Explanation:
Okay, so there's this problem about a class of 50 students. Some participate in MATHCOUNTS, some in the science club, and some in neither. The question is asking how many students participate in both. Let me try to figure this out step by step.

First, let me recall some set theory principles. I think this is a classic inclusion-exclusion problem. The formula for two sets is: the number of elements in either set A or set B is equal to the number in A plus the number in B minus the number in both. So, |A ∪ B| = |A| + |B| - |A ∩ B|.

Alright, so applying that here. Let me define the sets. Let me let M be the set of students in MATHCOUNTS, and S be the set of students in the science club. The total number of students is 50. We know that 6 students participate in neither, so that means the number of students who are in at least one of the two groups (MATHCOUNTS or science club) is 50 - 6 = 44.

So, |M ∪ S| = 44.

We also know that |M| is 28 and |S| is 21. So, plugging into the inclusion-exclusion formula:

|M ∪ S| = |M| + |S| - |M ∩ S|

So, 44 = 28 + 21 - |M ∩ S|

Let me compute 28 + 21. That's 49. So:

44 = 49 - |M ∩ S|

Hmm, so if I rearrange that equation, subtract 49 from both sides:

44 - 49 = - |M ∩ S|

Which is -5 = - |M ∩ S|

Multiply both sides by -1:

5 = |M ∩ S|

So, that would mean that 5 students are in both MATHCOUNTS and the science club. Let me check if that makes sense.

If there are 28 in MATHCOUNTS and 5 in both, then the number only in MATHCOUNTS is 28 - 5 = 23. Similarly, the number only in science club is 21 - 5 = 16. Then, adding those who are only in MATHCOUNTS, only in science club, and both, we should get the total number in at least one group. So 23 + 16 + 5 = 44. Which matches the 50 - 6 = 44. So that seems to check out.

Wait, let me verify again. Total students: 50. 6 in neither. So 44 in at least one. If 23 only in MATHCOUNTS, 16 only in science, and 5 in both, then total is 23 + 16 + 5 = 44. Yes, that's correct. So the number in both is 5.

But let me think again if I did the formula correctly. The formula says |M ∪ S| = |M| + |S| - |M ∩ S|. So solving for |M ∩ S|, it's |M| + |S| - |M ∪ S|. Which is 28 + 21 - 44 = 49 - 44 = 5. Yep, that's the same. So that seems correct.

Alternatively, maybe I can think of it as a Venn diagram. There's two circles overlapping. The total in the union is 44. The sum of the individual groups is 28 + 21 = 49. But since the union is only 44, the overlap is 49 - 44 = 5. So that's another way to think about it.

Is there another way to approach this? Maybe using a table or something? Let me try.

Total students: 50

Students in neither: 6

So students in at least one: 50 - 6 = 44

Let me make a table with rows for MATHCOUNTS and Science Club, and columns for Yes and No.

But maybe that's complicating. Alternatively, think of it as:

Number in MATHCOUNTS only: ?

Number in Science Club only: ?

Number in both: ?

Number in neither: 6

Total: 50

So, if I let x be the number in both, then:

Number in MATHCOUNTS only is 28 - x

Number in Science Club only is 21 - x

Then total in at least one is (28 - x) + (21 - x) + x = 28 - x + 21 - x + x = 28 + 21 - x = 49 - x

But we know that total in at least one is 44, so 49 - x = 44 => x = 5. Yep, same answer.

So that seems consistent. So I think 5 is the correct answer.

Wait, but let me check if there's any possibility that the numbers could be different. For example, if there was some overlap that's not considered? But no, the formula is pretty straightforward. The inclusion-exclusion principle is pretty fundamental here.

Alternatively, maybe someone might think that since 28 + 21 = 49, which is 5 more than 44, so the overlap is 5. That's exactly what we did. So that seems to make sense.

Another sanity check: if there were no overlap, the total number in at least one would be 28 + 21 = 49, but since there are only 44, that means there's an overlap of 5. So that's another way to think about it.

So, yeah, I think 5 is the right answer.

**Final Answer**
The number of students who participate in both MATHCOUNTS and science club is \boxed{5}.

To determine how many students participate in both MATHCOUNTS and the science club, we can use the **principle of inclusion and exclusion** from set theory.

---

### Step 1: Understand the Given Information

- Total number of students: **50**
- Students in **MATHCOUNTS**: **28**
- Students in **science club**: **21**
- Students in **neither**: **6**

From this, we can find the number of students who are in **at least one** of the two groups:

$$
\text{Students in at least one group} = 50 - 6 = 44
$$

---

### Step 2: Apply the Inclusion-Exclusion Principle

Let:
- $ |M| = 28 $: number of students in MATHCOUNTS
- $ |S| = 21 $: number of students in science club
- $ |M \cup S| = 44 $: number of students in at least one group
- $ |M \cap S| = x $: number of students in **both** groups (what we're solving for)

Using the inclusion-exclusion formula:

$$
|M \cup S| = |M| + |S| - |M \cap S|
$$

Substitute the known values:

$$
44 = 28 + 21 - x
$$

$$
44 = 49 - x
$$

$$
x = 49 - 44 = 5
$$

---

### Step 3: Interpret the Result

This means that **5 students** are participating in **both** MATHCOUNTS and the science club.

