$$

---

### Step 2: Rearranging the Equation

Subtract $9^x$ from both sides:

$$
10^x - 9^x = 19
$$

This is a transcendental equation involving exponential functions with different bases. It is not solvable algebraically in a straightforward way, so we consider numerical or analytical methods.

---

### Step 3: Try Integer Values

Let’s test integer values of $x$ to see if we can find a solution.

- For $x = 1$:
  $$
  10^1 - 9^1 = 10 - 9 = 1 \neq 19
  $$

- For $x = 2$:
  $$
  10^2 - 9^2 = 100 - 81 = 19
  $$

✅ This works!

---

### Step 4: Analyze the Function Behavior

Define the function:

$$
f(x) = 10^x - 9^x - 19
$$

We are solving $f(x) = 0$. Let’s analyze the behavior of $f(x)$:

- As $x \to \infty$, $10^x$ grows much faster than $9^x$, so $f(x) \to \infty$
- As $x \to -\infty$, both $10^x$ and $9^x$ approach 0, so $f(x) \to -19$
- The derivative of $f(x)$ is:

$$
f'(x) = \ln(10) \cdot 10^x - \ln(9) \cdot 9^x
$$

Since $\ln(10) > \ln(9)$ and $10^x > 9^x$ for $x > 0$, the derivative is positive for $x > 0$, meaning $f(x)$ is strictly increasing for $x > 0$.

Thus, the function $f(x)$ is strictly increasing for $x > 0$, and since $f(2) = 0$, there are **no other solutions** for $x > 0$.

For $x < 0$, $f(x) < 0$, so no solution exists there either.

---

### Step 5: Verify the Solution

Plug $x = 2$ back into the original equation:

$$
3^{2 \cdot 2} + 19 = 3^4 + 19 = 81 + 19 = 100
$$
$$
10^2 = 100
$$

✅ Both sides are equal, confirming that $x = 2$ is indeed a solution.

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation $3^{2x} + 19 = 10^x$. Let me think about how to approach this. Hmm, first, maybe I can simplify the equation a bit. Let me see... The term $3^{2x}$ can be rewritten as $(3^2)^x$, which is $9^x$. So the equation becomes $9^x + 19 = 10^x$. That seems a bit simpler.

Now, I have an equation with two exponential terms: $9^x$ and $10^x$. I remember that equations with different bases can sometimes be tricky. Maybe I can take logarithms? But I'm not sure if that would work directly here. Let me think. If I take the natural logarithm of both sides, I would get $\ln(9^x + 19) = \ln(10^x)$. But that doesn't seem helpful because the logarithm of a sum isn't easily simplified. Maybe there's another way.

Alternatively, maybe I can rearrange the equation to get all terms on one side. Let me try that. Subtract $9^x$ from both sides: $19 = 10^x - 9^x$. Hmm, so $10^x - 9^x = 19$. Not sure if that helps yet.

I wonder if there's an integer solution. Let me test some integer values for x. Let's try x=1: $10^1 - 9^1 = 10 - 9 = 1$, which is not 19. How about x=2: $10^2 - 9^2 = 100 - 81 = 19$. Oh! That works! So x=2 is a solution. Wait, so is that the only solution? Let me check x=3: $10^3 - 9^3 = 1000 - 729 = 271$, which is way bigger than 19. What about x=0: $10^0 - 9^0 = 1 - 1 = 0$, not 19. How about x=1.5? Let me calculate that.

First, 10^1.5 is sqrt(10^3) = sqrt(1000) ≈ 31.62. 9^1.5 is sqrt(9^3) = sqrt(729) = 27. So 31.62 - 27 ≈ 4.62, which is less than 19. So between x=1.5 and x=2, the value goes from ~4.62 to 19. Wait, but at x=2, it's exactly 19. What about x=1. Let me check again. At x=1, it's 1. So maybe there's another solution? Wait, but when x approaches infinity, which one grows faster? 10^x grows faster than 9^x, so 10^x - 9^x tends to infinity. But for negative x? Let me check x=-1: 10^{-1} - 9^{-1} = 1/10 - 1/9 ≈ -0.011, which is negative. So maybe there's another solution for some negative x? But the original equation was 3^{2x} + 19 = 10^x. Let me check for x negative. Let's say x=-1: 3^{-2} + 19 = 1/9 + 19 ≈ 19.11, and 10^{-1} = 0.1. So 19.11 ≈ 0.1? No, that's not equal. So maybe there's only one solution at x=2?

