Explanation:
Okay, so I need to figure out the fifth number in the list of positive integers that have exactly three positive divisors. Let me start by recalling some number theory concepts.

First, I remember that the number of divisors a number has depends on its prime factorization. If a number n can be expressed as n = p₁^a₁ * p₂^a₂ * ... * p_k^a_k, where p₁, p₂, ..., p_k are prime numbers and a₁, a₂, ..., a_k are their respective exponents, then the number of positive divisors of n is (a₁ + 1)*(a₂ + 1)*...*(a_k + 1).

So, if a number has exactly three positive divisors, then this product must equal 3. Since 3 is a prime number, the only way to write it as a product of integers greater than 1 is 3 = 3*1. Therefore, this implies that the number must be of the form p², where p is a prime number. Because if we have only one exponent, then (a₁ + 1) = 3, so a₁ = 2. Therefore, numbers with exactly three divisors are squares of primes.

Let me verify this with an example. Take the prime number 2. Its square is 4. The divisors of 4 are 1, 2, 4. That's three divisors. Similarly, take 3² = 9. Divisors are 1, 3, 9. Also three divisors. What about a composite number that isn't a square? Let's say 6. It's divisors are 1, 2, 3, 6. That's four divisors. So that doesn't fit. What about a cube of a prime? Let's take 8 (2³). Divisors are 1, 2, 4, 8. Four divisors again. So that's not three. Therefore, indeed, only squares of primes have exactly three divisors.

Therefore, the problem reduces to listing the squares of primes in ascending order and finding the fifth one. Let me list the primes first. The primes are 2, 3, 5, 7, 11, 13, etc. Then their squares are 4, 9, 25, 49, 121, 169, etc. So if we list them in order, the first number is 4, second is 9, third is 25, fourth is 49, fifth is 121. Therefore, the fifth number should be 121.

Wait, but hold on, is there any other number that might have exactly three divisors which is not a square of a prime? Let me think again. Suppose there is a number with three divisors. Let me suppose that such a number is not a square of a prime. Then, according to the divisor function formula, since the total number of divisors is three, which is prime, then the number must have a prime factorization with exponents such that when you add one to each exponent and multiply, you get three. Since three is prime, the only way is one exponent is 2 and the rest are zero. So the number must be a square of a prime. Therefore, there are no other numbers with exactly three divisors.

Therefore, the list is just the squares of primes. Hence, the fifth number would be the square of the fifth prime. Let me check the order of primes. First prime is 2, second is 3, third is 5, fourth is 7, fifth is 11. So 11 squared is 121. Hence, the fifth number is 121.

But just to make sure, let me list them out step by step:

1. 2² = 4
2. 3² = 9
3. 5² = 25
4. 7² = 49
5. 11² = 121

Yes, that seems correct. However, let me check if there are any other numbers between, say, 49 and 121 that might have exactly three divisors. For example, is there a number between 49 and 121 which is a square of a prime? Well, the next prime after 7 is 11, so 11² is 121. The primes between 7 and 11 are... well, 7, then 11. So there's no prime between 7 and 11. Wait, 7 is the fourth prime, then 11 is the fifth. So between 7²=49 and 11²=121, the next square would be 13²=169, which is larger. Therefore, between 49 and 121, there are no other squares of primes. Therefore, 121 is indeed the fifth number.

But just to be thorough, let me check if there are numbers which are not squares of primes but still have exactly three divisors. Suppose there's a number with three divisors. Suppose it's a product of two distinct primes. Let me think. Let's take 6, which is 2*3. Its divisors are 1, 2, 3, 6. Four divisors. If we take a cube of a prime, like 8=2³. Divisors are 1, 2, 4, 8. Four divisors again. If we take a product of three primes, like 30=2*3*5. Divisors are 1, 2, 3, 5, 6, 10, 15, 30. Eight divisors.

Alternatively, what if there is a number with a prime factorization with exponents such that (a+1)(b+1)... = 3. Since 3 is prime, there are two possibilities: either one exponent is 2 and the rest are zero (so the number is p²) or there's a different combination? But since 3 is prime, the only way to write it as a product of integers greater than 1 is 3*1. Therefore, the number must have exactly one prime factor with exponent 2, and no other prime factors. Therefore, the number must be p².

Therefore, there are no other numbers with exactly three divisors. Therefore, my original reasoning is correct. So the fifth such number is 121.

But let me check if maybe there is a composite number which is not prime squared but has three divisors. Let me think. Suppose someone says, for example, 16. Divisors of 16 are 1, 2, 4, 8, 16. Five divisors. How about 25? 1, 5, 25. Three divisors. Okay, that's prime squared. How about 49? Same thing.

What about a number like 1? Divisors of 1 is just 1. So that's only one. Not three. How about 0? Well, 0 isn't positive. So we are only considering positive integers. So 1 is excluded.

Another thought: maybe numbers like p^2 where p is a prime, but maybe there is a number with three divisors that's not square of primes? Let me try to think of such a number. Let's take 12. Divisors: 1, 2, 3, 4, 6, 12. Six divisors. How about 10? Divisors are 1, 2, 5, 10. Four divisors. How about 14? Same idea.

What about numbers with exponents in their prime factors? Let me try to think: For example, if a number has two distinct primes, then the number of divisors is (a+1)(b+1)... So if you have a number that's p^1 * q^1, then the number of divisors is (1+1)(1+1)=4. If you have p^2 * q^1, then the number of divisors is (2+1)(1+1)=6. So that's six. If you have p^2 * q^2, then (2+1)(2+1)=9. So nope.