---

### Final Answer

$$
\boxed{5}
$$

[Assistant]
Explanation:
Okay, so there's this problem about a class of 50 students. Some participate in MATHCOUNTS, some in the science club, and some in neither. The question is asking how many students participate in both. Let me try to figure this out step by step.

First, let me recall some set theory principles. I think this is a classic inclusion-exclusion problem. The formula for two sets is: the number of elements in either set A or set B is equal to the number in A plus the number in B minus the number in both. So, |A ∪ B| = |A| + |B| - |A ∩ B|.

Alright, so applying that here. Let me define the sets. Let me let M be the set of students in MATHCOUNTS, and S be the set of students in the science club. The total number of students is 50. We know that 6 students participate in neither, so that means the number of students who are in at least one of the two groups (MATHCOUNTS or science club) is 50 - 6 = 44.

So, |M ∪ S| = 44.

We also know that |M| is 28 and |S| is 21. So, plugging into the inclusion-exclusion formula:

|M ∪ S| = |M| + |S| - |M ∩ S|

So, 44 = 28 + 21 - |M ∩ S|

Let me compute 28 + 21. That's 49. So:

44 = 49 - |M ∩ S|

Hmm, so if I rearrange that equation, subtract 49 from both sides:

44 - 49 = - |M ∩ S|

Which is -5 = - |M ∩ S|

Multiply both sides by -1:

5 = |M ∩ S|

So, that would mean that 5 students are in both MATHCOUNTS and the science club. Let me check if that makes sense.

If there are 28 in MATHCOUNTS and 5 in both, then the number only in MATHCOUNTS is 28 - 5 = 23. Similarly, the number only in science club is 21 - 5 = 16. Then, adding those who are only in MATHCOUNTS, only in science club, and both, we should get the total number in at least one group. So 23 + 16 + 5 = 44. Which matches the 50 - 6 = 44. So that seems to check out.

Wait, let me verify again. Total students: 50. 6 in neither. So 44 in at least one. If 23 only in MATHCOUNTS, 16 only in science, and 5 in both, then total is 23 + 16 + 5 = 44. Yes, that's correct. So the number in both is 5.

But let me think again if I did the formula correctly. The formula says |M ∪ S| = |M| + |S| - |M ∩ S|. So solving for |M ∩ S|, it's |M| + |S| - |M ∪ S|. Which is 28 + 21 - 44 = 49 - 44 = 5. Yep, that's the same. So that seems correct.

Alternatively, maybe I can think of it as a Venn diagram. There's two circles overlapping. The total in the union is 44. The sum of the individual groups is 28 + 21 = 49. But since the union is only 44, the overlap is 49 - 44 = 5. So that's another way to think about it.

Is there another way to approach this? Maybe using a table or something? Let me try.

Total students: 50

Students in neither: 6

So students in at least one: 50 - 6 = 44

Let me make a table with rows for MATHCOUNTS and Science Club, and columns for Yes and No.

But maybe that's complicating. Alternatively, think of it as:

Number in MATHCOUNTS only: ?

Number in Science Club only: ?

Number in both: ?

Number in neither: 6

Total: 50

So, if I let x be the number in both, then:

Number in MATHCOUNTS only is 28 - x

Number in Science Club only is 21 - x

Then total in at least one is (28 - x) + (21 - x) + x = 28 - x + 21 - x + x = 28 + 21 - x = 49 - x

But we know that total in at least one is 44, so 49 - x = 44 => x = 5. Yep, same answer.

So that seems consistent. So I think 5 is the correct answer.

Wait, but let me check if there's any possibility that the numbers could be different. For example, if there was some overlap that's not considered? But no, the formula is pretty straightforward. The inclusion-exclusion principle is pretty fundamental here.

Alternatively, maybe someone might think that since 28 + 21 = 49, which is 5 more than 44, so the overlap is 5. That's exactly what we did. So that seems to make sense.

Another sanity check: if there were no overlap, the total number in at least one would be 28 + 21 = 49, but since there are only 44, that means there's an overlap of 5. So that's another way to think about it.

So, yeah, I think 5 is the right answer.

**Final Answer**
The number of students who participate in both MATHCOUNTS and science club is \boxed{5}.

To determine how many students participate in both MATHCOUNTS and the science club, we can use the **principle of inclusion and exclusion** from set theory.

