Explanation:
Okay, so there's this problem about a class of 50 students. Some participate in MATHCOUNTS, some in the science club, and 6 don't participate in either. The question is asking how many students are in both MATHCOUNTS and the science club. Alright, let me think through this step by step.

First, I remember that when dealing with problems involving overlapping groups, the principle of inclusion-exclusion is usually useful. Let me recall what that is. If I have two sets, the total number of elements in either set is equal to the sum of the sizes of the sets minus the size of their intersection. So, in formula terms, |A ∪ B| = |A| + |B| - |A ∩ B|. That seems right.

So applying that to this problem, let me define the sets. Let me let M be the set of students participating in MATHCOUNTS, and S be the set of students participating in the science club. The total number of students in the class is 50. But 6 students participate in neither, so the number of students participating in at least one of the two activities is 50 - 6 = 44. So |M ∪ S| = 44.

We are given that |M| = 28 and |S| = 21. We need to find |M ∩ S|, the number of students participating in both.

Using the inclusion-exclusion principle, we have:

|M ∪ S| = |M| + |S| - |M ∩ S|

Plugging in the numbers we have:

44 = 28 + 21 - |M ∩ S|

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

44 = 49 - |M ∩ S|

To solve for |M ∩ S|, I can rearrange the equation:

|M ∩ S| = 49 - 44

Which is 5.

Wait, so does that mean 5 students are in both? Let me check my steps again to make sure I didn't make a mistake.

Total students: 50. 6 in neither, so 50 - 6 = 44 in at least one. That seems right. Then |M| is 28, |S| is 21. Adding those gives 49. But since some students are counted in both, we subtract the overlap. So the formula says that the union is 49 minus overlap. But the union is 44, so 49 - overlap = 44. Therefore, overlap is 5. That seems correct.

Alternatively, maybe I can think about it in another way. Let me try using a Venn diagram approach. Imagine two overlapping circles, one for MATHCOUNTS and one for science club. The total number of students in the union is 44. The number of students only in MATHCOUNTS would be total in MATHCOUNTS minus those in both. Similarly, only in science club would be total in science club minus those in both. Then, adding those two and the intersection should total the union. So:

Only MATHCOUNTS: 28 - x

Only Science Club: 21 - x

Both: x

So total in union is (28 - x) + (21 - x) + x = 28 + 21 - x = 49 - x

Which equals 44. Therefore, 49 - x = 44 => x = 5. Yep, same result. So that seems to confirm it.

Alternatively, maybe plugging numbers. If 5 students are in both, then the number of students only in MATHCOUNTS is 28 - 5 = 23. The number only in science club is 21 - 5 = 16. Then, the total number of students in at least one is 23 + 16 + 5 = 44. Which matches the 50 - 6 = 44. So that works out.

Wait, but let me check if there's another way someone might approach this and maybe get a different answer. For example, if someone thought that 6 students are in neither, so they subtract 6 from the total, but maybe misapplied the formula? But no, the inclusion-exclusion seems straightforward here.

Let me think if there's any chance that the numbers don't add up? For instance, if the overlap was, say, 6, then 28 + 21 - 6 = 43, which is less than 44. Wait, but 43 isn't 44. Wait, maybe my calculation is off? Wait, no. Wait, 28 +21 is 49. If overlap is 5, 49 -5=44, which is correct. If overlap was 6, then 49 -6=43, which is less than 44, so that would not match. So 5 is correct.

Alternatively, someone might think that the total number of participants is 28 + 21 = 49, but since there are 50 students, and 6 are in neither, maybe 49 - 6 = 43? Wait, no, that's a different approach. But that would be wrong. Because the total number of participants in at least one is 44, not 49. So if you have 49 in total for the two groups, but subtract those who are in both, which is the overlap. So yeah, the correct way is 28 + 21 - overlap = 44.

Alternatively, if someone thought that the 6 students in neither are part of the total, but that's already accounted for. So the total number of students in either group is 50 - 6 = 44.

Another way: imagine total students = 50. Those in MATHCOUNTS: 28. Those in science club: 21. But some are in both. So if we add 28 + 21, we are counting the students in both twice. So to get the correct total, we need to subtract the overlap once. So the total number in either is 28 + 21 - overlap. Which is equal to 44, so overlap is 5.

Therefore, I think the answer is 5.

But let me check once again with actual numbers. Suppose 5 students are in both. Then:

- In MATHCOUNTS only: 28 - 5 = 23

- In Science club only: 21 - 5 = 16

- In both: 5

- In neither: 6

Total students: 23 + 16 + 5 + 6 = 50.

Yes, that adds up. 23 + 16 is 39, plus 5 is 44, plus 6 is 50. Perfect.

Alternatively, if I thought that maybe the overlap was different. Let's say someone thought that the overlap is 4. Then:

Only MATHCOUNTS: 24

Only Science: 17

Both: 4

Total in at least one: 24 +17 +4 = 45. Then total students would be 45 +6=51, which is more than 50. So that can't be. So 4 is too low. If overlap is 5, total is 44 +6=50. Correct.

If overlap is 6:

Only MATHCOUNTS: 22

Only Science: 15

Both: 6

Total in at least one: 22 +15 +6 = 43. Then total students would be 43 +6=49, which is less than 50. So that's not possible. Therefore, overlap can't be 6. Therefore, 5 is the only number that works.

Therefore, the answer must be 5. I don't think I made any mistake here.

But just to make sure, let me think if there's a possibility that the question is trying to trick me. For example, maybe "participate in both" means something different? But no, in standard set terms, that's the intersection. Or maybe there's a misunderstanding about the numbers? For example, maybe the 28 and 21 include those in neither? But no, the problem says 28 participate in MATHCOUNTS, meaning they are part of the class. Since 6 are in neither, the rest are in at least one. So the numbers 28 and 21 are counts of students in the respective clubs, not including the ones in neither. Therefore, the inclusion-exclusion applies as before.

Alternatively, could there be a case where the numbers are not possible? For instance, if the overlap was negative? But in this case, 5 is positive.

Alternatively, maybe the problem is in some other way? For example, maybe the total number of students in MATHCOUNTS and science club is more than the total class? But 28 +21 =49, which is less than 50. So that's possible.

Alternatively, if the overlap was larger? But we saw that if the overlap was 5, then the numbers check out.

Therefore, I think the answer is 5.

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

To find the number of students who participate in **both** MATHCOUNTS and the science club, we begin by understanding the total number of students and how they are distributed among the different groups.

---

### Step 1: Understand the Total and the Uninvolved

- The class has **50 students** in total.
- **6 students** participate in **neither** MATHCOUNTS nor the science club.
- That means the number of students involved in **at least one** of the two activities is:

$$
50 - 6 = 44
$$

So, we are dealing with a total of **44 students** that are in **either** MATHCOUNTS, the science club, **or both**.

---

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

Let:
- $ M $ = number of students in **MATHCOUNTS** = 28
- $ S $ = number of students in **science club** = 21
- $ |M \cup S| $ = number of students in **at least one** club = 44
- $ |M \cap S| $ = number of students in **both** MATHCOUNTS and science club (our goal)

According to the **Inclusion-Exclusion Principle**:

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

Substituting the known values:

$$
44 = 28 + 21 - |M \cap S|
$$

$$
44 = 49 - |M \cap S|
$$

$$
|M \cap S| = 49 - 44 = 5
$$

---

### Step 3: Verification (Optional)

To verify, we can break down the distribution:

- Students only in MATHCOUNTS: $ 28 - 5 = 23 $
- Students only in science club: $ 21 - 5 = 16 $
- Students in **both**: $ 5 $
- Students in **neither**: $ 6 $

Total students:

$$
23 + 16 + 5 + 6 = 50
$$

This confirms that the counts are consistent and the logic is sound.

---

### 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 6 don't participate in either. The question is asking how many students are in both MATHCOUNTS and the science club. Alright, let me think through this step by step.

First, I remember that when dealing with problems involving overlapping groups, the principle of inclusion-exclusion is usually useful. Let me recall what that is. If I have two sets, the total number of elements in either set is equal to the sum of the sizes of the sets minus the size of their intersection. So, in formula terms, |A ∪ B| = |A| + |B| - |A ∩ B|. That seems right.

So applying that to this problem, let me define the sets. Let me let M be the set of students participating in MATHCOUNTS, and S be the set of students participating in the science club. The total number of students in the class is 50. But 6 students participate in neither, so the number of students participating in at least one of the two activities is 50 - 6 = 44. So |M ∪ S| = 44.

We are given that |M| = 28 and |S| = 21. We need to find |M ∩ S|, the number of students participating in both.

Using the inclusion-exclusion principle, we have:

|M ∪ S| = |M| + |S| - |M ∩ S|

Plugging in the numbers we have:

44 = 28 + 21 - |M ∩ S|

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

44 = 49 - |M ∩ S|

To solve for |M ∩ S|, I can rearrange the equation:

|M ∩ S| = 49 - 44

Which is 5.

Wait, so does that mean 5 students are in both? Let me check my steps again to make sure I didn't make a mistake.

Total students: 50. 6 in neither, so 50 - 6 = 44 in at least one. That seems right. Then |M| is 28, |S| is 21. Adding those gives 49. But since some students are counted in both, we subtract the overlap. So the formula says that the union is 49 minus overlap. But the union is 44, so 49 - overlap = 44. Therefore, overlap is 5. That seems correct.

Alternatively, maybe I can think about it in another way. Let me try using a Venn diagram approach. Imagine two overlapping circles, one for MATHCOUNTS and one for science club. The total number of students in the union is 44. The number of students only in MATHCOUNTS would be total in MATHCOUNTS minus those in both. Similarly, only in science club would be total in science club minus those in both. Then, adding those two and the intersection should total the union. So:

Only MATHCOUNTS: 28 - x

Only Science Club: 21 - x

Both: x

So total in union is (28 - x) + (21 - x) + x = 28 + 21 - x = 49 - x

Which equals 44. Therefore, 49 - x = 44 => x = 5. Yep, same result. So that seems to confirm it.

Alternatively, maybe plugging numbers. If 5 students are in both, then the number of students only in MATHCOUNTS is 28 - 5 = 23. The number only in science club is 21 - 5 = 16. Then, the total number of students in at least one is 23 + 16 + 5 = 44. Which matches the 50 - 6 = 44. So that works out.

Wait, but let me check if there's another way someone might approach this and maybe get a different answer. For example, if someone thought that 6 students are in neither, so they subtract 6 from the total, but maybe misapplied the formula? But no, the inclusion-exclusion seems straightforward here.

Let me think if there's any chance that the numbers don't add up? For instance, if the overlap was, say, 6, then 28 + 21 - 6 = 43, which is less than 44. Wait, but 43 isn't 44. Wait, maybe my calculation is off? Wait, no. Wait, 28 +21 is 49. If overlap is 5, 49 -5=44, which is correct. If overlap was 6, then 49 -6=43, which is less than 44, so that would not match. So 5 is correct.