But let me think again. Let me consider the function f(x) = 10^x - 9^x - 19. We found that f(2) = 0. Let me check the behavior of f(x). For x approaching negative infinity, 10^x and 9^x both approach 0, so f(x) approaches -19. At x=0, f(0) = 1 - 1 -19 = -19. At x=1, f(1)=10 -9 -19= -18. At x=2, f(2)=100 -81 -19=0. For x>2, since 10^x grows faster than 9^x, f(x) will be positive and increasing. So maybe x=2 is the only solution?

But let me check between x=1 and x=2. At x=1, f(x)=-18, at x=2, f(x)=0. So it's increasing? Let me check the derivative to see if the function is increasing. The derivative of f(x) is f’(x)= ln(10)*10^x - ln(9)*9^x. Since ln(10) > ln(9) and 10^x > 9^x for x>0, so f’(x) is positive for x>0. Therefore, f(x) is increasing for x>0. Therefore, since f(2)=0 and it's increasing, there are no solutions for x>2. For x between 0 and 2, since f(0)=-19 and f(2)=0, and it's increasing, there's only one solution at x=2. For x<0, f(x) is always negative, as we saw. Therefore, x=2 is the only solution.

But let me check if there's another solution. Suppose someone says maybe x= something else? Let me try x= log base 10 of something? Wait, but maybe not. Let me think again.

Alternatively, maybe I can write the equation as 10^x - 9^x = 19. Let me consider substituting variables. Let me set y = x. Wait, that's not helpful. Maybe set t = something. Let me think. Let me take natural logs? But again, not sure.

Alternatively, maybe write 10^x = 9^x + 19. Let me divide both sides by 9^x: (10/9)^x = 1 + 19/(9^x). Let me denote t = (10/9)^x. Then, since 19/(9^x) = 19/(9^x) = 19*(1/9)^x = 19*( (1/9)^x ). But since t = (10/9)^x, then (1/9)^x = (1/10)^x * 10^x / 9^x? Wait, maybe not helpful. Let me think again.

Let me note that (10/9)^x = t. Then, 10^x = t * 9^x. Therefore, substituting into the equation 10^x = 9^x + 19, we get t*9^x = 9^x + 19. Then, divide both sides by 9^x: t = 1 + 19/(9^x). But t = (10/9)^x. Therefore, (10/9)^x = 1 + 19/(9^x). Hmm, not sure if that helps. Maybe take logarithms?

Alternatively, let me set u = 9^x. Then, since 10^x = (10^x). But 10^x can be written as (10/9)^x * 9^x = t * u. Wait, but that's similar to before. So, if u = 9^x, then 10^x = (10/9)^x * u. But then, the equation is u + 19 = (10/9)^x * u. Which is u + 19 = t * u, where t = (10/9)^x. But t is also equal to (10/9)^x. However, since u = 9^x, then taking natural logs: ln u = x ln 9, so x = (ln u)/ln 9. Therefore, t = (10/9)^x = e^{x ln(10/9)} = e^{(ln u / ln 9) * ln(10/9)}. This seems complicated. Maybe not helpful.

Alternatively, maybe think of the equation as 10^x - 9^x = 19. Let me consider that for x=2, it works. Since the function is strictly increasing for x>0, as we saw, and since for x approaching infinity, 10^x - 9^x approaches infinity, and at x=2 it's exactly 19, then x=2 is the only solution.

But maybe there's another solution? Let me check for x between 1 and 2. Let me try x=1.5 again. As before, 10^1.5 ≈ 31.62, 9^1.5 = 27, so difference is ~4.62. At x=1.8: Let me compute 10^1.8 and 9^1.8.

First, 10^1.8 = 10^(1 + 0.8) = 10 * 10^0.8. 10^0.8: Let me recall that ln(10^0.8) = 0.8 ln 10 ≈ 0.8 * 2.3026 ≈ 1.842. So e^1.842 ≈ 6.3 (since e^1.6 is about 4.95, e^1.8 is about 6.05, e^1.842 is a bit more). Let me check with calculator steps. Alternatively, approximate 10^0.8. Since 10^0.8 = e^{0.8 ln10} ≈ e^{0.8*2.302585} ≈ e^{1.842068} ≈ 6.3. So 10^1.8 ≈ 10 * 6.3 = 63.

Similarly, 9^1.8 = e^{1.8 ln9} ≈ e^{1.8 * 2.1972} ≈ e^{3.95496} ≈ 52. So 63 - 52 ≈ 11. Which is still less than 19.