---

### Step 1: Understand the Given Information

- Total number of students: **50**
- Students in **MATHCOUNTS**: **28**
- Students in **science club**: **21**
- Students in **neither**: **6**

From this, we can find the number of students who are in **at least one** of the two groups:

$$
\text{Students in at least one group} = 50 - 6 = 44
$$

---

### Step 2: Apply the Inclusion-Exclusion Principle

Let:
- $ |M| = 28 $: number of students in MATHCOUNTS
- $ |S| = 21 $: number of students in science club
- $ |M \cup S| = 44 $: number of students in at least one group
- $ |M \cap S| = x $: number of students in **both** groups (what we're solving for)

Using the inclusion-exclusion formula:

$$
|M \cup S| = |M| + |S| - |M \cap S|
$$

Substitute the known values:

$$
44 = 28 + 21 - x
$$

$$
44 = 49 - x
$$

$$
x = 49 - 44 = 5
$$

---

### Step 3: Interpret the Result

This means that **5 students** are participating in **both** MATHCOUNTS and the science club.

---

### Final Answer

$$
\boxed{5}
$$

Solution: Among the $50-6=44$ students that participate in either MATHCOUNTS or science club, $44-28=16$ students do not participate in MATHCOUNTS.  All 16 of these students only participate in science club.  The other $21-16=\boxed{5}$ science club participants participate in MATHCOUNTS as well.


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 do 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, subtracting, and then converting back to base 9 to see if it matches.

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 as before. So that seems correct.

But just to be thorough, let me think if there's any 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 would require borrowing. But in our case, 8 - 8 is straightforward. So, no problem.

Alternatively, if the problem had been 57_9 - 18_9, then in the units place, 7 - 8 would require borrowing. But since it's 8 - 8, no problem.

Another way to check is to think about the value of each digit. In base 9, each digit represents a power of 9. So, 58_9 is 5*9 + 8, which is 45 + 8 = 53. 18_9 is 1*9 + 8 = 9 + 8 = 17. 53 - 17 = 36. Then, converting 36 to base 9: 9*4 = 36, so 4*9 + 0 = 40_9. Yep, that's consistent.

Alternatively, if I wanted to do the subtraction in base 9 without converting to base 10, let me try that again. Let me write down the subtraction:

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

Starting from the rightmost digit: 8 - 8. Since 8 is equal to 8, we can subtract directly. So, 0. Then moving to the next digit: 5 - 1. That's 4. So, the result is 40_9.

But let me think if there's a possibility that I made a mistake in the alignment. For example, if the numbers were of different lengths, but here both numbers are two digits. So, the alignment is correct.

Alternatively, if I had to subtract a larger number from a smaller one, but in this case, 58_9 is larger than 18_9. Let me check that. Since 58_9 is 5*9 + 8 = 53, and 18_9 is 1*9 + 8 = 17. 53 is larger than 17, so the result is positive.

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 rightmost digit: 0 + 8. In base 9, 0 + 8 = 8. Then moving to the next digit: 4 + 1 = 5. So, the result is 58_9. Which matches. Therefore, the subtraction is correct.

Therefore, I can be confident that 58_9 - 18_9 is 40_9.

But just to make sure I didn't overlook anything, let me try another approach. Let me convert both numbers to base 10, subtract, then convert back. As I did before, 58_9 is 53, 18_9 is 17. 53 - 17 = 36. Now, converting 36 to base 9.

To convert 36 to base 9, divide by 9. 36 divided by 9 is 4 with a remainder of 0. So, the digits are 4 and 0, hence 40_9. Yep, same result.

Alternatively, if I didn't remember how to convert to base 9, I could think of it as follows: 9^0 is 1, 9^1 is 9, 9^2 is 81. Since 36 is less than 81, the highest power is 9^1. How many times does 9 go into 36? 4 times, because 4*9=36. So, 4*9^1 + 0*9^0 = 40_9.

Therefore, all methods lead to the same answer.

I think that's thorough enough. I don't see any mistakes in my reasoning. The key steps were aligning the numbers properly, performing the subtraction digit by digit, checking for borrowing (which wasn't needed here), and verifying by converting to base 10. All checks confirm that the answer is 40_9.

So, I think the final answer is 40 in base 9.

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

To compute $ 58_9 - 18_9 $, we will perform the subtraction directly in base 9, and then verify the result by converting the numbers to base 10 and back.

---

### Step 1: Align the Numbers Vertically

We align the numbers by place value:

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

---

### Step 2: Subtract Digit by Digit (Right to Left)

- **Units place**: $ 8_9 - 8_9 = 0_9 $
- **Nines place**: $ 5_9 - 1_9 = 4_9 $

So, the result is:

```
   4 0
```

Thus, the subtraction in base 9 gives us $ 40_9 $.