Alternatively, someone might think that the total number of participants is 28 + 21 = 49, but since there are 50 students, and 6 are in neither, maybe 49 - 6 = 43? Wait, no, that's a different approach. But that would be wrong. Because the total number of participants in at least one is 44, not 49. So if you have 49 in total for the two groups, but subtract those who are in both, which is the overlap. So yeah, the correct way is 28 + 21 - overlap = 44.

Alternatively, if someone thought that the 6 students in neither are part of the total, but that's already accounted for. So the total number of students in either group is 50 - 6 = 44.

Another way: imagine total students = 50. Those in MATHCOUNTS: 28. Those in science club: 21. But some are in both. So if we add 28 + 21, we are counting the students in both twice. So to get the correct total, we need to subtract the overlap once. So the total number in either is 28 + 21 - overlap. Which is equal to 44, so overlap is 5.

Therefore, I think the answer is 5.

But let me check once again with actual numbers. Suppose 5 students are in both. Then:

- In MATHCOUNTS only: 28 - 5 = 23

- In Science club only: 21 - 5 = 16

- In both: 5

- In neither: 6

Total students: 23 + 16 + 5 + 6 = 50.

Yes, that adds up. 23 + 16 is 39, plus 5 is 44, plus 6 is 50. Perfect.

Alternatively, if I thought that maybe the overlap was different. Let's say someone thought that the overlap is 4. Then:

Only MATHCOUNTS: 24

Only Science: 17

Both: 4

Total in at least one: 24 +17 +4 = 45. Then total students would be 45 +6=51, which is more than 50. So that can't be. So 4 is too low. If overlap is 5, total is 44 +6=50. Correct.

If overlap is 6:

Only MATHCOUNTS: 22

Only Science: 15

Both: 6

Total in at least one: 22 +15 +6 = 43. Then total students would be 43 +6=49, which is less than 50. So that's not possible. Therefore, overlap can't be 6. Therefore, 5 is the only number that works.

Therefore, the answer must be 5. I don't think I made any mistake here.

But just to make sure, let me think if there's a possibility that the question is trying to trick me. For example, maybe "participate in both" means something different? But no, in standard set terms, that's the intersection. Or maybe there's a misunderstanding about the numbers? For example, maybe the 28 and 21 include those in neither? But no, the problem says 28 participate in MATHCOUNTS, meaning they are part of the class. Since 6 are in neither, the rest are in at least one. So the numbers 28 and 21 are counts of students in the respective clubs, not including the ones in neither. Therefore, the inclusion-exclusion applies as before.

Alternatively, could there be a case where the numbers are not possible? For instance, if the overlap was negative? But in this case, 5 is positive.

Alternatively, maybe the problem is in some other way? For example, maybe the total number of students in MATHCOUNTS and science club is more than the total class? But 28 +21 =49, which is less than 50. So that's possible.

Alternatively, if the overlap was larger? But we saw that if the overlap was 5, then the numbers check out.

Therefore, I think the answer is 5.

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

To find the number of students who participate in **both** MATHCOUNTS and the science club, we begin by understanding the total number of students and how they are distributed among the different groups.

---

### Step 1: Understand the Total and the Uninvolved

- The class has **50 students** in total.
- **6 students** participate in **neither** MATHCOUNTS nor the science club.
- That means the number of students involved in **at least one** of the two activities is:

$$
50 - 6 = 44
$$

So, we are dealing with a total of **44 students** that are in **either** MATHCOUNTS, the science club, **or both**.

---

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

Let:
- $ M $ = number of students in **MATHCOUNTS** = 28
- $ S $ = number of students in **science club** = 21
- $ |M \cup S| $ = number of students in **at least one** club = 44
- $ |M \cap S| $ = number of students in **both** MATHCOUNTS and science club (our goal)

According to the **Inclusion-Exclusion Principle**:

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

Substituting the known values:

$$
44 = 28 + 21 - |M \cap S|
$$

$$
44 = 49 - |M \cap S|
$$

$$
|M \cap S| = 49 - 44 = 5
$$

---

### Step 3: Verification (Optional)

To verify, we can break down the distribution:

- Students only in MATHCOUNTS: $ 28 - 5 = 23 $
- Students only in science club: $ 21 - 5 = 16 $
- Students in **both**: $ 5 $
- Students in **neither**: $ 6 $

Total students:

$$
23 + 16 + 5 + 6 = 50
$$

This confirms that the counts are consistent and the logic is sound.

---

### 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, let me try to figure out how to subtract these two numbers in base 9. The problem is 58 base 9 minus 18 base 9, and I need to express the answer in base 9. Hmm, base 9... I remember that in different bases, the digits represent powers of that base. So, for example, in base 9, the rightmost digit is 9^0 (which is 1), then the next one is 9^1 (which is 9), then 9^2 (which is 81), and so on. But let me make sure I remember how subtraction works in different bases. It's similar to decimal subtraction, but with borrowing happening in multiples of 9 instead of 10.

First, maybe I should convert both numbers to decimal (base 10) to see what they are in decimal, subtract them, and then convert back to base 9. That might be easier for me to check my work. Let me try that approach.

Starting with 58 base 9. To convert this to decimal, I need to remember that each digit represents a power of 9. The rightmost digit is the 9^0 place, so let me write it out:

58_9 = 5 * 9^1 + 8 * 9^0

Calculating that:

5 * 9 = 45

8 * 1 = 8

Adding them together: 45 + 8 = 53. So 58 base 9 is 53 in decimal.

Now, 18 base 9. Let me convert that similarly:

18_9 = 1 * 9^1 + 8 * 9^0

Which is 1 * 9 + 8 * 1 = 9 + 8 = 17. So 18 base 9 is 17 in decimal.

Therefore, subtracting them in decimal: 53 - 17 = 36. Now, I need to convert 36 back to base 9. How do I do that?

To convert a decimal number to base 9, I divide the number by 9 and find the remainders. Let's do that step by step.

First, divide 36 by 9. 9 * 4 = 36, so the quotient is 4 and the remainder is 0. Since the quotient is 4, which is less than 9, we stop here. So the base 9 representation is the quotient followed by the remainder. Wait, but how does that work? Let me recall the process.

The standard method is:

Divide the number by 9, record the remainder.

Then divide the quotient by 9, record the remainder.

Continue until the quotient is 0.

Then the base 9 number is the remainders read in reverse order.

But in this case, 36 divided by 9 is 4 with remainder 0. Then 4 divided by 9 is 0 with remainder 4. So reading the remainders from last to first, it's 4 and 0. Therefore, 36 decimal is 40 in base 9. Let me check that.

40 base 9 is 4 * 9^1 + 0 * 9^0 = 36 + 0 = 36. Correct. So that seems right. Therefore, the answer is 40 base 9.

Wait, but maybe I should try doing the subtraction directly in base 9 without converting to decimal? Just to verify. Let me try that as well, to make sure I didn't make a mistake.

So, subtracting 18_9 from 58_9. Let me write it down vertically:

  5 8
- 1 8
--------
?

Starting from the rightmost digit. The digits are:

Rightmost digit: 8 (from 58) minus 8 (from 18). That's straightforward: 8 - 8 = 0. So the rightmost digit is 0.

Now, moving to the next digit: 5 (from 58) minus 1 (from 18). That's 5 - 1 = 4. So the result is 40_9, which matches the decimal conversion result. So that seems to check out. Wait, so why did I even bother converting? Well, maybe I made a mistake in the decimal conversion?

Wait, let me check again. Let me redo the decimal conversion.

Original numbers:

58_9: 5*9 + 8 = 45 + 8 = 53. Correct.

18_9: 1*9 + 8 = 9 + 8 = 17. Correct.

Subtraction: 53 - 17 = 36. Correct.

Convert 36 to base 9: 9*4 = 36, remainder 0. So 40_9. Correct. So both methods agree. Therefore, the answer is 40_9.

But let me try another way just to be thorough. Maybe I can do the subtraction in base 9 step by step, considering possible borrowing. Wait, in this case, there's no need for borrowing because in the rightmost digit, 8 - 8 is fine, and in the next digit, 5 - 1 is also fine. So no borrowing needed. So the result is indeed 40_9.

But let me think if there's a case where borrowing would be necessary, just to make sure I understand how that works in base 9.

For example, if the problem was something like 53_9 - 18_9. Wait, but that's not our problem, but just for understanding. Let's say if we had 52_9 - 18_9. Then the rightmost digit would be 2 - 8. Since 2 is less than 8, we need to borrow from the next digit.

In base 9, borrowing would mean taking 9 from the next higher place. So if we have 52_9, which is 5*9 + 2 = 47, and 18_9 is 17, so 47 - 17 = 30, which is 3*9 + 0 = 30_9. Let me do the subtraction in base 9:

Start with 52_9 - 18_9.

Rightmost digit: 2 - 8. Can't do that. So borrow 1 from the next digit. The next digit is 5, which becomes 4, and the borrowed 1 is worth 9 in the current digit. So the current digit becomes 2 + 9 = 11 (in decimal). Then subtract 8: 11 - 8 = 3. So rightmost digit is 3.

Next digit: 4 (since we borrowed 1 earlier) - 1 = 3. So the result is 33_9. Which is correct because 33_9 is 3*9 + 3 = 30. So that works. So in that case, borrowing is needed.

But in our original problem, no borrowing is needed. So that's straightforward. Therefore, 58_9 - 18_9 is 40_9.

But just to be absolutely sure, let me verify once again with another method. Let me write out the numbers in base 9 and do the subtraction step by step.

So:

First, write down 58_9 and 18_9:

  5 8
- 1 8
--------
?

Starting from the rightmost digit:

8 (units place) minus 8 (units place). 8 - 8 = 0. So units digit is 0.

Next, the 9's place: 5 - 1 = 4. So the 9's digit is 4.

Therefore, the result is 40_9. Which is the same as before.

Alternatively, if I think of base 9 numbers as numbers in base 10 but with digits limited to 0-8, and subtraction is done similarly. So, as in base 10, but with base 9. So since both digits in the units place are 8 and 8, subtracting gives 0, then the next digits are 5 and 1, giving 4. So 40_9.

Alternatively, if I convert both numbers to base 10, subtract, and convert back, as done before, also gives 40_9. So that seems consistent.

Therefore, I think the answer is 40 base 9.

But just to make sure I didn't have a miscalculation in converting to decimal, let me check once more.

For 58_9: 5 is in the 9^1 place, which is 9, so 5*9=45. 8 is in the 9^0 place, so 8*1=8. Total 45+8=53. Correct.

For 18_9: 1*9 + 8*1 = 9 + 8 = 17. Correct.

Subtraction: 53 - 17. Let me do 53 - 10 = 43, then 43 - 7 = 36. Correct.

Convert 36 to base 9: 9*4 = 36, so 4*9 + 0 = 36. Therefore, 40_9. Yep, that's right.

Therefore, I think the answer is definitely 40 base 9.

But maybe there's another way to think about it? Let me try adding 18_9 to 40_9 and see if it gives 58_9. Because if 58_9 - 18_9 = 40_9, then adding 18_9 to 40_9 should give 58_9. Let me check that.