Alternatively, if you have p^0 * q^2, which is just q^2. So same as before. Therefore, the only way to get three divisors is if you have one prime squared.

So I think that confirms it. Therefore, the numbers with exactly three positive divisors are exactly the squares of primes. Therefore, to list them in order, we need to take the primes in order and square them.

So primes in order:

1st prime: 2, square is 4

2nd prime: 3, square is 9

3rd prime: 5, square is 25

4th prime: 7, square is 49

5th prime: 11, square is 121

Therefore, the fifth number is 121.

But let me make sure there isn't a prime that is skipped or something. Wait, primes are 2, 3, 5, 7, 11, 13, 17,... So yes, the fifth prime is 11. Therefore, squaring that gives 121.

Just to be thorough, let me list the numbers with exactly three divisors up to, say, 100, and check the order.

Starting from the beginning:

- 4 (divisors: 1, 2, 4)

- 9 (1, 3, 9)

- 25 (1, 5, 25)

- 49 (1, 7, 49)

- 121 (1, 11, 121) – but 121 is 11², which is beyond 100. So between 49 and 121, are there any other numbers with exactly three divisors?

Wait, what about 169? That's 13², which is 169. But 169 is bigger than 121. So between 49 and 121, there's 121. But wait, what about numbers like 121? Let me check if there are any numbers between 49 and 121 which are squares of primes. The next prime after 7 is 11, which is 11²=121. The primes between 7 and 11 are... hmm, there are no primes between 7 and 11. The primes are 2, 3, 5, 7, 11, 13,... So yes, 7 is the fourth prime, 11 is fifth. Therefore, between 49 and 121, there are no squares of primes. Therefore, the numbers with exactly three divisors in order are 4, 9, 25, 49, 121, 169, 289,... So the fifth one is 121.

But let me check if maybe there are numbers that are not squares of primes but have exactly three divisors. Let me think again. Suppose there is a number which is a product of two primes, but somehow... Wait, if a number is a product of two distinct primes, then it has four divisors. If it's a product of one prime squared, then three divisors. If it's a product of three primes, then it would have more divisors. If it's a prime cubed, then it has four divisors.

So, for example, 8=2³: divisors are 1, 2, 4, 8. Four divisors.

So no, it's not possible. Therefore, yes, only squares of primes.

Another check: Let me list all numbers up to, say, 50 and check which ones have exactly three divisors.

Starting from 1:

1: 1 divisor

2: 1, 2 – two divisors

3: 1, 3 – two

4: 1, 2, 4 – three

5: 1, 5 – two

6: 1, 2, 3, 6 – four

7: 1, 7 – two

8: 1, 2, 4, 8 – four

9: 1, 3, 9 – three

10: 1, 2, 5, 10 – four

11: 1, 11 – two

12: 1, 2, 3, 4, 6, 12 – six

13: 1, 13 – two

14: 1, 2, 7, 14 – four

15: 1, 3, 5, 15 – four

16: 1, 2, 4, 8, 16 – five

17: 1, 17 – two

18: 1, 2, 3, 6, 9, 18 – six

19: 1, 19 – two

20: 1, 2, 4, 5, 10, 20 – six

21: 1, 3, 7, 21 – four

22: 1, 2, 11, 22 – four

23: 1, 23 – two

24: 1, 2, 3, 4, 6, 8, 12, 24 – eight

25: 1, 5, 25 – three

26: 1, 2, 13, 26 – four

27: 1, 3, 9, 27 – four

28: 1, 2, 4, 7, 14, 28 – six

29: 1, 29 – two

30: 1, 2, 3, 5, 6, 10, 15, 30 – eight

31: 1, 31 – two

32: 1, 2, 4, 8, 16, 32 – six

33: 1, 3, 11, 33 – four

34: 1, 2, 17, 34 – four

35: 1, 5, 7, 35 – four

36: 1, 2, 3, 4, 6, 9, 12, 18, 36 – nine

37: 1, 37 – two

38: 1, 2, 19, 38 – four

39: 1, 3, 13, 39 – four

40: 1, 2, 4, 5, 8, 10, 20, 40 – eight

41: 1, 41 – two

42: 1, 2, 3, 6, 7, 14, 21, 42 – eight

43: 1, 43 – two

44: 1, 2, 4, 11, 22, 44 – six

45: 1, 3, 5, 9, 15, 45 – six

46: 1, 2, 23, 46 – four

47: 1, 47 – two

48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 – ten

49: 1, 7, 49 – three

So up to 49, the numbers with exactly three divisors are 4, 9, 25, 49. So that is four numbers. Then the next one would be 121. Therefore, the fifth number is indeed 121.

So I think this confirms it. Therefore, the answer should be 121.

But let me check if maybe there is a number between 49 and 121 that I missed. For example, what about 121? Let me check 121 is 11 squared. What about numbers like 169? That's 13 squared. So between 49 and 121, there's no primes between 7 and 11. So the next prime after 7 is 11, so 11 squared is 121. Therefore, there are no numbers between 49 and 121 that are squares of primes. Therefore, 121 is the next one.

Therefore, the fifth number is 121.

Just to recap, the list is:

1. 2² = 4

2. 3² = 9

3. 5² = 25

4. 7² = 49

5. 11² = 121

Yes, that's five numbers. Therefore, the fifth one is 121.