---

### Step 3: Verification by Converting to Base 10

Let’s verify the result by converting both numbers to base 10:

- $ 58_9 = 5 \times 9^1 + 8 \times 9^0 = 45 + 8 = 53 $
- $ 18_9 = 1 \times 9^1 + 8 \times 9^0 = 9 + 8 = 17 $

Now subtract in base 10:

$$
53 - 17 = 36
$$

Convert $ 36 $ back to base 9:

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

This confirms our earlier result.

---

### Final Answer

$$
\boxed{40}
$$

[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 do 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, subtracting, and then converting back to base 9 to see if it matches.

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 as before. So that seems correct.

But just to be thorough, let me think if there's any 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 would require borrowing. But in our case, 8 - 8 is straightforward. So, no problem.

Alternatively, if the problem had been 57_9 - 18_9, then in the units place, 7 - 8 would require borrowing. But since it's 8 - 8, no problem.

Another way to check is to think about the value of each digit. In base 9, each digit represents a power of 9. So, 58_9 is 5*9 + 8, which is 45 + 8 = 53. 18_9 is 1*9 + 8 = 9 + 8 = 17. 53 - 17 = 36. Then, converting 36 to base 9: 9*4 = 36, so 4*9 + 0 = 40_9. Yep, that's consistent.

Alternatively, if I wanted to do the subtraction in base 9 without converting to base 10, let me try that again. Let me write down the subtraction:

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

Starting from the rightmost digit: 8 - 8. Since 8 is equal to 8, we can subtract directly. So, 0. Then moving to the next digit: 5 - 1. That's 4. So, the result is 40_9.

But let me think if there's a possibility that I made a mistake in the alignment. For example, if the numbers were of different lengths, but here both numbers are two digits. So, the alignment is correct.

Alternatively, if I had to subtract a larger number from a smaller one, but in this case, 58_9 is larger than 18_9. Let me check that. Since 58_9 is 5*9 + 8 = 53, and 18_9 is 1*9 + 8 = 17. 53 is larger than 17, so the result is positive.

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 rightmost digit: 0 + 8. In base 9, 0 + 8 = 8. Then moving to the next digit: 4 + 1 = 5. So, the result is 58_9. Which matches. Therefore, the subtraction is correct.

Therefore, I can be confident that 58_9 - 18_9 is 40_9.

But just to make sure I didn't overlook anything, let me try another approach. Let me convert both numbers to base 10, subtract, then convert back. As I did before, 58_9 is 53, 18_9 is 17. 53 - 17 = 36. Now, converting 36 to base 9.

To convert 36 to base 9, divide by 9. 36 divided by 9 is 4 with a remainder of 0. So, the digits are 4 and 0, hence 40_9. Yep, same result.

Alternatively, if I didn't remember how to convert to base 9, I could think of it as follows: 9^0 is 1, 9^1 is 9, 9^2 is 81. Since 36 is less than 81, the highest power is 9^1. How many times does 9 go into 36? 4 times, because 4*9=36. So, 4*9^1 + 0*9^0 = 40_9.

Therefore, all methods lead to the same answer.

I think that's thorough enough. I don't see any mistakes in my reasoning. The key steps were aligning the numbers properly, performing the subtraction digit by digit, checking for borrowing (which wasn't needed here), and verifying by converting to base 10. All checks confirm that the answer is 40_9.

So, I think the final answer is 40 in base 9.

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

To compute $ 58_9 - 18_9 $, we will perform the subtraction directly in base 9, and then verify the result by converting the numbers to base 10 and back.

---

### Step 1: Align the Numbers Vertically

We align the numbers by place value:

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

---

### Step 2: Subtract Digit by Digit (Right to Left)

- **Units place**: $ 8_9 - 8_9 = 0_9 $
- **Nines place**: $ 5_9 - 1_9 = 4_9 $

So, the result is:

```
   4 0
```

Thus, the subtraction in base 9 gives us $ 40_9 $.

---

### Step 3: Verification by Converting to Base 10

Let’s verify the result by converting both numbers to base 10:

- $ 58_9 = 5 \times 9^1 + 8 \times 9^0 = 45 + 8 = 53 $
- $ 18_9 = 1 \times 9^1 + 8 \times 9^0 = 9 + 8 = 17 $

Now subtract in base 10:

$$
53 - 17 = 36
$$

Convert $ 36 $ back to base 9:

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

This confirms our earlier result.

---

### Final Answer

$$
\boxed{40}
$$

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 when r = 3.

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

Surface area: 4π*(3)² = 4π*9 = 36π

Yes, they are equal. So that checks out.

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.

Wait, but let me think again if there's a possibility of another solution. For example, if we consider negative radii? But radius can't be negative, so r must be positive. So no other solutions.