Adding 40_9 + 18_9:

Starting with the rightmost digit: 0 + 8 = 8.

Next digit: 4 + 1 = 5.

So the sum is 58_9, which is exactly the original number. Therefore, this confirms that the subtraction was done correctly.

So that's another verification. Therefore, I can be confident that 58_9 - 18_9 equals 40_9.

Alternatively, if I had done the subtraction in base 9 with borrowing, but in this case, no borrowing was needed. However, just for thoroughness, let me try a similar problem where borrowing is needed and see if my technique holds. Suppose we have 53_9 - 18_9. Let's do that.

First, converting to decimal:

53_9 is 5*9 + 3 = 45 + 3 = 48.

18_9 is 1*9 + 8 = 17.

Subtraction: 48 - 17 = 31.

Converting 31 to base 9: 3*9 + 4 = 31? Wait, 3*9 is 27, plus 4 is 31. So 34_9. Let me check by base 9 subtraction.

Subtracting 18_9 from 53_9:

  5 3
- 1 8
--------
?

Starting from the rightmost digit: 3 - 8. Cannot do that, so need to borrow.

The next digit is 5 in the 9's place. So borrow 1 from the 5, making it 4, and add 9 to the current digit (the units place). So 3 + 9 = 12 (in decimal). Now subtract 8: 12 - 8 = 4. So the units digit is 4.

Then the next digit is 4 (since we borrowed 1) minus 1 = 3. So the result is 34_9. Which matches the decimal conversion. Therefore, the method works. So if we had a similar case here, we could do the same. But in our original problem, no borrowing was needed.

Therefore, in the original problem, since there was no need for borrowing, the answer is straightforward.

Another check: Let me think about the place values. In base 9, each digit's place is 9^0, 9^1, etc. So the number 58_9 is 5*9 + 8, which is 53, and 18_9 is 1*9 + 8 = 17, so 53 -17 = 36. Then 36 in base 9 is 4*9 + 0, so 40_9.

Alternatively, if I think about the numbers in base 9:

58_9 is like 5*9 + 8, so if I subtract 1*9 + 8, it's 5*9 + 8 - 1*9 - 8 = (5-1)*9 + (8 -8) = 4*9 + 0 = 40_9. Which is exactly the same as above.

So this algebraic approach in base 9 also gives the same result.

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

Is there any chance I misread the problem? Let me check again. The problem says compute 58_9 minus 18_9. Yes. Both numbers are in base 9, and the answer should be in base 9.

So, I think I've verified it multiple ways: converting to decimal, subtracting, converting back, doing the subtraction directly in base 9, adding to check, and algebraic manipulation. All methods confirm the result is 40 base 9.

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

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

To compute $58_9 - 18_9$ and express the result in base 9, we can proceed with the following step-by-step approach:

---

### **1. Understanding Base 9 Subtraction**

In base 9, each digit represents a power of 9 (rightmost digit is $9^0$, next is $9^1$, etc.). When subtracting, we follow the same rules as in base 10, but with borrowing in chunks of 9.

We’ll also verify the result through decimal conversion for accuracy.

---

### **2. Convert Both Numbers to Decimal (Base 10)**

- $58_9 = 5 \times 9^1 + 8 \times 9^0 = 45 + 8 = 53_{10}$
- $18_9 = 1 \times 9^1 + 8 \times 9^0 = 9 + 8 = 17_{10}$

Now subtract:
$$
53_{10} - 17_{10} = 36_{10}
$$

---

### **3. Convert the Result Back to Base 9**

To convert $36_{10}$ to base 9:

1. Divide 36 by 9:
   $$
   36 \div 9 = 4 \text{ remainder } 0
   $$

2. The quotient is 4, which is already less than 9, so no further steps are needed.

So, the base 9 representation of 36 is:
$$
40_9
$$

---

### **4. Direct Base 9 Subtraction (Optional Verification)**

Perform the subtraction directly in base 9:

$$
\begin{array}{c@{}c@{}c}
 & 5 & 8 \\
- & 1 & 8 \\
\hline
 & 4 & 0 \\
\end{array}
$$

- Rightmost digit: $8 - 8 = 0$
- Next digit: $5 - 1 = 4$

So, the result is again $40_9$

---

### **5. Verify by Addition (Reverse Check)**

Add $40_9$ and $18_9$ to see if we get back the original number $58_9$:

$$
40_9 + 18_9 = (4 \times 9^1 + 0 \times 9^0) + (1 \times 9^1 + 8 \times 9^0) = 4 \times 9 + 1 \times 9 + 8 = 5 \times 9 + 8 = 58_9
$$

The reverse check confirms our result.

---

### **Final Answer**

$$
\boxed{40}
$$

[Assistant]
Explanation:
Okay, let me try to figure out how to subtract these two numbers in base 9. The problem is 58 base 9 minus 18 base 9, and I need to express the answer in base 9. Hmm, base 9... I remember that in different bases, the digits represent powers of that base. So, for example, in base 9, the rightmost digit is 9^0 (which is 1), then the next one is 9^1 (which is 9), then 9^2 (which is 81), and so on. But let me make sure I remember how subtraction works in different bases. It's similar to decimal subtraction, but with borrowing happening in multiples of 9 instead of 10.

First, maybe I should convert both numbers to decimal (base 10) to see what they are in decimal, subtract them, and then convert back to base 9. That might be easier for me to check my work. Let me try that approach.

Starting with 58 base 9. To convert this to decimal, I need to remember that each digit represents a power of 9. The rightmost digit is the 9^0 place, so let me write it out:

58_9 = 5 * 9^1 + 8 * 9^0

Calculating that:

5 * 9 = 45

8 * 1 = 8

Adding them together: 45 + 8 = 53. So 58 base 9 is 53 in decimal.

Now, 18 base 9. Let me convert that similarly:

18_9 = 1 * 9^1 + 8 * 9^0

Which is 1 * 9 + 8 * 1 = 9 + 8 = 17. So 18 base 9 is 17 in decimal.

Therefore, subtracting them in decimal: 53 - 17 = 36. Now, I need to convert 36 back to base 9. How do I do that?

To convert a decimal number to base 9, I divide the number by 9 and find the remainders. Let's do that step by step.

First, divide 36 by 9. 9 * 4 = 36, so the quotient is 4 and the remainder is 0. Since the quotient is 4, which is less than 9, we stop here. So the base 9 representation is the quotient followed by the remainder. Wait, but how does that work? Let me recall the process.

The standard method is:

Divide the number by 9, record the remainder.

Then divide the quotient by 9, record the remainder.

Continue until the quotient is 0.

Then the base 9 number is the remainders read in reverse order.

But in this case, 36 divided by 9 is 4 with remainder 0. Then 4 divided by 9 is 0 with remainder 4. So reading the remainders from last to first, it's 4 and 0. Therefore, 36 decimal is 40 in base 9. Let me check that.

40 base 9 is 4 * 9^1 + 0 * 9^0 = 36 + 0 = 36. Correct. So that seems right. Therefore, the answer is 40 base 9.

Wait, but maybe I should try doing the subtraction directly in base 9 without converting to decimal? Just to verify. Let me try that as well, to make sure I didn't make a mistake.

So, subtracting 18_9 from 58_9. Let me write it down vertically:

  5 8
- 1 8
--------
?

Starting from the rightmost digit. The digits are:

Rightmost digit: 8 (from 58) minus 8 (from 18). That's straightforward: 8 - 8 = 0. So the rightmost digit is 0.

Now, moving to the next digit: 5 (from 58) minus 1 (from 18). That's 5 - 1 = 4. So the result is 40_9, which matches the decimal conversion result. So that seems to check out. Wait, so why did I even bother converting? Well, maybe I made a mistake in the decimal conversion?

Wait, let me check again. Let me redo the decimal conversion.

Original numbers:

58_9: 5*9 + 8 = 45 + 8 = 53. Correct.

18_9: 1*9 + 8 = 9 + 8 = 17. Correct.

Subtraction: 53 - 17 = 36. Correct.

Convert 36 to base 9: 9*4 = 36, remainder 0. So 40_9. Correct. So both methods agree. Therefore, the answer is 40_9.

But let me try another way just to be thorough. Maybe I can do the subtraction in base 9 step by step, considering possible borrowing. Wait, in this case, there's no need for borrowing because in the rightmost digit, 8 - 8 is fine, and in the next digit, 5 - 1 is also fine. So no borrowing needed. So the result is indeed 40_9.

But let me think if there's a case where borrowing would be necessary, just to make sure I understand how that works in base 9.

For example, if the problem was something like 53_9 - 18_9. Wait, but that's not our problem, but just for understanding. Let's say if we had 52_9 - 18_9. Then the rightmost digit would be 2 - 8. Since 2 is less than 8, we need to borrow from the next digit.

In base 9, borrowing would mean taking 9 from the next higher place. So if we have 52_9, which is 5*9 + 2 = 47, and 18_9 is 17, so 47 - 17 = 30, which is 3*9 + 0 = 30_9. Let me do the subtraction in base 9:

Start with 52_9 - 18_9.

Rightmost digit: 2 - 8. Can't do that. So borrow 1 from the next digit. The next digit is 5, which becomes 4, and the borrowed 1 is worth 9 in the current digit. So the current digit becomes 2 + 9 = 11 (in decimal). Then subtract 8: 11 - 8 = 3. So rightmost digit is 3.

Next digit: 4 (since we borrowed 1 earlier) - 1 = 3. So the result is 33_9. Which is correct because 33_9 is 3*9 + 3 = 30. So that works. So in that case, borrowing is needed.

But in our original problem, no borrowing is needed. So that's straightforward. Therefore, 58_9 - 18_9 is 40_9.

But just to be absolutely sure, let me verify once again with another method. Let me write out the numbers in base 9 and do the subtraction step by step.

So:

First, write down 58_9 and 18_9:

  5 8
- 1 8
--------
?

Starting from the rightmost digit:

8 (units place) minus 8 (units place). 8 - 8 = 0. So units digit is 0.

Next, the 9's place: 5 - 1 = 4. So the 9's digit is 4.

Therefore, the result is 40_9. Which is the same as before.

Alternatively, if I think of base 9 numbers as numbers in base 10 but with digits limited to 0-8, and subtraction is done similarly. So, as in base 10, but with base 9. So since both digits in the units place are 8 and 8, subtracting gives 0, then the next digits are 5 and 1, giving 4. So 40_9.

Alternatively, if I convert both numbers to base 10, subtract, and convert back, as done before, also gives 40_9. So that seems consistent.

Therefore, I think the answer is 40 base 9.

But just to make sure I didn't have a miscalculation in converting to decimal, let me check once more.

For 58_9: 5 is in the 9^1 place, which is 9, so 5*9=45. 8 is in the 9^0 place, so 8*1=8. Total 45+8=53. Correct.

For 18_9: 1*9 + 8*1 = 9 + 8 = 17. Correct.

Subtraction: 53 - 17. Let me do 53 - 10 = 43, then 43 - 7 = 36. Correct.