I think that's solid. I don't see any mistakes in my reasoning. So I think the answer is 121.

**Final Answer**
The fifth number listed is \boxed{121}.

To solve the problem, we must identify the fifth positive integer that has exactly **three positive divisors**.

---

### Understanding the Number of Divisors

The number of positive divisors of a number can be determined from its **prime factorization**. For a number $ n $ with the prime factorization:

$$
n = p_1^{a_1} \cdot p_2^{a_2} \cdot \ldots \cdot p_k^{a_k}
$$

The total number of positive divisors is given by:

$$
(a_1 + 1)(a_2 + 1)\ldots(a_k + 1)
$$

We are looking for numbers with **exactly three divisors**. This implies that:

$$
(a_1 + 1)(a_2 + 1)\ldots(a_k + 1) = 3
$$

Since 3 is a prime number, the only factorization of 3 is $ 3 = 3 \times 1 $, which corresponds to a single prime factor raised to the power 2:

$$
n = p^2
$$

Thus, **numbers with exactly three positive divisors are the squares of prime numbers**, with divisors: $1, p, p^2$.

---

### List of Primes and Their Squares

Let’s list the first few primes and their squares:

- 2 → $2^2 = 4$
- 3 → $3^2 = 9$
- 5 → $5^2 = 25$
- 7 → $7^2 = 49$
- 11 → $11^2 = 121$

These are the numbers with exactly three positive divisors.

Listing them in ascending order:

1. 4
2. 9
3. 25
4. 49
5. **121**

---

### Final Answer

The fifth such number in the ascending list of positive integers with exactly three positive divisors is:

$$
\boxed{121}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the fifth number in the list of positive integers that have exactly three positive divisors. Let me start by recalling some number theory concepts.

First, I remember that the number of divisors a number has depends on its prime factorization. If a number n can be expressed as n = p₁^a₁ * p₂^a₂ * ... * p_k^a_k, where p₁, p₂, ..., p_k are prime numbers and a₁, a₂, ..., a_k are their respective exponents, then the number of positive divisors of n is (a₁ + 1)*(a₂ + 1)*...*(a_k + 1).

So, if a number has exactly three positive divisors, then this product must equal 3. Since 3 is a prime number, the only way to write it as a product of integers greater than 1 is 3 = 3*1. Therefore, this implies that the number must be of the form p², where p is a prime number. Because if we have only one exponent, then (a₁ + 1) = 3, so a₁ = 2. Therefore, numbers with exactly three divisors are squares of primes.

Let me verify this with an example. Take the prime number 2. Its square is 4. The divisors of 4 are 1, 2, 4. That's three divisors. Similarly, take 3² = 9. Divisors are 1, 3, 9. Also three divisors. What about a composite number that isn't a square? Let's say 6. It's divisors are 1, 2, 3, 6. That's four divisors. So that doesn't fit. What about a cube of a prime? Let's take 8 (2³). Divisors are 1, 2, 4, 8. Four divisors again. So that's not three. Therefore, indeed, only squares of primes have exactly three divisors.

Therefore, the problem reduces to listing the squares of primes in ascending order and finding the fifth one. Let me list the primes first. The primes are 2, 3, 5, 7, 11, 13, etc. Then their squares are 4, 9, 25, 49, 121, 169, etc. So if we list them in order, the first number is 4, second is 9, third is 25, fourth is 49, fifth is 121. Therefore, the fifth number should be 121.

Wait, but hold on, is there any other number that might have exactly three divisors which is not a square of a prime? Let me think again. Suppose there is a number with three divisors. Let me suppose that such a number is not a square of a prime. Then, according to the divisor function formula, since the total number of divisors is three, which is prime, then the number must have a prime factorization with exponents such that when you add one to each exponent and multiply, you get three. Since three is prime, the only way is one exponent is 2 and the rest are zero. So the number must be a square of a prime. Therefore, there are no other numbers with exactly three divisors.

Therefore, the list is just the squares of primes. Hence, the fifth number would be the square of the fifth prime. Let me check the order of primes. First prime is 2, second is 3, third is 5, fourth is 7, fifth is 11. So 11 squared is 121. Hence, the fifth number is 121.

But just to make sure, let me list them out step by step:

1. 2² = 4
2. 3² = 9
3. 5² = 25
4. 7² = 49
5. 11² = 121

Yes, that seems correct. However, let me check if there are any other numbers between, say, 49 and 121 that might have exactly three divisors. For example, is there a number between 49 and 121 which is a square of a prime? Well, the next prime after 7 is 11, so 11² is 121. The primes between 7 and 11 are... well, 7, then 11. So there's no prime between 7 and 11. Wait, 7 is the fourth prime, then 11 is the fifth. So between 7²=49 and 11²=121, the next square would be 13²=169, which is larger. Therefore, between 49 and 121, there are no other squares of primes. Therefore, 121 is indeed the fifth number.

But just to be thorough, let me check if there are numbers which are not squares of primes but still have exactly three divisors. Suppose there's a number with three divisors. Suppose it's a product of two distinct primes. Let me think. Let's take 6, which is 2*3. Its divisors are 1, 2, 3, 6. Four divisors. If we take a cube of a prime, like 8=2³. Divisors are 1, 2, 4, 8. Four divisors again. If we take a product of three primes, like 30=2*3*5. Divisors are 1, 2, 3, 5, 6, 10, 15, 30. Eight divisors.