Alternatively, maybe I can think about units? But the problem says they are numerically equal, so units are already considered. The question says "numerically equal," so we don't have to worry about units conflicting.

So yeah, seems like r = 3 is the answer.

But just to make sure, let me try plugging in another number. Let's say r = 1. Volume is (4/3)π, surface area is 4π. Not equal. r = 2: Volume is (4/3)π*(8) = 32π/3 ≈ 10.666π, surface area is 4π*4 = 16π. Not equal. r = 3: 36π and 36π. Equal. r = 4: Volume is (4/3)π*64 = 256π/3 ≈ 85.333π, surface area is 4π*16 = 64π. Not equal. So yeah, only at r = 3 they are equal.

Therefore, I think 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 when $ r = 3 $, the volume and surface area are indeed 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 values are equal, confirming the solution is correct.

---

### 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 when r = 3.

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

Surface area: 4π*(3)² = 4π*9 = 36π

Yes, they are equal. So that checks out.

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.

Wait, but let me think again if there's a possibility of another solution. For example, if we consider negative radii? But radius can't be negative, so r must be positive. So no other solutions.

Alternatively, maybe I can think about units? But the problem says they are numerically equal, so units are already considered. The question says "numerically equal," so we don't have to worry about units conflicting.

So yeah, seems like r = 3 is the answer.

But just to make sure, let me try plugging in another number. Let's say r = 1. Volume is (4/3)π, surface area is 4π. Not equal. r = 2: Volume is (4/3)π*(8) = 32π/3 ≈ 10.666π, surface area is 4π*4 = 16π. Not equal. r = 3: 36π and 36π. Equal. r = 4: Volume is (4/3)π*64 = 256π/3 ≈ 85.333π, surface area is 4π*16 = 64π. Not equal. So yeah, only at r = 3 they are equal.

Therefore, I think 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 when $ r = 3 $, the volume and surface area are indeed 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 values are equal, confirming the solution is correct.

---

### 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 the question didn't specify the form, both are correct, but $\cot x$ is more concise.

Therefore, I think the answer 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 expression

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

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

This 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 the question didn't specify the form, both are correct, but $\cot x$ is more concise.

Therefore, I think the answer 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 expression

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

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

This 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's phrasing. The problem says "the domain of the function log x² is x < a or x > b". If the domain is x < a or x > b, then that would mean that the function is defined for all x less than a and all x greater than b. If the domain is x ≠ 0, then that would be equivalent to x < 0 or x > 0. Therefore, comparing to x < a or x > b, we have a = 0 and b = 0. Therefore, a + b = 0.

But maybe the problem is written differently? For example, if the function was log(x)², which is (log x)², then the domain would be 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 in a way where the logarithm is defined only for positive arguments, but if we consider that x² is positive except at x = 0, so the domain is x ≠ 0. Therefore, the answer is 0 + 0 = 0.

But maybe the problem is from a source where the answer is different? Wait, let me think again. Maybe there's a mistake in my understanding of the function. Let me check with an example. Suppose x is positive, then x² is positive, so log(x²) is defined. If x is negative, x² is still positive, so log(x²) is defined. If x is zero, x² is zero, and log(0) is undefined. Therefore, the domain is indeed all real numbers except zero.

Therefore, 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? Let me check if there's another interpretation. For example, 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 base? No, the base doesn't affect the domain.

Alternatively, maybe the problem is in a different notation where log x² is meant to be log base 2 or something else? No, that still doesn't change the domain.

Alternatively, maybe the problem is considering the logarithm of x squared as a function that's defined for x² > 0, but maybe there's a mistake in the problem's phrasing? For example, if the function was log(x)², but that's not the case here.

Alternatively, maybe the problem is written in a way where the domain is x < a or x > b, but with a and b being different numbers? Wait, if the domain was 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 all real numbers except between -1 and 1. But that's not the case here.

Alternatively, if the problem had said that the domain is x < a or x > b, and the function was log(x^2) with some other condition? Wait, maybe the problem is in a different context where the logarithm is defined only for positive arguments, but maybe there's a restriction?

Wait, maybe the problem is written in a way where the function is log(x^2) but with x being a real number, so the domain is x ≠ 0. Therefore, the answer is a = 0 and b = 0.

Alternatively, maybe the problem is expecting me to consider that log(x²) can be written as 2 log |x|, so the domain is x ≠ 0. Therefore, same conclusion.

Alternatively, maybe the problem is from a textbook or a course where they have a different definition? For example, sometimes in some contexts, people might consider the domain of log(x²) as x > 0, but that would be incorrect. Because log(x²) is defined for all x except zero.

Alternatively, maybe the problem is written incorrectly? For example, if the function was log(x)², then the domain is x > 0, but that's not in the form x < a or x > b.