Convert 36 to base 9: 9*4 = 36, so 4*9 + 0 = 36. Therefore, 40_9. Yep, that's right.

Therefore, I think the answer is definitely 40 base 9.

But maybe there's another way to think about it? Let me try adding 18_9 to 40_9 and see if it gives 58_9. Because if 58_9 - 18_9 = 40_9, then adding 18_9 to 40_9 should give 58_9. Let me check that.

Adding 40_9 + 18_9:

Starting with the rightmost digit: 0 + 8 = 8.

Next digit: 4 + 1 = 5.

So the sum is 58_9, which is exactly the original number. Therefore, this confirms that the subtraction was done correctly.

So that's another verification. Therefore, I can be confident that 58_9 - 18_9 equals 40_9.

Alternatively, if I had done the subtraction in base 9 with borrowing, but in this case, no borrowing was needed. However, just for thoroughness, let me try a similar problem where borrowing is needed and see if my technique holds. Suppose we have 53_9 - 18_9. Let's do that.

First, converting to decimal:

53_9 is 5*9 + 3 = 45 + 3 = 48.

18_9 is 1*9 + 8 = 17.

Subtraction: 48 - 17 = 31.

Converting 31 to base 9: 3*9 + 4 = 31? Wait, 3*9 is 27, plus 4 is 31. So 34_9. Let me check by base 9 subtraction.

Subtracting 18_9 from 53_9:

  5 3
- 1 8
--------
?

Starting from the rightmost digit: 3 - 8. Cannot do that, so need to borrow.

The next digit is 5 in the 9's place. So borrow 1 from the 5, making it 4, and add 9 to the current digit (the units place). So 3 + 9 = 12 (in decimal). Now subtract 8: 12 - 8 = 4. So the units digit is 4.

Then the next digit is 4 (since we borrowed 1) minus 1 = 3. So the result is 34_9. Which matches the decimal conversion. Therefore, the method works. So if we had a similar case here, we could do the same. But in our original problem, no borrowing was needed.

Therefore, in the original problem, since there was no need for borrowing, the answer is straightforward.

Another check: Let me think about the place values. In base 9, each digit's place is 9^0, 9^1, etc. So the number 58_9 is 5*9 + 8, which is 53, and 18_9 is 1*9 + 8 = 17, so 53 -17 = 36. Then 36 in base 9 is 4*9 + 0, so 40_9.

Alternatively, if I think about the numbers in base 9:

58_9 is like 5*9 + 8, so if I subtract 1*9 + 8, it's 5*9 + 8 - 1*9 - 8 = (5-1)*9 + (8 -8) = 4*9 + 0 = 40_9. Which is exactly the same as above.

So this algebraic approach in base 9 also gives the same result.

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

Is there any chance I misread the problem? Let me check again. The problem says compute 58_9 minus 18_9. Yes. Both numbers are in base 9, and the answer should be in base 9.

So, I think I've verified it multiple ways: converting to decimal, subtracting, converting back, doing the subtraction directly in base 9, adding to check, and algebraic manipulation. All methods confirm the result is 40 base 9.

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

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

To compute $58_9 - 18_9$ and express the result in base 9, we can proceed with the following step-by-step approach:

---

### **1. Understanding Base 9 Subtraction**

In base 9, each digit represents a power of 9 (rightmost digit is $9^0$, next is $9^1$, etc.). When subtracting, we follow the same rules as in base 10, but with borrowing in chunks of 9.

We’ll also verify the result through decimal conversion for accuracy.

---

### **2. Convert Both Numbers to Decimal (Base 10)**

- $58_9 = 5 \times 9^1 + 8 \times 9^0 = 45 + 8 = 53_{10}$
- $18_9 = 1 \times 9^1 + 8 \times 9^0 = 9 + 8 = 17_{10}$

Now subtract:
$$
53_{10} - 17_{10} = 36_{10}
$$

---

### **3. Convert the Result Back to Base 9**

To convert $36_{10}$ to base 9:

1. Divide 36 by 9:
   $$
   36 \div 9 = 4 \text{ remainder } 0
   $$

2. The quotient is 4, which is already less than 9, so no further steps are needed.

So, the base 9 representation of 36 is:
$$
40_9
$$

---

### **4. Direct Base 9 Subtraction (Optional Verification)**

Perform the subtraction directly in base 9:

$$
\begin{array}{c@{}c@{}c}
 & 5 & 8 \\
- & 1 & 8 \\
\hline
 & 4 & 0 \\
\end{array}
$$

- Rightmost digit: $8 - 8 = 0$
- Next digit: $5 - 1 = 4$

So, the result is again $40_9$

---

### **5. Verify by Addition (Reverse Check)**

Add $40_9$ and $18_9$ to see if we get back the original number $58_9$:

$$
40_9 + 18_9 = (4 \times 9^1 + 0 \times 9^0) + (1 \times 9^1 + 8 \times 9^0) = 4 \times 9 + 1 \times 9 + 8 = 5 \times 9 + 8 = 58_9
$$

The reverse check confirms our 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, 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 gives:

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

Simplify the left side: The 4π 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

Multiply both sides by 3:

r = 3

Wait, so is the radius 3 units? Let me check that. Let me compute the volume and surface area with r=3 and see if they are equal.

Volume: (4/3)π*(3)³ = (4/3)π*27 = 36π

Surface area: 4π*(3)² = 4π*9 = 36π

Yes! They are equal. So the radius is 3 units. That seems straightforward. But let me make sure I didn't skip any steps or make any mistakes in my algebra.

Starting again: (4/3)πr³ = 4πr². Let me subtract 4πr² from both sides to set the equation to zero:

(4/3)πr³ - 4πr² = 0

Factor out common terms. Let's factor out 4πr²:

4πr²( (1/3)r - 1 ) = 0

So, the equation is 4πr²( (1/3)r - 1 ) = 0.

Now, since 4πr² is always non-negative (and for a sphere, r can't be zero because that would make the volume and surface area zero, which is trivial but probably not considered here), so the only solution is when (1/3)r - 1 = 0. Solving that:

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

Same answer. So that's correct.

But let me check if there could be another solution. For example, if we consider r = 0, but then both volume and surface area are zero. But the problem says "a sphere", so radius can't be zero. So r = 3 is the only valid solution.

Alternatively, maybe there's another approach. Let me try solving the original equation again:

(4/3)πr³ = 4πr²

Divide both sides by π: (4/3)r³ = 4r²

Divide both sides by 4r² (assuming r ≠ 0): (1/3)r = 1

Multiply both sides by 3: r = 3. Yep, same result.

So I think that's solid. The radius is 3 units. Let me check again with numbers. If r = 3:

Volume = (4/3)π*(27) = 36π
Surface area = 4π*(9) = 36π
Yes, equal.

Is there any other possible radius? Let me think. Suppose, hypothetically, that there is another value of r where (4/3)πr³ = 4πr². But solving the equation leads uniquely to r = 3. Because if you rearrange:

(4/3)r³ = 4r²
Divide both sides by r² (r ≠ 0): (4/3)r = 4
Multiply both sides by 3: 4r = 12
Divide by 4: r = 3.

Same answer. So no other solutions. Therefore, the radius is definitely 3.

Wait, but just to think about units. The problem says the volume and surface area are numerically equal. So, they have the same number, but different units? Wait, the problem says "numerically equal", so it's just the numerical values, ignoring the units. So, for example, if volume was 36π cubic units and surface area was 36π square units, then numerically they are both 36π, so they are equal.

But in the problem statement, they are saying "numerically equal", so the numbers are the same, even though the units are different. So yes, that's correct. So, as long as the numerical value is the same, regardless of units, which is what we did.

Therefore, I think my answer is correct. The radius is 3 units.

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

To solve the problem, we are asked to find the radius of a sphere where the **numerical values** of its **volume** and **surface area** are equal. This means we will **not** consider units in the equality, but rather treat the expressions as purely numerical values.

---

### Step 1: Recall the Formulas

The volume $ V $ and surface area $ S $ of a sphere are given by:

$$
V = \frac{4}{3}\pi r^3 \quad \text{and} \quad S = 4\pi r^2
$$

We are told these are **numerically equal**:

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

---

### Step 2: Solve the Equation

Start by simplifying the equation:

Divide both sides by $ 4\pi r^2 $ (note: $ r \neq 0 $) to cancel terms:

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

Simplify the left-hand side:

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

Solve for $ r $:

$$
r = 3
$$

---

### Step 3: Verify the Solution

Let’s plug $ r = 3 $ back into both formulas:

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

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

Both expressions yield $ 36\pi $, confirming the numerical values are indeed equal.

---

### Final Answer

$$
\boxed{3}
$$

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

[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, 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 gives:

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

Simplify the left side: The 4π 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

Multiply both sides by 3:

r = 3

Wait, so is the radius 3 units? Let me check that. Let me compute the volume and surface area with r=3 and see if they are equal.

Volume: (4/3)π*(3)³ = (4/3)π*27 = 36π

Surface area: 4π*(3)² = 4π*9 = 36π

Yes! They are equal. So the radius is 3 units. That seems straightforward. But let me make sure I didn't skip any steps or make any mistakes in my algebra.

Starting again: (4/3)πr³ = 4πr². Let me subtract 4πr² from both sides to set the equation to zero:

(4/3)πr³ - 4πr² = 0

Factor out common terms. Let's factor out 4πr²:

4πr²( (1/3)r - 1 ) = 0

So, the equation is 4πr²( (1/3)r - 1 ) = 0.

Now, since 4πr² is always non-negative (and for a sphere, r can't be zero because that would make the volume and surface area zero, which is trivial but probably not considered here), so the only solution is when (1/3)r - 1 = 0. Solving that:

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

Same answer. So that's correct.

But let me check if there could be another solution. For example, if we consider r = 0, but then both volume and surface area are zero. But the problem says "a sphere", so radius can't be zero. So r = 3 is the only valid solution.

Alternatively, maybe there's another approach. Let me try solving the original equation again:

(4/3)πr³ = 4πr²

Divide both sides by π: (4/3)r³ = 4r²

Divide both sides by 4r² (assuming r ≠ 0): (1/3)r = 1

Multiply both sides by 3: r = 3. Yep, same result.

So I think that's solid. The radius is 3 units. Let me check again with numbers. If r = 3:

Volume = (4/3)π*(27) = 36π
Surface area = 4π*(9) = 36π
Yes, equal.

Is there any other possible radius? Let me think. Suppose, hypothetically, that there is another value of r where (4/3)πr³ = 4πr². But solving the equation leads uniquely to r = 3. Because if you rearrange:

(4/3)r³ = 4r²
Divide both sides by r² (r ≠ 0): (4/3)r = 4
Multiply both sides by 3: 4r = 12
Divide by 4: r = 3.

Same answer. So no other solutions. Therefore, the radius is definitely 3.

Wait, but just to think about units. The problem says the volume and surface area are numerically equal. So, they have the same number, but different units? Wait, the problem says "numerically equal", so it's just the numerical values, ignoring the units. So, for example, if volume was 36π cubic units and surface area was 36π square units, then numerically they are both 36π, so they are equal.