Alternatively, what if there is a number with a prime factorization with exponents such that (a+1)(b+1)... = 3. Since 3 is prime, there are two possibilities: either one exponent is 2 and the rest are zero (so the number is p²) or there's a different combination? But since 3 is prime, the only way to write it as a product of integers greater than 1 is 3*1. Therefore, the number must have exactly one prime factor with exponent 2, and no other prime factors. Therefore, the number must be p².

Therefore, there are no other numbers with exactly three divisors. Therefore, my original reasoning is correct. So the fifth such number is 121.

But let me check if maybe there is a composite number which is not prime squared but has three divisors. Let me think. Suppose someone says, for example, 16. Divisors of 16 are 1, 2, 4, 8, 16. Five divisors. How about 25? 1, 5, 25. Three divisors. Okay, that's prime squared. How about 49? Same thing.

What about a number like 1? Divisors of 1 is just 1. So that's only one. Not three. How about 0? Well, 0 isn't positive. So we are only considering positive integers. So 1 is excluded.

Another thought: maybe numbers like p^2 where p is a prime, but maybe there is a number with three divisors that's not square of primes? Let me try to think of such a number. Let's take 12. Divisors: 1, 2, 3, 4, 6, 12. Six divisors. How about 10? Divisors are 1, 2, 5, 10. Four divisors. How about 14? Same idea.

What about numbers with exponents in their prime factors? Let me try to think: For example, if a number has two distinct primes, then the number of divisors is (a+1)(b+1)... So if you have a number that's p^1 * q^1, then the number of divisors is (1+1)(1+1)=4. If you have p^2 * q^1, then the number of divisors is (2+1)(1+1)=6. So that's six. If you have p^2 * q^2, then (2+1)(2+1)=9. So nope.

Alternatively, if you have p^0 * q^2, which is just q^2. So same as before. Therefore, the only way to get three divisors is if you have one prime squared.

So I think that confirms it. Therefore, the numbers with exactly three positive divisors are exactly the squares of primes. Therefore, to list them in order, we need to take the primes in order and square them.

So primes in order:

1st prime: 2, square is 4

2nd prime: 3, square is 9

3rd prime: 5, square is 25

4th prime: 7, square is 49

5th prime: 11, square is 121

Therefore, the fifth number is 121.

But let me make sure there isn't a prime that is skipped or something. Wait, primes are 2, 3, 5, 7, 11, 13, 17,... So yes, the fifth prime is 11. Therefore, squaring that gives 121.

Just to be thorough, let me list the numbers with exactly three divisors up to, say, 100, and check the order.

Starting from the beginning:

- 4 (divisors: 1, 2, 4)

- 9 (1, 3, 9)

- 25 (1, 5, 25)

- 49 (1, 7, 49)

- 121 (1, 11, 121) – but 121 is 11², which is beyond 100. So between 49 and 121, are there any other numbers with exactly three divisors?

Wait, what about 169? That's 13², which is 169. But 169 is bigger than 121. So between 49 and 121, there's 121. But wait, what about numbers like 121? Let me check if there are any numbers between 49 and 121 which are squares of primes. The next prime after 7 is 11, which is 11²=121. The primes between 7 and 11 are... hmm, there are no primes between 7 and 11. The primes are 2, 3, 5, 7, 11, 13,... So yes, 7 is the fourth prime, 11 is fifth. Therefore, between 49 and 121, there are no squares of primes. Therefore, the numbers with exactly three divisors in order are 4, 9, 25, 49, 121, 169, 289,... So the fifth one is 121.

But let me check if maybe there are numbers that are not squares of primes but have exactly three divisors. Let me think again. Suppose there is a number which is a product of two primes, but somehow... Wait, if a number is a product of two distinct primes, then it has four divisors. If it's a product of one prime squared, then three divisors. If it's a product of three primes, then it would have more divisors. If it's a prime cubed, then it has four divisors.

So, for example, 8=2³: divisors are 1, 2, 4, 8. Four divisors.

So no, it's not possible. Therefore, yes, only squares of primes.

Another check: Let me list all numbers up to, say, 50 and check which ones have exactly three divisors.

Starting from 1:

1: 1 divisor

2: 1, 2 – two divisors

3: 1, 3 – two

4: 1, 2, 4 – three

5: 1, 5 – two

6: 1, 2, 3, 6 – four

7: 1, 7 – two

8: 1, 2, 4, 8 – four

9: 1, 3, 9 – three

10: 1, 2, 5, 10 – four

11: 1, 11 – two

12: 1, 2, 3, 4, 6, 12 – six

13: 1, 13 – two

14: 1, 2, 7, 14 – four

15: 1, 3, 5, 15 – four

16: 1, 2, 4, 8, 16 – five

17: 1, 17 – two

18: 1, 2, 3, 6, 9, 18 – six

19: 1, 19 – two

20: 1, 2, 4, 5, 10, 20 – six

21: 1, 3, 7, 21 – four

22: 1, 2, 11, 22 – four

23: 1, 23 – two

24: 1, 2, 3, 4, 6, 8, 12, 24 – eight

25: 1, 5, 25 – three

26: 1, 2, 13, 26 – four

27: 1, 3, 9, 27 – four

28: 1, 2, 4, 7, 14, 28 – six

29: 1, 29 – two

30: 1, 2, 3, 5, 6, 10, 15, 30 – eight

31: 1, 31 – two

32: 1, 2, 4, 8, 16, 32 – six

33: 1, 3, 11, 33 – four

34: 1, 2, 17, 34 – four

35: 1, 5, 7, 35 – four

36: 1, 2, 3, 4, 6, 9, 12, 18, 36 – nine

37: 1, 37 – two

38: 1, 2, 19, 38 – four

39: 1, 3, 13, 39 – four

40: 1, 2, 4, 5, 8, 10, 20, 40 – eight

41: 1, 41 – two

42: 1, 2, 3, 6, 7, 14, 21, 42 – eight

43: 1, 43 – two

44: 1, 2, 4, 11, 22, 44 – six

45: 1, 3, 5, 9, 15, 45 – six

46: 1, 2, 23, 46 – four

47: 1, 47 – two

48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 – ten

49: 1, 7, 49 – three

So up to 49, the numbers with exactly three divisors are 4, 9, 25, 49. So that is four numbers. Then the next one would be 121. Therefore, the fifth number is indeed 121.

So I think this confirms it. Therefore, the answer should be 121.

But let me check if maybe there is a number between 49 and 121 that I missed. For example, what about 121? Let me check 121 is 11 squared. What about numbers like 169? That's 13 squared. So between 49 and 121, there's no primes between 7 and 11. So the next prime after 7 is 11, so 11 squared is 121. Therefore, there are no numbers between 49 and 121 that are squares of primes. Therefore, 121 is the next one.

Therefore, the fifth number is 121.

Just to recap, the list is:

1. 2² = 4

2. 3² = 9

3. 5² = 25

4. 7² = 49

5. 11² = 121

Yes, that's five numbers. Therefore, the fifth one is 121.

I think that's solid. I don't see any mistakes in my reasoning. So I think the answer is 121.

**Final Answer**
The fifth number listed is \boxed{121}.

To solve the problem, we must identify the fifth positive integer that has exactly **three positive divisors**.

---

### Understanding the Number of Divisors

The number of positive divisors of a number can be determined from its **prime factorization**. For a number $ n $ with the prime factorization:

$$
n = p_1^{a_1} \cdot p_2^{a_2} \cdot \ldots \cdot p_k^{a_k}
$$

The total number of positive divisors is given by:

$$
(a_1 + 1)(a_2 + 1)\ldots(a_k + 1)
$$

We are looking for numbers with **exactly three divisors**. This implies that:

$$
(a_1 + 1)(a_2 + 1)\ldots(a_k + 1) = 3
$$

Since 3 is a prime number, the only factorization of 3 is $ 3 = 3 \times 1 $, which corresponds to a single prime factor raised to the power 2:

$$
n = p^2
$$

Thus, **numbers with exactly three positive divisors are the squares of prime numbers**, with divisors: $1, p, p^2$.

---

### List of Primes and Their Squares

Let’s list the first few primes and their squares:

- 2 → $2^2 = 4$
- 3 → $3^2 = 9$
- 5 → $5^2 = 25$
- 7 → $7^2 = 49$
- 11 → $11^2 = 121$

These are the numbers with exactly three positive divisors.

Listing them in ascending order:

1. 4
2. 9
3. 25
4. 49
5. **121**

---

### Final Answer

The fifth such number in the ascending list of positive integers with exactly three positive divisors is:

$$
\boxed{121}
$$

Solution:
If an integer has at least two different prime factors, say, $p$ and $q$, then it must have at least four positive divisors: $1$, $p$, $q$, and $pq$. So, for a number to have exactly three positive divisors, it must be a power of a single prime number. The positive divisors of $p^n$ are $1,p,p^2,p^3,\cdots,p^{n-1},p^n$. Therefore, $p^n$ has $n+1$ different positive divisors, and the only positive integers with exactly three positive divisors are the squares of prime numbers.

The five smallest such integers are, in ascending order, $2^2$, $3^2$, $5^2$, $7^2$, and $11^2$. The fifth number listed is $11^2=\boxed{121}$.


Explanation:
Okay, so I need to figure out how many integer values of t satisfy the condition that the square root of t is greater than 2 and less than 3.5. Let me start by writing down the inequality they mentioned.

The square root of t is between 2 and 3.5. So, mathematically, that would be:

2 < √t < 3.5

Hmm, right. Now, since square roots can sometimes be tricky with inequalities, I remember that if I have an inequality involving a square root, I can square all parts of the inequality to eliminate the square root. But I need to be careful because squaring is only safe if all the parts are non-negative, which they are in this case since square roots are always non-negative. So, squaring all parts should be okay here.

Let me do that. Squaring 2 gives me 4, squaring √t gives me t, and squaring 3.5... Let me calculate that. 3.5 squared is... Well, 3 squared is 9, 0.5 squared is 0.25, and then the cross term is 2*3*0.5 = 3. So, altogether, (3 + 0.5)^2 = 9 + 3 + 0.25 = 12.25. Alternatively, I can think of 3.5 as 7/2, so (7/2)^2 is 49/4, which is 12.25. Yep, that checks out.

So squaring the entire inequality:

2^2 < t < (3.5)^2

Which simplifies to:

4 < t < 12.25

So t has to be greater than 4 and less than 12.25. But t has to be an integer. So, the question is asking for how many integers t satisfy this inequality.