Alternatively, maybe the problem is referring to the domain of log(x^2) in a different way? For example, if the function is log(x^2) and the problem is considering the principal branch or something else?

Alternatively, maybe the problem is in complex numbers? But the question mentions x < a or x > b, which are real numbers, so it's definitely in real numbers.

Therefore, I think my conclusion is correct. The domain is x ≠ 0, so x < 0 or x > 0, so a = 0 and b = 0, so a + b = 0.

But maybe the problem is expecting me to write a and b as different numbers? Wait, if the domain was x < a or x > b, and the domain is x ≠ 0, then a and b must be such that the intervals x < a and x > b cover all real numbers except zero. So, if a = 0 and b = 0, then x < 0 or x > 0. If a and b are different, say a = 1 and b = -1, then the domain would be x < 1 or x > -1, which is all real numbers except between -1 and 1. But that's not the case here.

Alternatively, if a and b are both zero, then the domain is x < 0 or x > 0. Therefore, the answer is 0 + 0 = 0.

Alternatively, maybe the problem is written in a way where the domain is x < a or x > b, and the function is log(x^2) with some other constraints? For example, maybe the problem is in a context where x is positive? But no, the domain is x ≠ 0.

Alternatively, maybe the problem is a trick question? For example, if someone mistakenly thinks that log(x²) is only defined for x > 0, then they might say the domain is x > 0, so a and b would be... but that's not in the form x < a or x > b.

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 inverse function? No, the question is about the domain of log(x²).

Alternatively, maybe the problem is in a different language, and "log x²" is meant to be something else? For example, in some contexts, log x² could be interpreted as log(x) squared, but that's not standard.

Alternatively, maybe the problem is from a calculus textbook where they consider the domain of log(x²) as x > 0, but that's incorrect.

Wait, let me check with an example. Let me plug in x = 1: 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 zero.

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

But maybe the problem is expecting me to write a and b as different numbers? For example, if the domain was x < a or x > b, and the function was log(x^2) with some other condition? Wait, unless there's a typo in the problem.

Alternatively, maybe the original problem was log(x)², which is [log(x)]², which is defined for x > 0, but then the domain is x > 0, which is not in the form x < a or x > b.

Alternatively, maybe the problem is log(x^2) with x being a complex number? But again, the question mentions x < a or x > b, which are real numbers.

Alternatively, maybe the problem is referring to the domain of the function log(x^2) in a restricted context, such as in a specific textbook chapter where they consider only certain cases?

Alternatively, maybe the problem is written incorrectly, and the function is log(x)², but that's just speculation.

Alternatively, maybe the problem is from a source where the answer is different? For example, if the problem was log(x^2) with x being a real number, but the answer is different?

Wait, let me think again. If the domain is x < a or x > b, then the function is defined for all x less than a and all x greater than b. If the domain is x ≠ 0, then the function is defined for all x < 0 and all x > 0. Therefore, comparing to x < a or x > b, we have a = 0 and b = 0. Therefore, a + b = 0.

Therefore, unless there's a misunderstanding in the problem's phrasing, I think the answer is 0.

But maybe the problem is expecting me to consider that log(x²) is equivalent to 2 log |x|, so the domain is x ≠ 0, which is the same as before.

Alternatively, maybe the problem is in a context where the logarithm is defined only for positive real numbers, and the function log(x²) is considered as log(x^2) where x^2 is positive, so x ≠ 0.

Therefore, I think my answer is correct. The domain is x ≠ 0, so a = 0 and b = 0, so a + b = 0.

But maybe the problem is expecting me to write a and b as different numbers? For example, if the domain was x < a or x > b, and the function was log(x^2) with some other condition?

Wait, unless there's a mistake in the problem statement. For example, if the function was log(x)², then the domain is x > 0, but that's not in the form x < a or x > b. If the function was log(x^2) with x being a positive number, then the domain is x > 0, but again, not in the required form.

Alternatively, if the function was log(x^2) with x being a negative number, then the domain is x < 0, but again, not in the required form.

Alternatively, maybe the problem is referring to the range? But the question says domain.

Alternatively, maybe the problem is written incorrectly, and the function is log(x)², which is [log(x)]², and the domain is x > 0, but then the answer would be... but the question says the domain is x < a or x > b.

Alternatively, maybe the problem is referring to the domain of the inverse function?

Alternatively, maybe the problem is in a different language, and "log x²" is meant to be something else?

Alternatively, maybe the problem is from a calculus course where they consider the domain of log(x²) as x > 0, but that's incorrect.

Alternatively, maybe the problem is expecting me to consider that log(x²) is defined for x² > 0, which is x ≠ 0, but then the answer is a = 0 and b = 0.

Alternatively, maybe the problem is expecting me to write a and b as 0 and 0, so a + b = 0.