But in the problem statement, they are saying "numerically equal", so the numbers are the same, even though the units are different. So yes, that's correct. So, as long as the numerical value is the same, regardless of units, which is what we did.

Therefore, I think my answer is correct. The radius is 3 units.

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

To solve the problem, we are asked to find the radius of a sphere where the **numerical values** of its **volume** and **surface area** are equal. This means we will **not** consider units in the equality, but rather treat the expressions as purely numerical values.

---

### Step 1: Recall the Formulas

The volume $ V $ and surface area $ S $ of a sphere are given by:

$$
V = \frac{4}{3}\pi r^3 \quad \text{and} \quad S = 4\pi r^2
$$

We are told these are **numerically equal**:

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

---

### Step 2: Solve the Equation

Start by simplifying the equation:

Divide both sides by $ 4\pi r^2 $ (note: $ r \neq 0 $) to cancel terms:

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

Simplify the left-hand side:

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

Solve for $ r $:

$$
r = 3
$$

---

### Step 3: Verify the Solution

Let’s plug $ r = 3 $ back into both formulas:

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

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

Both expressions yield $ 36\pi $, confirming the numerical values are indeed equal.

---

### Final Answer

$$
\boxed{3}
$$

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

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 the expression $\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}$. Hmm, let me think about how to approach this. First, I remember that secant is the reciprocal of cosine, so maybe I can rewrite sec x in terms of cosine. Let me try that.

Starting with the first term: $\frac{\sec x}{\sin x}$. Since $\sec x = \frac{1}{\cos x}$, substituting that in gives me $\frac{1}{\cos x \cdot \sin x}$. Wait, is that right? Let me check. Yes, because dividing by sin x would be multiplying by 1/sin x, so $\frac{1}{\cos x} \times \frac{1}{\sin x} = \frac{1}{\cos x \sin x}$. Okay, so the first term simplifies to $\frac{1}{\sin x \cos x}$.

Now the second term is $\frac{\sin x}{\cos x}$, which I know is tangent x. But maybe I should keep it as $\frac{\sin x}{\cos x}$ for now to see if there's a common denominator or something. So the expression becomes $\frac{1}{\sin x \cos x} - \frac{\sin x}{\cos x}$.

Hmm, to combine these two terms, they need a common denominator. The first term has $\sin x \cos x$ in the denominator, and the second term has $\cos x$. So if I multiply the numerator and denominator of the second term by sin x, that would make the denominators match. Let me try that.

So, rewrite the second term: $\frac{\sin x}{\cos x} = \frac{\sin x \cdot \sin x}{\cos x \cdot \sin x} = \frac{\sin^2 x}{\sin x \cos x}$. Okay, so now both terms have the same denominator. That means I can subtract them:

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

Wait, that's combining the numerators. Now, I remember that $1 - \sin^2 x$ is a Pythagorean identity. Let me recall: $\sin^2 x + \cos^2 x = 1$, so $1 - \sin^2 x = \cos^2 x$. Yes! So substituting that in:

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

Now, simplifying the fraction. The numerator is $\cos^2 x$ and the denominator is $\sin x \cos x$. I can cancel out one $\cos x$ from numerator and denominator:

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

Which is, of course, $\cot x$. So the simplified form of the original expression is $\cot x$. Let me just check my steps again to make sure I didn't make a mistake.

Starting with $\sec x / \sin x - \sin x / \cos x$. Convert sec x to 1/cos x, so first term is 1/(sin x cos x). Second term is sin x / cos x. Then common denominator is sin x cos x, so second term becomes sin²x / (sin x cos x). Subtracting gives (1 - sin²x)/ (sin x cos x) which is cos²x / (sin x cos x) simplifies to cos x / sin x = cot x. Seems correct.

Alternatively, maybe there's another way to approach it? Let me try another method for verification.

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.

So, $\frac{\sec x}{\sin x}$ is $\frac{1/\cos x}{\sin x}$ which is $\frac{1}{\sin x \cos x}$, same as before. Then subtract $\frac{\sin x}{\cos x}$. So same steps. So yes, same result.

Alternatively, maybe factor something out? Let me see. Not sure. Alternatively, think about combining the terms differently.

Alternatively, let's think about the original expression:

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

Let me write sec x as 1/cos x:

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

Alternatively, factor out 1/cos x:

$$
\frac{1}{\cos x} \left( \frac{1}{\sin x} - \sin x \right )
$$

Yes, that's another way. Let me try that. So factoring out 1/cos x:

$$
\frac{1}{\cos x} \left( \frac{1 - \sin^2 x}{\sin x} \right )
$$

Because $\frac{1}{\sin x} - \sin x$ is equal to $\frac{1 - \sin^2 x}{\sin x}$. Then again, $1 - \sin^2 x = \cos^2 x$, so substituting that:

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

Same result. So that's a good check. So regardless of the method, it seems that the answer is cotangent x. Let me check with specific values to make sure.

Let me pick x = 45 degrees (π/4 radians). Let's compute original expression and cot x.

First, compute original expression:

sec(45°)/sin(45°) - sin(45°)/cos(45°).

We know that sin(45°) = cos(45°) = √2/2 ≈ 0.7071.

sec(45°) is 1/cos(45°) = √2 ≈ 1.4142.

So first term: sec(45°)/sin(45°) = √2 / (√2/2) = √2 * 2/√2 = 2.

Second term: sin(45°)/cos(45°) = (√2/2)/(√2/2) = 1.

Therefore, original expression is 2 - 1 = 1.

Now, cot(45°) = cos(45°)/sin(45°) = 1. So that matches. Good.

Another test: x = 60° (π/3 radians). Let's check.

Original expression:

sec(60°)/sin(60°) - sin(60°)/cos(60°).

sec(60°) = 1/cos(60°) = 1/(0.5) = 2.

sin(60°) = √3/2 ≈ 0.8660.

cos(60°) = 0.5.

First term: 2 / (√3/2) = 2 * 2/√3 = 4/√3 ≈ 2.3094.

Second term: (√3/2)/0.5 = (√3/2) / (1/2) = √3 ≈ 1.732.

So original expression is 4/√3 - √3. Let me compute that:

Convert to same denominator: 4/√3 - √3 = 4/√3 - (√3 * √3)/√3 = 4/√3 - 3/√3 = (4 - 3)/√3 = 1/√3 ≈ 0.577.

Now, cot(60°) = cos(60°)/sin(60°) = (0.5)/(√3/2) = (0.5 * 2)/√3 = 1/√3 ≈ 0.577. Which matches. So that checks out.

Another test: x = 30° (π/6). Let's check.

Original expression:

sec(30°)/sin(30°) - sin(30°)/cos(30°).

sec(30°) = 1/cos(30°) = 1/(√3/2) = 2/√3 ≈ 1.1547.

sin(30°) = 0.5, cos(30°) = √3/2 ≈ 0.8660.

First term: (2/√3)/0.5 = (2/√3) / (1/2) = 4/√3 ≈ 2.3094.

Second term: 0.5 / (√3/2) = (0.5 * 2)/√3 = 1/√3 ≈ 0.577.

Original expression is 4/√3 - 1/√3 = 3/√3 = √3 ≈ 1.732.

cot(30°) = cos(30°)/sin(30°) = (√3/2)/0.5 = √3/2 / 1/2 = √3 ≈ 1.732. Correct again.

So all test cases check out. Therefore, my simplification seems to be correct.

Therefore, the simplified form of the given expression 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 expressing all trigonometric functions in terms of sine and cosine for easier manipulation:

- Recall that $\sec x = \frac{1}{\cos x}$.

Substituting this into the first term:

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

Now rewrite the full expression:

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

To combine these two terms, they must have a common denominator. The first term has $\sin x \cos x$ in the denominator, while the second has $\cos x$. Multiply the second term’s numerator and denominator by $\sin x$:

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

Now combine the fractions:

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

Recall the Pythagorean identity:

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

So the expression becomes:

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

Simplify by canceling one $\cos x$ in the numerator and denominator:

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

This is the definition of the cotangent function:

$$
\boxed{\cot x}
$$

---

### Final Answer:

$$
\boxed{\cot x}
$$

[Assistant]
Explanation:
Okay, so I need to simplify the expression $\frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}$. Hmm, let me think about how to approach this. First, I remember that secant is the reciprocal of cosine, so maybe I can rewrite sec x in terms of cosine. Let me try that.

Starting with the first term: $\frac{\sec x}{\sin x}$. Since $\sec x = \frac{1}{\cos x}$, substituting that in gives me $\frac{1}{\cos x \cdot \sin x}$. Wait, is that right? Let me check. Yes, because dividing by sin x would be multiplying by 1/sin x, so $\frac{1}{\cos x} \times \frac{1}{\sin x} = \frac{1}{\cos x \sin x}$. Okay, so the first term simplifies to $\frac{1}{\sin x \cos x}$.

Now the second term is $\frac{\sin x}{\cos x}$, which I know is tangent x. But maybe I should keep it as $\frac{\sin x}{\cos x}$ for now to see if there's a common denominator or something. So the expression becomes $\frac{1}{\sin x \cos x} - \frac{\sin x}{\cos x}$.

Hmm, to combine these two terms, they need a common denominator. The first term has $\sin x \cos x$ in the denominator, and the second term has $\cos x$. So if I multiply the numerator and denominator of the second term by sin x, that would make the denominators match. Let me try that.

So, rewrite the second term: $\frac{\sin x}{\cos x} = \frac{\sin x \cdot \sin x}{\cos x \cdot \sin x} = \frac{\sin^2 x}{\sin x \cos x}$. Okay, so now both terms have the same denominator. That means I can subtract them:

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

Wait, that's combining the numerators. Now, I remember that $1 - \sin^2 x$ is a Pythagorean identity. Let me recall: $\sin^2 x + \cos^2 x = 1$, so $1 - \sin^2 x = \cos^2 x$. Yes! So substituting that in:

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

Now, simplifying the fraction. The numerator is $\cos^2 x$ and the denominator is $\sin x \cos x$. I can cancel out one $\cos x$ from numerator and denominator:

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

Which is, of course, $\cot x$. So the simplified form of the original expression is $\cot x$. Let me just check my steps again to make sure I didn't make a mistake.

Starting with $\sec x / \sin x - \sin x / \cos x$. Convert sec x to 1/cos x, so first term is 1/(sin x cos x). Second term is sin x / cos x. Then common denominator is sin x cos x, so second term becomes sin²x / (sin x cos x). Subtracting gives (1 - sin²x)/ (sin x cos x) which is cos²x / (sin x cos x) simplifies to cos x / sin x = cot x. Seems correct.

Alternatively, maybe there's another way to approach it? Let me try another method for verification.

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.

So, $\frac{\sec x}{\sin x}$ is $\frac{1/\cos x}{\sin x}$ which is $\frac{1}{\sin x \cos x}$, same as before. Then subtract $\frac{\sin x}{\cos x}$. So same steps. So yes, same result.