At x=1.9: 10^1.9 = 10^1 * 10^0.9 ≈ 10 * 10^0.9. 10^0.9 ≈ e^{0.9 * 2.3026} ≈ e^{2.0723} ≈ 7.94. So 10^1.9 ≈ 10 * 7.94 ≈ 79.4.

9^1.9 = e^{1.9 * ln9} ≈ e^{1.9 * 2.1972} ≈ e^{4.1747} ≈ 64.8. So difference is 79.4 - 64.8 ≈ 14.6. Still less than 19.

At x=1.95: Let me compute 10^1.95 and 9^1.95.

First, 10^1.95 = 10^(2 - 0.05) = 10^2 / 10^0.05 ≈ 100 / 1.122 ≈ 89.1.

9^1.95 = 9^(2 - 0.05) = 81 / 9^0.05. Let me compute 9^0.05. ln(9^0.05) = 0.05 * ln9 ≈ 0.05 * 2.1972 ≈ 0.10986. So e^0.10986 ≈ 1.116. Therefore, 9^1.95 ≈ 81 / 1.116 ≈ 72.6. Therefore, difference is 89.1 - 72.6 ≈ 16.5. Closer to 19, but still not there.

At x=1.98: Let me try.

10^1.98 = 10^(2 - 0.02) = 100 / 10^0.02. 10^0.02 ≈ e^{0.02 * ln10} ≈ e^{0.04605} ≈ 1.047. Therefore, 10^1.98 ≈ 100 / 1.047 ≈ 95.5.

9^1.98 = 9^(2 - 0.02) = 81 / 9^0.02. 9^0.02 = e^{0.02 * ln9} ≈ e^{0.04394} ≈ 1.045. Therefore, 9^1.98 ≈ 81 / 1.045 ≈ 77.5. So difference is 95.5 - 77.5 ≈ 18.

Almost 19. So at x=1.98, difference is ~18. Let me try x=1.99.

10^1.99 = 100 / 10^0.01 ≈ 100 / 1.023 ≈ 97.7.

9^1.99 = 81 / 9^0.01 ≈ 81 / 1.021 ≈ 79.3. Difference ≈ 97.7 - 79.3 ≈ 18.4.

Wait, but earlier at x=1.98, difference was ~18, and at x=2, it's 19. So maybe between x=1.98 and x=2, the difference increases from 18 to 19. So maybe there's another solution? Wait, but earlier I thought that the function is increasing. Wait, but according to the derivative, f’(x) = ln(10)*10^x - ln(9)*9^x. For x>0, since ln(10) > ln(9) and 10^x > 9^x, so f’(x) is positive. Therefore, the function is strictly increasing for x>0. Therefore, since at x=2, f(x)=0, and it's increasing, there can't be another solution. But according to my approximate calculations, at x=1.98, f(x)=~18, and at x=2, f(x)=19. Wait, but that would mean that f(x) increases from 18 to 19 as x goes from 1.98 to 2. But that would mean that the function crosses 19 at x=2. But if f(x) is increasing, then there's only one solution at x=2.

Wait, but earlier when I tried x=1.98, I thought the difference was ~18, but maybe my approximations were rough. Let me check more accurately.

Alternatively, maybe I can use linear approximation. Let me consider that between x=1.98 and x=2, the function increases from 18 to 19. So if I assume that the function is approximately linear in that interval, then to reach 19, which is 1 unit above 18, and since the derivative at x=2 is f’(2) = ln(10)*10^2 - ln(9)*9^2 ≈ 2.3026*100 - 2.1972*81 ≈ 230.26 - 177.97 ≈ 52.29. So the slope is about 52.29. Therefore, to increase by 1, the change in x would be approximately 1 / 52.29 ≈ 0.019. So x ≈ 1.98 + 0.019 ≈ 1.999. Which is very close to 2. Therefore, the solution is exactly at x=2.

Therefore, my initial thought that x=2 is the only solution seems correct.

But let me check if there's another solution for x < 0. Suppose x is negative. Let me try x=-1: 10^{-1} - 9^{-1} = 1/10 - 1/9 ≈ -0.011, which is negative. So f(x) = -19 + something? Wait, original equation was 10^x - 9^x = 19. For x negative, 10^x and 9^x are both positive, but since 10^x < 9^x when x is negative (because 10 < 9 when raised to negative exponents), so 10^x - 9^x is negative. Therefore, f(x) = negative number, which can't be 19. Therefore, no solution for x < 0.

Therefore, the only solution is x=2.

But let me check if there's another solution for x>2. For example, suppose x=3, as before, 10^3 - 9^3 = 1000 - 729 = 271, which is greater than 19. Since the function is increasing, it will keep increasing, so no other solutions.