Therefore, after all this thinking, I think the answer is 0.

But to make sure, let me check with another approach. Let me consider the inequality x² > 0. Solving this inequality, we get x ≠ 0. Therefore, the domain is all real numbers except zero.

Expressed as intervals, this is (-∞, 0) ∪ (0, ∞).

Comparing this to the given form x < a or x > b, we can see that (-∞, 0) corresponds to x < a, so a = 0, and (0, ∞) corresponds to x > b, so b = 0. Therefore, a = 0 and b = 0, so a + b = 0 + 0 = 0.

Therefore, the answer is 0.

I think that's solid. I don't see any other way to interpret the problem. Unless there's a typo or a misinterpretation, but given the problem as stated, this seems correct.

**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 $. That is, the expression inside the logarithm must be **positive**.

In this case, the function is $ \log(x^2) $, so we need to find the values of $ x $ for which:

$$
x^2 > 0
$$

---

### 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 $
- $ x^2 > 0 $ for all $ x \ne 0 $

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

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

---

### Step 3: Match the Domain to 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:

- $ a = 0 $
- $ 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's phrasing. The problem says "the domain of the function log x² is x < a or x > b". If the domain is x < a or x > b, then that would mean that the function is defined for all x less than a and all x greater than b. If the domain is x ≠ 0, then that would be equivalent to x < 0 or x > 0. Therefore, comparing to x < a or x > b, we have a = 0 and b = 0. Therefore, a + b = 0.

But maybe the problem is written differently? For example, if the function was log(x)², which is (log x)², then the domain would be 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 in a way where the logarithm is defined only for positive arguments, but if we consider that x² is positive except at x = 0, so the domain is x ≠ 0. Therefore, the answer is 0 + 0 = 0.

But maybe the problem is from a source where the answer is different? Wait, let me think again. Maybe there's a mistake in my understanding of the function. Let me check with an example. Suppose x is positive, then x² is positive, so log(x²) is defined. If x is negative, x² is still positive, so log(x²) is defined. If x is zero, x² is zero, and log(0) is undefined. Therefore, the domain is indeed all real numbers except zero.

Therefore, 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? Let me check if there's another interpretation. For example, 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 base? No, the base doesn't affect the domain.

Alternatively, maybe the problem is in a different notation where log x² is meant to be log base 2 or something else? No, that still doesn't change the domain.

Alternatively, maybe the problem is considering the logarithm of x squared as a function that's defined for x² > 0, but maybe there's a mistake in the problem's phrasing? For example, if the function was log(x)², but that's not the case here.

Alternatively, maybe the problem is written in a way where the domain is x < a or x > b, but with a and b being different numbers? Wait, if the domain was 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 all real numbers except between -1 and 1. But that's not the case here.

Alternatively, if the problem had said that the domain is x < a or x > b, and the function was log(x^2) with some other condition? Wait, maybe the problem is in a different context where the logarithm is defined only for positive arguments, but maybe there's a restriction?

Wait, maybe the problem is written in a way where the function is log(x^2) but with x being a real number, so the domain is x ≠ 0. Therefore, the answer is a = 0 and b = 0.

Alternatively, maybe the problem is expecting me to consider that log(x²) can be written as 2 log |x|, so the domain is x ≠ 0. Therefore, same conclusion.

Alternatively, maybe the problem is from a textbook or a course where they have a different definition? For example, sometimes in some contexts, people might consider the domain of log(x²) as x > 0, but that would be incorrect. Because log(x²) is defined for all x except zero.

Alternatively, maybe the problem is written incorrectly? For example, if the function was log(x)², then the domain is x > 0, but that's not in the form x < a or x > b.

Alternatively, maybe the problem is referring to the domain of log(x^2) in a different way? For example, if the function is log(x^2) and the problem is considering the principal branch or something else?

Alternatively, maybe the problem is in complex numbers? But the question mentions x < a or x > b, which are real numbers, so it's definitely in real numbers.

Therefore, I think my conclusion is correct. The domain is x ≠ 0, so x < 0 or x > 0, so a = 0 and b = 0, so a + b = 0.

But maybe the problem is expecting me to write a and b as different numbers? Wait, if the domain was x < a or x > b, and the domain is x ≠ 0, then a and b must be such that the intervals x < a and x > b cover all real numbers except zero. So, if a = 0 and b = 0, then x < 0 or x > 0. If a and b are different, say a = 1 and b = -1, then the domain would be x < 1 or x > -1, which is all real numbers except between -1 and 1. But that's not the case here.

Alternatively, if a and b are both zero, then the domain is x < 0 or x > 0. Therefore, the answer is 0 + 0 = 0.

Alternatively, maybe the problem is written in a way where the domain is x < a or x > b, and the function is log(x^2) with some other constraints? For example, maybe the problem is in a context where x is positive? But no, the domain is x ≠ 0.