Alternatively, maybe factor something out? Let me see. Not sure. Alternatively, think about combining the terms differently.

Alternatively, let's think about the original expression:

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

Let me write sec x as 1/cos x:

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

Alternatively, factor out 1/cos x:

$$
\frac{1}{\cos x} \left( \frac{1}{\sin x} - \sin x \right )
$$

Yes, that's another way. Let me try that. So factoring out 1/cos x:

$$
\frac{1}{\cos x} \left( \frac{1 - \sin^2 x}{\sin x} \right )
$$

Because $\frac{1}{\sin x} - \sin x$ is equal to $\frac{1 - \sin^2 x}{\sin x}$. Then again, $1 - \sin^2 x = \cos^2 x$, so substituting that:

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

Same result. So that's a good check. So regardless of the method, it seems that the answer is cotangent x. Let me check with specific values to make sure.

Let me pick x = 45 degrees (π/4 radians). Let's compute original expression and cot x.

First, compute original expression:

sec(45°)/sin(45°) - sin(45°)/cos(45°).

We know that sin(45°) = cos(45°) = √2/2 ≈ 0.7071.

sec(45°) is 1/cos(45°) = √2 ≈ 1.4142.

So first term: sec(45°)/sin(45°) = √2 / (√2/2) = √2 * 2/√2 = 2.

Second term: sin(45°)/cos(45°) = (√2/2)/(√2/2) = 1.

Therefore, original expression is 2 - 1 = 1.

Now, cot(45°) = cos(45°)/sin(45°) = 1. So that matches. Good.

Another test: x = 60° (π/3 radians). Let's check.

Original expression:

sec(60°)/sin(60°) - sin(60°)/cos(60°).

sec(60°) = 1/cos(60°) = 1/(0.5) = 2.

sin(60°) = √3/2 ≈ 0.8660.

cos(60°) = 0.5.

First term: 2 / (√3/2) = 2 * 2/√3 = 4/√3 ≈ 2.3094.

Second term: (√3/2)/0.5 = (√3/2) / (1/2) = √3 ≈ 1.732.

So original expression is 4/√3 - √3. Let me compute that:

Convert to same denominator: 4/√3 - √3 = 4/√3 - (√3 * √3)/√3 = 4/√3 - 3/√3 = (4 - 3)/√3 = 1/√3 ≈ 0.577.

Now, cot(60°) = cos(60°)/sin(60°) = (0.5)/(√3/2) = (0.5 * 2)/√3 = 1/√3 ≈ 0.577. Which matches. So that checks out.

Another test: x = 30° (π/6). Let's check.

Original expression:

sec(30°)/sin(30°) - sin(30°)/cos(30°).

sec(30°) = 1/cos(30°) = 1/(√3/2) = 2/√3 ≈ 1.1547.

sin(30°) = 0.5, cos(30°) = √3/2 ≈ 0.8660.

First term: (2/√3)/0.5 = (2/√3) / (1/2) = 4/√3 ≈ 2.3094.

Second term: 0.5 / (√3/2) = (0.5 * 2)/√3 = 1/√3 ≈ 0.577.

Original expression is 4/√3 - 1/√3 = 3/√3 = √3 ≈ 1.732.

cot(30°) = cos(30°)/sin(30°) = (√3/2)/0.5 = √3/2 / 1/2 = √3 ≈ 1.732. Correct again.

So all test cases check out. Therefore, my simplification seems to be correct.

Therefore, the simplified form of the given expression 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 expressing all trigonometric functions in terms of sine and cosine for easier manipulation:

- Recall that $\sec x = \frac{1}{\cos x}$.

Substituting this into the first term:

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

Now rewrite the full expression:

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

To combine these two terms, they must have a common denominator. The first term has $\sin x \cos x$ in the denominator, while the second has $\cos x$. Multiply the second term’s numerator and denominator by $\sin x$:

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

Now combine the fractions:

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

Recall the Pythagorean identity:

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

So the expression becomes:

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

Simplify by canceling one $\cos x$ in the numerator and denominator:

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

This is the definition of the cotangent function:

$$
\boxed{\cot x}
$$

---

### 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. Alright, let me start by recalling what the domain of a logarithmic function is.

The logarithm function log(y) is defined only when y is positive. So, for log(x²), the argument inside the log, which is x², must be greater than zero. That makes sense. So, I need to solve the inequality x² > 0.

Wait, but let me think. Is there anything else I need to consider here? The logarithm function itself is only defined for positive arguments. So, yes, the key here is that x² has to be positive. But x² is always non-negative, right? Because squaring any real number gives a non-negative result. So, x² is zero when x is zero, and positive otherwise.

Therefore, the inequality x² > 0 is equivalent to x ≠ 0. So, the domain of log(x²) should be all real numbers except zero. That would mean x can be any real number except 0. But the problem states that the domain is x < a or x > b. Hmm, so maybe there's something I'm missing here?

Wait, let me check again. If we write log(x²), is there a different way to interpret this? Like, is it (log x)² or log(x²)? The original problem says log x², which is more likely log(x²), not (log x)². Because if it were (log x)², the domain would be x > 0, since log x is defined only for x > 0. But here, since it's log of x squared, which is different.

So, let me confirm: log(x²) is defined when x² > 0, so x ≠ 0. Therefore, the domain is all real numbers except zero, which would be (-∞, 0) union (0, ∞). So, in the problem statement, they mention the domain is x < a or x > b. Comparing that with my conclusion, that would be x < 0 or x > 0. So, in this case, a would be 0 and b would be 0. Therefore, a + b would be 0 + 0 = 0.

Wait, but maybe the question is written differently? Let me check again. The problem says "the domain of the function log x² is x < a or x > b, for some a and b". So, maybe they have different values? But according to my reasoning, the domain is x ≠ 0, so it's x < 0 or x > 0. Therefore, a is 0 and b is 0. Hence, a + b = 0.

But maybe there is a mistake here. Let me think again. Is there a possibility that the original function is log(x)^2, which is different? If that were the case, then log(x) squared would have domain x > 0, but the problem states that the domain is x < a or x > b, which would not make sense for log(x)^2. So, that's probably not the case.

Alternatively, maybe the problem is in a different base? But 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 only depends on the argument being positive.

Wait, maybe there is a misinterpretation here. Let me check the original problem again: "the domain of the function log x² is x < a or x > b". So, the question is about log(x²). So, log(x²) is defined when x² > 0.

Another thought: maybe the problem is in a context where log is defined only for positive arguments, but perhaps there's an absolute value or something else? Wait, no. If you have log(x²), then x² is positive except when x is zero. So, unless there's a restriction on x that I'm missing.

Alternatively, maybe the problem is written in a different notation where log x² is interpreted as log(x) squared? But in standard notation, log(x²) is the logarithm of x squared, not the square of log x.

But to make sure, let me verify with an example. If I take x = 1, log(1²) = log(1) = 0, which is fine. If x = -1, log((-1)^2) = log(1) = 0, which is also fine. If x = 0, log(0²) = log(0), which is undefined. So, indeed, the domain is all real numbers except 0.

So, then, if the problem says the domain is x < a or x > b, then comparing with the domain being (-∞, 0) ∪ (0, ∞), so that would mean a is 0 and b is 0. Therefore, a + b is 0.

But maybe there's something else here. Wait, could the original problem have a typo? Or perhaps I need to consider complex numbers? But no, the domain of a logarithm function is typically considered in real numbers unless specified otherwise.

Alternatively, maybe the problem is referring to a different logarithm function, like log base 10 or natural log, but regardless of the base, the domain is still the same.

Wait, but the problem says "the domain of the function log x² is x < a or x > b". If I take the domain as all real numbers except 0, then it's x < 0 or x > 0, so a and b are both zero. Therefore, a + b is 0.

But the problem says "for some a and b". Maybe the problem is expecting different a and b? Maybe I made a mistake here? Let me think again.

Alternatively, perhaps the problem is written incorrectly, and they meant log(x) squared? Let's check. If it's [log(x)]², then the domain would be x > 0, which is not x < a or x > b. So, that doesn't fit.

Alternatively, maybe it's log base 2 of x² or something else? No, the base doesn't affect the domain here.

Alternatively, maybe the problem is about the inverse function? Wait, no.

Wait, maybe there's a misunderstanding in notation. In some contexts, log x² could be interpreted as log(x) squared, but that's not standard. Usually, log x² is log(x²). Also, in calculus, sometimes people write log x² as 2 log |x|, but again, that's just an identity.

So, if we consider that log(x²) is equivalent to 2 log |x|, then the domain of 2 log |x| is still x ≠ 0, because |x| is positive unless x is 0. So, same domain.

So, again, that would mean that the domain is x < 0 or x > 0. Therefore, a = 0 and b = 0. Hence, a + b = 0.

But maybe the question is expecting me to write something else? Let me check if there's another approach.

Wait, suppose someone incorrectly solves the inequality log(x²) > 0? But the problem is about the domain, not about solving log(x²) > 0. The domain is when the argument is positive.

Alternatively, maybe someone confuses the domain of log(x²) with solving log(x²) = something? But no, the domain is about where the function is defined.

Alternatively, maybe the problem is from a textbook or a course where they define the domain differently? For example, in some contexts, maybe they consider log(x²) as log(x^2) only if x is positive? But that doesn't make sense. The logarithm of a positive number is defined regardless of the sign of x, as long as x² is positive.

Alternatively, maybe there is a mistake in the problem statement? If the domain was supposed to be, say, log(x^2) is defined for x < a or x > b, but maybe a and b are different?

Wait, but if the domain is x ≠ 0, that's the same as x < 0 or x > 0. So, a = 0 and b = 0. Therefore, a + b = 0.

Alternatively, maybe the problem is referring to something else. Let me check again.

Wait, if you have log(x^2), then depending on the base, if the base is between 0 and 1, then the function is decreasing, but that still doesn't affect the domain. The domain is determined by the argument being positive.

Alternatively, maybe the problem is written in a different language, and "log x²" is misinterpreted? For example, in some countries, comma is used as decimal separator, but here it's not an issue.

Alternatively, maybe the original problem is about the function log(x)^2, and there was a formatting issue?

If that were the case, then log(x)^2 would have domain x > 0, but the problem states that the domain is x < a or x > b, which would not fit.

Alternatively, if the problem was log(x^2) with some restrictions?

Wait, unless there's a typo and the function is log(x)², but again, same issue.

Alternatively, maybe the problem is referring to the natural logarithm? No, that doesn't change anything.

Alternatively, perhaps the problem is considering complex logarithms? But in complex analysis, the logarithm is multi-valued, and the domain would be different, but the problem seems to be in the context of real numbers.

Alternatively, maybe the problem is from a calculus textbook where they define log x² as the principal branch? But again, the domain would still be x ≠ 0.

Wait, unless there's a restriction on x due to the logarithm's properties? For example, if log(x²) is considered as 2 log|x|, then maybe someone could think that the domain is x ≠ 0, but still.