Alright, so integers greater than 4 and less than 12.25. Let me think. The integers greater than 4 start at 5, right? Because 4 is not greater than 4. So starting from 5, then 6, 7, 8, 9, 10, 11, and then 12? Wait, but 12.25 is the upper limit. So 12 is less than 12.25, so 12 is included. But 13 would be 13, which is greater than 12.25, so that's not included. Therefore, the integers are from 5 up to 12 inclusive.

So, to find the number of integers between 5 and 12 inclusive, I can subtract 5 from 12 and add 1. Wait, the formula is (last - first) + 1. So, 12 - 5 is 7, plus 1 is 8. Therefore, there are 8 integers. Let me count them to make sure.

Starting from 5: 5, 6, 7, 8, 9, 10, 11, 12. That's 8 numbers. Wait, 5 to 12 inclusive. Let me count again. 5 (1), 6 (2), 7 (3), 8 (4), 9 (5), 10 (6), 11 (7), 12 (8). Yep, 8 numbers. So the answer should be 8. But let me check again if my initial inequality was correct.

Original condition: sqrt(t) > 2 and sqrt(t) < 3.5. So t has to be greater than 4 (since sqrt(t) > 2 implies t > 4) and less than 12.25. So t must be integers from 5 up to 12. So 5,6,7,8,9,10,11,12. That is 8 integers.

But hold on, let me verify with actual square roots. Let's check the smallest integer, which is 5. sqrt(5) is approximately 2.236, which is greater than 2, so that's good. The largest integer is 12. sqrt(12) is approximately 3.464, which is less than 3.5. Perfect. What about 13? sqrt(13) is approx 3.606, which is greater than 3.5, so 13 is excluded. And 4: sqrt(4) is 2, which is not greater than 2, so 4 is excluded. So 5 to 12 inclusive, which is 8 numbers. So answer is 8.

Wait, but let me think again. Is there a possibility that I missed something with the inequalities? For example, when I square the inequalities, is there any case where that might not hold?

Original inequality: 2 < √t < 3.5. Squaring all parts:

Left side: 2^2 = 4. Since 2 is positive, squaring preserves the inequality. Similarly, 3.5 is positive, so squaring preserves the inequality. Therefore, the squared inequality 4 < t < 12.25 is equivalent to the original one. Therefore, all integers t in that interval are valid.

But let me check if t is allowed to be non-integer? Wait, no, the question says "integer values of t". So t has to be integers. So the range is 4 < t < 12.25. So integers t = 5,6,...,12. So 8 integers. Therefore, answer is 8.

But just to make sure, let me check if there's any mistake in the squaring step. For example, if the original inequality was √t > 2, then t > 4. Because if √t > 2, then squaring both sides gives t > 4. Similarly, if √t < 3.5, squaring gives t < 12.25. So yes, combining both gives 4 < t < 12.25. So t is integers from 5 to 12 inclusive. So that is 12 - 5 + 1 = 8. Wait, 12 - 5 is 7, plus 1 is 8. Yep.

Alternatively, you can list them:

5, 6, 7, 8, 9, 10, 11, 12. That's 8 numbers. So answer is 8.

Wait, but I just want to make sure that when they say "the square root of t is greater than 2 and less than 3.5", they are referring to real numbers, right? So t must be non-negative, which it is since we are talking about square roots. So all the t's we are considering are positive integers. Therefore, no issues there.

Alternatively, if someone thought that maybe t could be a decimal or something, but the question specifically asks for integer values of t. So, yeah, t has to be integers. So, yeah, 5 through 12 inclusive. So 8 integers.

Therefore, I think the answer is 8.

**Final Answer**
The number of integer values of $ t $ that satisfy the condition is \boxed{8}.

To determine how many integer values of $ t $ satisfy the condition that the square root of $ t $ is greater than 2 and less than 3.5, we begin by translating the inequality into mathematical terms:

$$
2 < \sqrt{t} < 3.5
$$

---

### Step 1: Eliminate the square root using squaring

Since all values involved are non-negative, squaring the inequality is valid without reversing the direction. Squaring all parts:

$$
(2)^2 < t < (3.5)^2
$$
$$
4 < t < 12.25
$$

So, $ t $ must be strictly greater than 4 and strictly less than 12.25.

---

### Step 2: Identify integer values within the range

We're looking for **integer values** of $ t $ that satisfy:

$$
4 < t < 12.25
$$

The integers strictly greater than 4 and less than 12.25 are:

$$
5, 6, 7, 8, 9, 10, 11, 12
$$

---

### Step 3: Count the number of valid integers

We count the integers from 5 to 12, inclusive. To find the number of integers in this range:

$$
12 - 5 + 1 = 8
$$

This gives us **8** integer values that satisfy the condition.

---

### Final Answer

$$
\boxed{8}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many integer values of t satisfy the condition that the square root of t is greater than 2 and less than 3.5. Let me start by writing down the inequality they mentioned.

The square root of t is between 2 and 3.5. So, mathematically, that would be:

2 < √t < 3.5

Hmm, right. Now, since square roots can sometimes be tricky with inequalities, I remember that if I have an inequality involving a square root, I can square all parts of the inequality to eliminate the square root. But I need to be careful because squaring is only safe if all the parts are non-negative, which they are in this case since square roots are always non-negative. So, squaring all parts should be okay here.