Alternatively, maybe the problem is a trick question? For example, if someone mistakenly thinks that log(x²) is only defined for x > 0, then they might say the domain is x > 0, so a and b would be... but that's not in the form x < a or x > b.

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 inverse function? No, the question is about the domain of log(x²).

Alternatively, maybe the problem is in a different language, and "log x²" is meant to be something else? For example, in some contexts, log x² could be interpreted as log(x) squared, but that's not standard.

Alternatively, maybe the problem is from a calculus textbook where they consider the domain of log(x²) as x > 0, but that's incorrect.

Wait, let me check with an example. Let me plug in x = 1: 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 zero.

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

But maybe the problem is expecting me to write a and b as different numbers? For example, if the domain was x < a or x > b, and the function was log(x^2) with some other condition? Wait, unless there's a typo in the problem.

Alternatively, maybe the original problem was log(x)², which is [log(x)]², which is defined for x > 0, but then the domain is x > 0, which is not in the form x < a or x > b.

Alternatively, maybe the problem is log(x^2) with x being a complex number? But again, the question mentions x < a or x > b, which are real numbers.

Alternatively, maybe the problem is referring to the domain of the function log(x^2) in a restricted context, such as in a specific textbook chapter where they consider only certain cases?

Alternatively, maybe the problem is written incorrectly, and the function is log(x)², but that's just speculation.

Alternatively, maybe the problem is from a source where the answer is different? For example, if the problem was log(x^2) with x being a real number, but the answer is different?

Wait, let me think again. If the domain is x < a or x > b, then the function is defined for all x less than a and all x greater than b. If the domain is x ≠ 0, then the function is defined for all x < 0 and all x > 0. Therefore, comparing to x < a or x > b, we have a = 0 and b = 0. Therefore, a + b = 0.

Therefore, unless there's a misunderstanding in the problem's phrasing, I think the answer is 0.

But maybe the problem is expecting me to consider that log(x²) is equivalent to 2 log |x|, so the domain is x ≠ 0, which is the same as before.

Alternatively, maybe the problem is in a context where the logarithm is defined only for positive real numbers, and the function log(x²) is considered as log(x^2) where x^2 is positive, so x ≠ 0.

Therefore, I think my answer is correct. The domain is x ≠ 0, so a = 0 and b = 0, so a + b = 0.

But maybe the problem is expecting me to write a and b as different numbers? For example, if the domain was x < a or x > b, and the function was log(x^2) with some other condition?

Wait, unless there's a mistake in the problem statement. For example, if the function was log(x)², then the domain is x > 0, but that's not in the form x < a or x > b. If the function was log(x^2) with x being a positive number, then the domain is x > 0, but again, not in the required form.

Alternatively, if the function was log(x^2) with x being a negative number, then the domain is x < 0, but again, not in the required form.

Alternatively, maybe the problem is referring to the range? But the question says domain.

Alternatively, maybe the problem is written incorrectly, and the function is log(x)², which is [log(x)]², and the domain is x > 0, but then the answer would be... but the question says the domain is x < a or x > b.

Alternatively, maybe the problem is referring to the domain of the inverse function?

Alternatively, maybe the problem is in a different language, and "log x²" is meant to be something else?

Alternatively, maybe the problem is from a calculus course where they consider the domain of log(x²) as x > 0, but that's incorrect.

Alternatively, maybe the problem is expecting me to consider that log(x²) is defined for x² > 0, which is x ≠ 0, but then the answer is a = 0 and b = 0.

Alternatively, maybe the problem is expecting me to write a and b as 0 and 0, so a + b = 0.

Therefore, after all this thinking, I think the answer is 0.

But to make sure, let me check with another approach. Let me consider the inequality x² > 0. Solving this inequality, we get x ≠ 0. Therefore, the domain is all real numbers except zero.

Expressed as intervals, this is (-∞, 0) ∪ (0, ∞).

Comparing this to the given form x < a or x > b, we can see that (-∞, 0) corresponds to x < a, so a = 0, and (0, ∞) corresponds to x > b, so b = 0. Therefore, a = 0 and b = 0, so a + b = 0 + 0 = 0.

Therefore, the answer is 0.

I think that's solid. I don't see any other way to interpret the problem. Unless there's a typo or a misinterpretation, but given the problem as stated, this seems correct.

**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 $. That is, the expression inside the logarithm must be **positive**.

In this case, the function is $ \log(x^2) $, so we need to find the values of $ x $ for which:

$$
x^2 > 0
$$

---

### 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 $
- $ x^2 > 0 $ for all $ x \ne 0 $

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

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

---

### Step 3: Match the Domain to 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:

- $ a = 0 $
- $ 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 $0 + 0 = \boxed{0}$.