Alternatively, perhaps the problem is in a context where the logarithm is only defined for positive real numbers, but for log(x²), it's defined when x² is positive, which is x ≠ 0.

Therefore, unless there's a misunderstanding here, the answer should be a = 0 and b = 0, so a + b = 0.

But let me check with specific values. If x = 1, log(1²) = log(1) = 0, which is defined. If x = -1, log((-1)^2) = log(1) = 0, also defined. If x = 0, log(0²) = log(0), which is undefined. If x = 2, log(4) is defined. If x = -2, log(4) is defined. So, yes, the domain is indeed all real numbers except zero.

Therefore, the domain is x < 0 or x > 0, which would imply a = 0 and b = 0. Hence, a + b = 0.

But the problem says "for some a and b". Maybe they expect different values? Maybe the problem is actually about log(x) with some restriction? Or maybe the problem was mistyped?

Wait, let me think again. Maybe the original problem is not log(x²), but log(x^2) with some other notation. Wait, suppose the problem is written as log x^2, and if someone misinterprets it as (log x)^2, then the domain would be x > 0, but the problem says x < a or x > b. So, maybe there's a confusion here?

Alternatively, maybe the original problem was about the function log(x^2 + something). But given the problem as stated is log x², I think my reasoning is correct.

Alternatively, perhaps the problem is expecting me to write the domain in terms of inequalities with a and b where a and b are not zero?

Wait, but if the domain is x ≠ 0, then how can we write it as x < a or x > b? It's possible only if a and b are both zero. Because if a and b are not zero, then x < a or x > b would not cover all real numbers except zero. For example, if a = 1 and b = -1, then x < 1 or x > -1 would be all real numbers except... Wait, no, x < 1 or x > -1 would actually be all real numbers except the interval [-1, 1). But that's not the case here.

Wait, so if the domain is x < 0 or x > 0, then that's all real numbers except zero. So, if we need to write it as x < a or x > b, then a must be 0 and b must be 0. Because if a is 0, then x < 0 is the left part, and x > 0 is the right part.

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

Alternatively, maybe the problem was written differently, like log(x^2) is defined for x < a or x > b, but maybe there are more constraints?

Wait, let me think again. Is there a case where log(x²) is not defined for some x < 0 or x > 0? For example, if x is negative, then x² is positive, so log(x²) is defined. If x is positive, same thing. Only at x = 0 is it undefined. So, yes, the domain is indeed all real numbers except zero.

Therefore, unless there is a trick here that I'm missing, the answer should be 0.

But maybe the question is from a source where the answer is different? Let me see. Wait, maybe the problem is about the function log(x^2) in some restricted domain? For example, if we consider the logarithm to have a certain range or something else?

Alternatively, perhaps the problem is in the context of solving an equation? For example, if log(x²) is defined, then... but no, the problem specifically talks about the domain.

Alternatively, maybe the problem is written in a way that log x^2 is interpreted as log(x^2), but with some other constraints. Wait, no.

Alternatively, maybe the person who wrote the problem made a mistake and intended it to be log(x) squared? If that's the case, then the domain is x > 0, which can't be expressed as x < a or x > b unless a is something else, but then it's not possible.

Alternatively, maybe it's log base 10 of x squared, but again same thing.

Alternatively, maybe the question is referring to the natural logarithm, but again, same result.

Alternatively, maybe the problem is in another language, and "domain" is translated differently?

Wait, perhaps the problem is from a calculus problem where they have a function log(x²) with some other condition? For example, maybe the function is log(x²) and they have another condition? But the problem doesn't say anything else.

Alternatively, maybe the problem is from a textbook where they define the domain of log(x²) as x < a or x > b, and they want to find a and b. But according to standard math, it's x ≠ 0, so a = 0 and b = 0.

Alternatively, maybe there's a misinterpretation of the function's domain. For example, someone might think that log(x²) is defined only for x > 0, but that's not correct. Because x can be negative as long as x² is positive.

Wait, so if I think of log(x²) as 2 log |x|, then 2 log |x| is defined when |x| > 0, which again is x ≠ 0.

So, again, the domain is x ≠ 0.

Therefore, I think that my conclusion that a = 0 and b = 0 is correct, so a + b = 0.

But maybe the answer is different? Let me check with another approach. Suppose someone tried to solve log(x²) is defined. So, x² > 0. The solution is x ≠ 0. So, the domain is (-infty, 0) union (0, infty). Therefore, written as x < 0 or x > 0. So, comparing to x < a or x > b, then a is 0 and b is 0. Therefore, a + b = 0.

Alternatively, if the problem had said the domain is x < a or x > b, and they want a and b such that... maybe there is no such a and b? But no, since we can write it as x < 0 or x > 0.

Alternatively, maybe the problem is in a context where a and b are supposed to be different? For example, if the domain was x < a or x > b, with a ≠ b. But in this case, a and b are both zero.

Alternatively, maybe the problem was supposed to say that the domain is x < a or x > b, and find a + b given that the function has some other property. But as stated, the problem is as given.

So, perhaps the answer is 0.

Alternatively, maybe the problem is expecting me to consider the logarithm's range or something else? But the domain is purely about the argument being positive.

Alternatively, maybe the question is a trick question. For example, if log x² is defined for all real numbers except 0, but someone might think that x must be positive because log(x) is only defined for positive x, but that's not the case here. The log of x squared is defined even if x is negative, because x squared is positive. So, the domain is all real numbers except zero.

Therefore, I think the answer is 0.

But just to be thorough, let me check online if there's any standard problem like this. Wait, I can't actually browse, but I can recall. For example, in calculus, when we have log(x²), it's often rewritten as 2 log|x|, which is defined for x ≠ 0. So, the domain is x ≠ 0.

Alternatively, maybe the problem considers log(x²) as log(x) squared, but that's non-standard.

Alternatively, maybe there's a different interpretation. Suppose the original problem is written as log x^2, and there is ambiguity in the notation. If someone interprets it as log(x^2) versus (log x)^2, but as discussed, the former is different.

Alternatively, maybe the problem is from a source where log is defined only for positive numbers, but the domain of log(x²) is actually x > 0, because x² is positive for all real x except zero, but maybe the problem assumes that x is positive? But that seems arbitrary.

Wait, but if you have log(x²), and you consider that x² is positive, but maybe the problem is in a context where x is allowed to be negative? But the domain is still all real numbers except zero.

Therefore, I think my conclusion is correct.

Therefore, the value of a + b is 0.

**Final Answer**
\boxed{0}

To determine the domain of the function $\log x^2$, we begin by recalling that the **logarithmic function** is defined **only for positive real numbers**. That is, the expression $\log y$ is defined if and only if $y > 0$.

---

### Step 1: Understand the Argument of the Logarithm

The argument of the logarithm in this case is not $x$, but rather $x^2$. Therefore, we need to determine for which real numbers $x$, the expression
$$
x^2 > 0
$$
holds.

---

### Step 2: Solve the Inequality

We solve the inequality:
$$
x^2 > 0
$$

This is true for all real numbers $x$ **except** $x = 0$, because $x^2 = 0$ only when $x = 0$. Thus, the logarithm $\log x^2$ is **defined for all $x \ne 0$**.

So, the domain of the function is:
$$
x \in (-\infty, 0) \cup (0, \infty)
$$

---

### Step 3: Match to the Given Domain Form and Find $a + b$

The problem states that the domain can be written as:
$$
x < a \quad \text{or} \quad x > b
$$

We compare this with our result:
$$
x < 0 \quad \text{or} \quad x > 0
$$

This means that:
- $a = 0$
- $b = 0$

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

---

### 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. Alright, let me start by recalling what the domain of a logarithmic function is.

The logarithm function log(y) is defined only when y is positive. So, for log(x²), the argument inside the log, which is x², must be greater than zero. That makes sense. So, I need to solve the inequality x² > 0.

Wait, but let me think. Is there anything else I need to consider here? The logarithm function itself is only defined for positive arguments. So, yes, the key here is that x² has to be positive. But x² is always non-negative, right? Because squaring any real number gives a non-negative result. So, x² is zero when x is zero, and positive otherwise.

Therefore, the inequality x² > 0 is equivalent to x ≠ 0. So, the domain of log(x²) should be all real numbers except zero. That would mean x can be any real number except 0. But the problem states that the domain is x < a or x > b. Hmm, so maybe there's something I'm missing here?

Wait, let me check again. If we write log(x²), is there a different way to interpret this? Like, is it (log x)² or log(x²)? The original problem says log x², which is more likely log(x²), not (log x)². Because if it were (log x)², the domain would be x > 0, since log x is defined only for x > 0. But here, since it's log of x squared, which is different.

So, let me confirm: log(x²) is defined when x² > 0, so x ≠ 0. Therefore, the domain is all real numbers except zero, which would be (-∞, 0) union (0, ∞). So, in the problem statement, they mention the domain is x < a or x > b. Comparing that with my conclusion, that would be x < 0 or x > 0. So, in this case, a would be 0 and b would be 0. Therefore, a + b would be 0 + 0 = 0.

Wait, but maybe the question is written differently? Let me check again. The problem says "the domain of the function log x² is x < a or x > b, for some a and b". So, maybe they have different values? But according to my reasoning, the domain is x ≠ 0, so it's x < 0 or x > 0. Therefore, a is 0 and b is 0. Hence, a + b = 0.

But maybe there is a mistake here. Let me think again. Is there a possibility that the original function is log(x)^2, which is different? If that were the case, then log(x) squared would have domain x > 0, but the problem states that the domain is x < a or x > b, which would not make sense for log(x)^2. So, that's probably not the case.

Alternatively, maybe the problem is in a different base? But 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 only depends on the argument being positive.

Wait, maybe there is a misinterpretation here. Let me check the original problem again: "the domain of the function log x² is x < a or x > b". So, the question is about log(x²). So, log(x²) is defined when x² > 0.

Another thought: maybe the problem is in a context where log is defined only for positive arguments, but perhaps there's an absolute value or something else? Wait, no. If you have log(x²), then x² is positive except when x is zero. So, unless there's a restriction on x that I'm missing.

Alternatively, maybe the problem is written in a different notation where log x² is interpreted as log(x) squared? But in standard notation, log(x²) is the logarithm of x squared, not the square of log x.

But to make sure, let me verify with an example. If I take x = 1, log(1²) = log(1) = 0, which is fine. If x = -1, log((-1)^2) = log(1) = 0, which is also fine. If x = 0, log(0²) = log(0), which is undefined. So, indeed, the domain is all real numbers except 0.

So, then, if the problem says the domain is x < a or x > b, then comparing with the domain being (-∞, 0) ∪ (0, ∞), so that would mean a is 0 and b is 0. Therefore, a + b is 0.

But maybe there's something else here. Wait, could the original problem have a typo? Or perhaps I need to consider complex numbers? But no, the domain of a logarithm function is typically considered in real numbers unless specified otherwise.

Alternatively, maybe the problem is referring to a different logarithm function, like log base 10 or natural log, but regardless of the base, the domain is still the same.