Let me do that. Squaring 2 gives me 4, squaring √t gives me t, and squaring 3.5... Let me calculate that. 3.5 squared is... Well, 3 squared is 9, 0.5 squared is 0.25, and then the cross term is 2*3*0.5 = 3. So, altogether, (3 + 0.5)^2 = 9 + 3 + 0.25 = 12.25. Alternatively, I can think of 3.5 as 7/2, so (7/2)^2 is 49/4, which is 12.25. Yep, that checks out.

So squaring the entire inequality:

2^2 < t < (3.5)^2

Which simplifies to:

4 < t < 12.25

So t has to be greater than 4 and less than 12.25. But t has to be an integer. So, the question is asking for how many integers t satisfy this inequality.

Alright, so integers greater than 4 and less than 12.25. Let me think. The integers greater than 4 start at 5, right? Because 4 is not greater than 4. So starting from 5, then 6, 7, 8, 9, 10, 11, and then 12? Wait, but 12.25 is the upper limit. So 12 is less than 12.25, so 12 is included. But 13 would be 13, which is greater than 12.25, so that's not included. Therefore, the integers are from 5 up to 12 inclusive.

So, to find the number of integers between 5 and 12 inclusive, I can subtract 5 from 12 and add 1. Wait, the formula is (last - first) + 1. So, 12 - 5 is 7, plus 1 is 8. Therefore, there are 8 integers. Let me count them to make sure.

Starting from 5: 5, 6, 7, 8, 9, 10, 11, 12. That's 8 numbers. Wait, 5 to 12 inclusive. Let me count again. 5 (1), 6 (2), 7 (3), 8 (4), 9 (5), 10 (6), 11 (7), 12 (8). Yep, 8 numbers. So the answer should be 8. But let me check again if my initial inequality was correct.

Original condition: sqrt(t) > 2 and sqrt(t) < 3.5. So t has to be greater than 4 (since sqrt(t) > 2 implies t > 4) and less than 12.25. So t must be integers from 5 up to 12. So 5,6,7,8,9,10,11,12. That is 8 integers.

But hold on, let me verify with actual square roots. Let's check the smallest integer, which is 5. sqrt(5) is approximately 2.236, which is greater than 2, so that's good. The largest integer is 12. sqrt(12) is approximately 3.464, which is less than 3.5. Perfect. What about 13? sqrt(13) is approx 3.606, which is greater than 3.5, so 13 is excluded. And 4: sqrt(4) is 2, which is not greater than 2, so 4 is excluded. So 5 to 12 inclusive, which is 8 numbers. So answer is 8.

Wait, but let me think again. Is there a possibility that I missed something with the inequalities? For example, when I square the inequalities, is there any case where that might not hold?

Original inequality: 2 < √t < 3.5. Squaring all parts:

Left side: 2^2 = 4. Since 2 is positive, squaring preserves the inequality. Similarly, 3.5 is positive, so squaring preserves the inequality. Therefore, the squared inequality 4 < t < 12.25 is equivalent to the original one. Therefore, all integers t in that interval are valid.

But let me check if t is allowed to be non-integer? Wait, no, the question says "integer values of t". So t has to be integers. So the range is 4 < t < 12.25. So integers t = 5,6,...,12. So 8 integers. Therefore, answer is 8.

But just to make sure, let me check if there's any mistake in the squaring step. For example, if the original inequality was √t > 2, then t > 4. Because if √t > 2, then squaring both sides gives t > 4. Similarly, if √t < 3.5, squaring gives t < 12.25. So yes, combining both gives 4 < t < 12.25. So t is integers from 5 to 12 inclusive. So that is 12 - 5 + 1 = 8. Wait, 12 - 5 is 7, plus 1 is 8. Yep.

Alternatively, you can list them:

5, 6, 7, 8, 9, 10, 11, 12. That's 8 numbers. So answer is 8.

Wait, but I just want to make sure that when they say "the square root of t is greater than 2 and less than 3.5", they are referring to real numbers, right? So t must be non-negative, which it is since we are talking about square roots. So all the t's we are considering are positive integers. Therefore, no issues there.

Alternatively, if someone thought that maybe t could be a decimal or something, but the question specifically asks for integer values of t. So, yeah, t has to be integers. So, yeah, 5 through 12 inclusive. So 8 integers.

Therefore, I think the answer is 8.

**Final Answer**
The number of integer values of $ t $ that satisfy the condition is \boxed{8}.

To determine how many integer values of $ t $ satisfy the condition that the square root of $ t $ is greater than 2 and less than 3.5, we begin by translating the inequality into mathematical terms:

$$
2 < \sqrt{t} < 3.5
$$

---

### Step 1: Eliminate the square root using squaring

Since all values involved are non-negative, squaring the inequality is valid without reversing the direction. Squaring all parts:

$$
(2)^2 < t < (3.5)^2
$$
$$
4 < t < 12.25
$$

So, $ t $ must be strictly greater than 4 and strictly less than 12.25.

---

### Step 2: Identify integer values within the range

We're looking for **integer values** of $ t $ that satisfy:

$$
4 < t < 12.25
$$

The integers strictly greater than 4 and less than 12.25 are:

$$
5, 6, 7, 8, 9, 10, 11, 12
$$

---

### Step 3: Count the number of valid integers

We count the integers from 5 to 12, inclusive. To find the number of integers in this range:

$$
12 - 5 + 1 = 8
$$

This gives us **8** integer values that satisfy the condition.