Therefore, the only solution is x=2.

But let me check if there's a way to solve this algebraically, instead of just testing values. Maybe taking logarithms? Let me try again.

Original equation: 9^x + 19 = 10^x. Let me rearrange it as 10^x - 9^x = 19.

Let me consider taking logarithms, but since it's a difference, it's not straightforward. However, if I assume that x is an integer, which we saw x=2 works, but maybe there's another way.

Alternatively, maybe write 10^x = 9^x + 19. Let me divide both sides by 9^x: (10/9)^x = 1 + 19/(9^x). Let me denote y = (10/9)^x. Then, since 9^x = (9^x), but also note that (10/9)^x = y, so 9^x = 10^x / y. Wait, maybe not helpful.

Alternatively, take natural logs of both sides of the equation 10^x = 9^x + 19. But again, log(9^x + 19) = x log 10. Not helpful.

Alternatively, maybe set t = x, and think of it as a transcendental equation, which might not have an algebraic solution. Therefore, maybe the only way is to find that x=2 is the solution by testing.

Alternatively, maybe consider that 10^x - 9^x = 19. Let me think about the difference between 10^x and 9^x. For x=2, it's 19. For x=1, it's 1. For x=0, it's 0. For x= -1, it's negative. So since the function is strictly increasing, there's only one solution.

Therefore, the answer is x=2.

But just to make sure, let me plug x=2 back into the original equation.

Left side: 3^{2*2} + 19 = 3^4 + 19 = 81 + 19 = 100.

Right side: 10^2 = 100.

Yes, they are equal. Therefore, x=2 is indeed the solution.

I think that's it. So the solution is x=2.

**Final Answer**
The solution is \boxed{2}.

To solve the equation:

$$
3^{2x} + 19 = 10^x
$$

---

### Step 1: Simplify the Equation

We begin by simplifying the left-hand side using exponent rules:

$$
3^{2x} = (3^2)^x = 9^x
$$

So the equation becomes:

$$
9^x + 19 = 10^x
$$

---

### Step 2: Rearranging the Equation

Subtract $9^x$ from both sides:

$$
10^x - 9^x = 19
$$

This is a transcendental equation involving exponential functions with different bases. It is not solvable algebraically in a straightforward way, so we consider numerical or analytical methods.

---

### Step 3: Try Integer Values

Let’s test integer values of $x$ to see if we can find a solution.

- For $x = 1$:
  $$
  10^1 - 9^1 = 10 - 9 = 1 \neq 19
  $$

- For $x = 2$:
  $$
  10^2 - 9^2 = 100 - 81 = 19
  $$

✅ This works!

---

### Step 4: Analyze the Function Behavior

Define the function:

$$
f(x) = 10^x - 9^x - 19
$$

We are solving $f(x) = 0$. Let’s analyze the behavior of $f(x)$:

- As $x \to \infty$, $10^x$ grows much faster than $9^x$, so $f(x) \to \infty$
- As $x \to -\infty$, both $10^x$ and $9^x$ approach 0, so $f(x) \to -19$
- The derivative of $f(x)$ is:

$$
f'(x) = \ln(10) \cdot 10^x - \ln(9) \cdot 9^x
$$

Since $\ln(10) > \ln(9)$ and $10^x > 9^x$ for $x > 0$, the derivative is positive for $x > 0$, meaning $f(x)$ is strictly increasing for $x > 0$.

Thus, the function $f(x)$ is strictly increasing for $x > 0$, and since $f(2) = 0$, there are **no other solutions** for $x > 0$.

For $x < 0$, $f(x) < 0$, so no solution exists there either.

---

### Step 5: Verify the Solution

Plug $x = 2$ back into the original equation:

$$
3^{2 \cdot 2} + 19 = 3^4 + 19 = 81 + 19 = 100
$$
$$
10^2 = 100
$$

✅ Both sides are equal, confirming that $x = 2$ is indeed a solution.

---

### Final Answer

$$
\boxed{2}
$$

Solution:
Rewrite $3^{2x}$ as $(3^2)^x=9^x$, and subtract $9^x$ from both sides to obtain $19=10^x-9^x$.  There are no solutions of this equation for $x\leq 0$, because neither $10^x$ nor $9^x$ is greater than 1 if $x\leq 0$.  Trying $x=1$, $x=2$, and $x=3$, we see that $10^x-9^x$ is increasing for $x>0$, and it equals 19 when $x=\boxed{2}$.

Note: Using calculus, we could prove that $10^x-9^x$ is monotonically increasing for $x>0$, which would prove that the solution we found is unique.