Wait, but the problem says "the domain of the function log x² is x < a or x > b". If I take the domain as all real numbers except 0, then it's x < 0 or x > 0, so a and b are both zero. Therefore, a + b is 0.

But the problem says "for some a and b". Maybe the problem is expecting different a and b? Maybe I made a mistake here? Let me think again.

Alternatively, perhaps the problem is written incorrectly, and they meant log(x) squared? Let's check. If it's [log(x)]², then the domain would be x > 0, which is not x < a or x > b. So, that doesn't fit.

Alternatively, maybe it's log base 2 of x² or something else? No, the base doesn't affect the domain here.

Alternatively, maybe the problem is about the inverse function? Wait, no.

Wait, maybe there's a misunderstanding in notation. In some contexts, log x² could be interpreted as log(x) squared, but that's not standard. Usually, log x² is log(x²). Also, in calculus, sometimes people write log x² as 2 log |x|, but again, that's just an identity.

So, if we consider that log(x²) is equivalent to 2 log |x|, then the domain of 2 log |x| is still x ≠ 0, because |x| is positive unless x is 0. So, same domain.

So, again, that would mean that the domain is x < 0 or x > 0. Therefore, a = 0 and b = 0. Hence, a + b = 0.

But maybe the question is expecting me to write something else? Let me check if there's another approach.

Wait, suppose someone incorrectly solves the inequality log(x²) > 0? But the problem is about the domain, not about solving log(x²) > 0. The domain is when the argument is positive.

Alternatively, maybe someone confuses the domain of log(x²) with solving log(x²) = something? But no, the domain is about where the function is defined.

Alternatively, maybe the problem is from a textbook or a course where they define the domain differently? For example, in some contexts, maybe they consider log(x²) as log(x^2) only if x is positive? But that doesn't make sense. The logarithm of a positive number is defined regardless of the sign of x, as long as x² is positive.

Alternatively, maybe there is a mistake in the problem statement? If the domain was supposed to be, say, log(x^2) is defined for x < a or x > b, but maybe a and b are different?

Wait, but if the domain is x ≠ 0, that's the same as x < 0 or x > 0. So, a = 0 and b = 0. Therefore, a + b = 0.

Alternatively, maybe the problem is referring to something else. Let me check again.

Wait, if you have log(x^2), then depending on the base, if the base is between 0 and 1, then the function is decreasing, but that still doesn't affect the domain. The domain is determined by the argument being positive.

Alternatively, maybe the problem is written in a different language, and "log x²" is misinterpreted? For example, in some countries, comma is used as decimal separator, but here it's not an issue.

Alternatively, maybe the original problem is about the function log(x)^2, and there was a formatting issue?

If that were the case, then log(x)^2 would have domain x > 0, but the problem states that the domain is x < a or x > b, which would not fit.

Alternatively, if the problem was log(x^2) with some restrictions?

Wait, unless there's a typo and the function is log(x)², but again, same issue.

Alternatively, maybe the problem is referring to the natural logarithm? No, that doesn't change anything.

Alternatively, perhaps the problem is considering complex logarithms? But in complex analysis, the logarithm is multi-valued, and the domain would be different, but the problem seems to be in the context of real numbers.

Alternatively, maybe the problem is from a calculus textbook where they define log x² as the principal branch? But again, the domain would still be x ≠ 0.

Wait, unless there's a restriction on x due to the logarithm's properties? For example, if log(x²) is considered as 2 log|x|, then maybe someone could think that the domain is x ≠ 0, but still.

Alternatively, perhaps the problem is in a context where the logarithm is only defined for positive real numbers, but for log(x²), it's defined when x² is positive, which is x ≠ 0.

Therefore, unless there's a misunderstanding here, the answer should be a = 0 and b = 0, so a + b = 0.

But let me check with specific values. If x = 1, log(1²) = log(1) = 0, which is defined. If x = -1, log((-1)^2) = log(1) = 0, also defined. If x = 0, log(0²) = log(0), which is undefined. If x = 2, log(4) is defined. If x = -2, log(4) is defined. So, yes, the domain is indeed all real numbers except zero.

Therefore, the domain is x < 0 or x > 0, which would imply a = 0 and b = 0. Hence, a + b = 0.

But the problem says "for some a and b". Maybe they expect different values? Maybe the problem is actually about log(x) with some restriction? Or maybe the problem was mistyped?

Wait, let me think again. Maybe the original problem is not log(x²), but log(x^2) with some other notation. Wait, suppose the problem is written as log x^2, and if someone misinterprets it as (log x)^2, then the domain would be x > 0, but the problem says x < a or x > b. So, maybe there's a confusion here?

Alternatively, maybe the original problem was about the function log(x^2 + something). But given the problem as stated is log x², I think my reasoning is correct.

Alternatively, perhaps the problem is expecting me to write the domain in terms of inequalities with a and b where a and b are not zero?

Wait, but if the domain is x ≠ 0, then how can we write it as x < a or x > b? It's possible only if a and b are both zero. Because if a and b are not zero, then x < a or x > b would not cover all real numbers except zero. For example, if a = 1 and b = -1, then x < 1 or x > -1 would be all real numbers except... Wait, no, x < 1 or x > -1 would actually be all real numbers except the interval [-1, 1). But that's not the case here.

Wait, so if the domain is x < 0 or x > 0, then that's all real numbers except zero. So, if we need to write it as x < a or x > b, then a must be 0 and b must be 0. Because if a is 0, then x < 0 is the left part, and x > 0 is the right part.

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

Alternatively, maybe the problem was written differently, like log(x^2) is defined for x < a or x > b, but maybe there are more constraints?

Wait, let me think again. Is there a case where log(x²) is not defined for some x < 0 or x > 0? For example, if x is negative, then x² is positive, so log(x²) is defined. If x is positive, same thing. Only at x = 0 is it undefined. So, yes, the domain is indeed all real numbers except zero.

Therefore, unless there is a trick here that I'm missing, the answer should be 0.

But maybe the question is from a source where the answer is different? Let me see. Wait, maybe the problem is about the function log(x^2) in some restricted domain? For example, if we consider the logarithm to have a certain range or something else?

Alternatively, perhaps the problem is in the context of solving an equation? For example, if log(x²) is defined, then... but no, the problem specifically talks about the domain.

Alternatively, maybe the problem is written in a way that log x^2 is interpreted as log(x^2), but with some other constraints. Wait, no.

Alternatively, maybe the person who wrote the problem made a mistake and intended it to be log(x) squared? If that's the case, then the domain is x > 0, which can't be expressed as x < a or x > b unless a is something else, but then it's not possible.

Alternatively, maybe it's log base 10 of x squared, but again same thing.

Alternatively, maybe the question is referring to the natural logarithm, but again, same result.

Alternatively, maybe the problem is in another language, and "domain" is translated differently?

Wait, perhaps the problem is from a calculus problem where they have a function log(x²) with some other condition? For example, maybe the function is log(x²) and they have another condition? But the problem doesn't say anything else.

Alternatively, maybe the problem is from a textbook where they define the domain of log(x²) as x < a or x > b, and they want to find a and b. But according to standard math, it's x ≠ 0, so a = 0 and b = 0.

Alternatively, maybe there's a misinterpretation of the function's domain. For example, someone might think that log(x²) is defined only for x > 0, but that's not correct. Because x can be negative as long as x² is positive.

Wait, so if I think of log(x²) as 2 log |x|, then 2 log |x| is defined when |x| > 0, which again is x ≠ 0.

So, again, the domain is x ≠ 0.

Therefore, I think that my conclusion that a = 0 and b = 0 is correct, so a + b = 0.

But maybe the answer is different? Let me check with another approach. Suppose someone tried to solve log(x²) is defined. So, x² > 0. The solution is x ≠ 0. So, the domain is (-infty, 0) union (0, infty). Therefore, written as x < 0 or x > 0. So, comparing to x < a or x > b, then a is 0 and b is 0. Therefore, a + b = 0.

Alternatively, if the problem had said the domain is x < a or x > b, and they want a and b such that... maybe there is no such a and b? But no, since we can write it as x < 0 or x > 0.

Alternatively, maybe the problem is in a context where a and b are supposed to be different? For example, if the domain was x < a or x > b, with a ≠ b. But in this case, a and b are both zero.

Alternatively, maybe the problem was supposed to say that the domain is x < a or x > b, and find a + b given that the function has some other property. But as stated, the problem is as given.

So, perhaps the answer is 0.

Alternatively, maybe the problem is expecting me to consider the logarithm's range or something else? But the domain is purely about the argument being positive.

Alternatively, maybe the question is a trick question. For example, if log x² is defined for all real numbers except 0, but someone might think that x must be positive because log(x) is only defined for positive x, but that's not the case here. The log of x squared is defined even if x is negative, because x squared is positive. So, the domain is all real numbers except zero.

Therefore, I think the answer is 0.

But just to be thorough, let me check online if there's any standard problem like this. Wait, I can't actually browse, but I can recall. For example, in calculus, when we have log(x²), it's often rewritten as 2 log|x|, which is defined for x ≠ 0. So, the domain is x ≠ 0.

Alternatively, maybe the problem considers log(x²) as log(x) squared, but that's non-standard.

Alternatively, maybe there's a different interpretation. Suppose the original problem is written as log x^2, and there is ambiguity in the notation. If someone interprets it as log(x^2) versus (log x)^2, but as discussed, the former is different.

Alternatively, maybe the problem is from a source where log is defined only for positive numbers, but the domain of log(x²) is actually x > 0, because x² is positive for all real x except zero, but maybe the problem assumes that x is positive? But that seems arbitrary.

Wait, but if you have log(x²), and you consider that x² is positive, but maybe the problem is in a context where x is allowed to be negative? But the domain is still all real numbers except zero.

Therefore, I think my conclusion is correct.

Therefore, the value of a + b is 0.

**Final Answer**
\boxed{0}

To determine the domain of the function $\log x^2$, we begin by recalling that the **logarithmic function** is defined **only for positive real numbers**. That is, the expression $\log y$ is defined if and only if $y > 0$.

---

### Step 1: Understand the Argument of the Logarithm

The argument of the logarithm in this case is not $x$, but rather $x^2$. Therefore, we need to determine for which real numbers $x$, the expression
$$
x^2 > 0
$$
holds.

---

### Step 2: Solve the Inequality

We solve the inequality:
$$
x^2 > 0
$$

This is true for all real numbers $x$ **except** $x = 0$, because $x^2 = 0$ only when $x = 0$. Thus, the logarithm $\log x^2$ is **defined for all $x \ne 0$**.

So, the domain of the function is:
$$
x \in (-\infty, 0) \cup (0, \infty)
$$

---

### Step 3: Match to the Given Domain Form and Find $a + b$

The problem states that the domain can be written as:
$$
x < a \quad \text{or} \quad x > b
$$

We compare this with our result:
$$
x < 0 \quad \text{or} \quad x > 0
$$

This means that:
- $a = 0$
- $b = 0$

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

---

### 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 $0 + 0 = \boxed{0}$.