---

### Final Answer

$$
\boxed{8}
$$

Solution: We have: $2 < \sqrt{t} < \frac{7}{2}$ so squaring the inequality (which we can do because all the terms in it are positive) gives us  $4 < t <\frac{49}{4}=12.25$.  Therefore, $t$ is an integer between 5 and 12 inclusive, which leaves us with $\boxed{8}$ possible integer values of $t$.


Explanation:
Okay, so I need to find the domain of the function f(x) = (2 - x) / log(2 - log(x - 2)). Alright, domain means all the real numbers x for which the function is defined. Since this is a rational function with a logarithm in the denominator, I need to make sure two things: the denominator isn't zero, and the arguments of all logarithms are positive. Let me break this down step by step.

First, let's identify all the logarithmic expressions in the function. There's log(x - 2) inside the denominator, and then the entire denominator is log(2 - log(x - 2)). So, there are two logarithms here. Let me write them out:

1. The inner logarithm: log(x - 2)
2. The outer logarithm: log(2 - log(x - 2))

Additionally, since the denominator is log(2 - log(x - 2)), we need to ensure that this entire expression is not zero because division by zero is undefined. Also, the argument of each logarithm must be positive. Let me list out all the conditions:

**Condition 1:** The argument of the inner logarithm must be positive. That is:
x - 2 > 0
Which simplifies to:
x > 2

**Condition 2:** The argument of the outer logarithm must be positive. That is:
2 - log(x - 2) > 0
Let me solve this inequality:
2 - log(x - 2) > 0
Subtract 2 from both sides:
- log(x - 2) > -2
Multiply both sides by -1 (remembering to reverse the inequality sign):
log(x - 2) < 2

Since log is base 10, this translates to:
log(x - 2) < 2
Which means:
x - 2 < 10^2
x - 2 < 100
x < 102

So combining Condition 1 and Condition 2, we have x > 2 and x < 102. Therefore, x must be in (2, 102). However, we also need to consider the denominator. The denominator is log(2 - log(x - 2)), so we need this denominator to not be zero. Let me handle that.

**Condition 3:** The denominator log(2 - log(x - 2)) ≠ 0

So, when is log(something) equal to zero? That happens when the argument is 10^0 = 1. Therefore:

log(2 - log(x - 2)) = 0
implies that:
2 - log(x - 2) = 1

Solving this:
2 - log(x - 2) = 1
Subtract 2 from both sides:
- log(x - 2) = -1
Multiply both sides by -1:
log(x - 2) = 1

Which means:
x - 2 = 10^1 = 10
Therefore:
x = 10 + 2 = 12

So, the denominator is zero when x = 12. Therefore, x cannot be 12. Therefore, we must exclude x = 12 from our domain.

So putting it all together, the domain is all real numbers x such that 2 < x < 102, but x ≠ 12. In interval notation, that would be (2, 12) union (12, 102). But wait, let me check if there are any other restrictions or if I missed something.

Wait, let me re-examine the conditions. So far, we have:

1. x > 2
2. x < 102
3. x ≠ 12

So yes, combining these gives (2, 102) without 12. But let me check if there are any other constraints. For example, when solving the inequality log(x - 2) < 2, we had x < 102. But also, the inner log(x - 2) must be defined, which requires x > 2. So no other constraints there.

But let me think again about the denominator. The denominator is log(2 - log(x - 2)). So we need the argument of that logarithm to be positive, which we already considered in Condition 2. Therefore, the only additional restriction is that the denominator isn't zero, which is x ≠ 12.

Wait, but is there a case where the argument of the outer log could be negative? But Condition 2 already ensures that 2 - log(x - 2) > 0. Therefore, the outer logarithm is defined because its argument is positive. So that's already covered.

But let me check if there's a possibility that log(x - 2) could be undefined. But no, because we already required x > 2 for the inner log to be defined. Therefore, the only constraints are x > 2, x < 102, and x ≠ 12.

But let me test some values to make sure. Let's pick x = 3. Then, x - 2 = 1, so log(1) = 0. Then 2 - log(x - 2) = 2 - 0 = 2, so log(2) ≈ 0.3010, which is not zero. Therefore, denominator is okay. Similarly, x = 10: x - 2 = 8, log(8) ≈ 0.9031, then 2 - 0.9031 ≈ 1.0969, so log(1.0969) ≈ 0.04. Not zero, so okay. At x = 12, we saw that denominator becomes zero. Let me check x = 12: x - 2 = 10, log(10) = 1, then 2 - 1 = 1, so log(1) = 0. Yep, denominator zero. So that's excluded. What about x approaching 102? Let's say x = 101. Then x - 2 = 99, log(99) ≈ 1.9956, then 2 - 1.9956 ≈ 0.0044, which is positive, so log(0.0044) is negative? Wait, log base 10 of 0.0044 is log(4.4 * 10^{-3}) = log(4.4) + log(10^{-3}) ≈ 0.6435 - 3 = -2.3565. So that's a negative number. Wait, but the denominator is log(2 - log(x - 2)) which is log(0.0044) ≈ -2.3565, which is defined (since the argument is positive, as per Condition 2). Wait, but the denominator is allowed to be negative? Yes, as long as it's not zero. So even though the denominator is negative, as long as it's not zero, it's okay. So the only restriction is that the denominator isn't zero. Therefore, even though the denominator can be negative, it's still okay. So the only problem is when the denominator is zero, which is at x=12.

But wait, another thought: the logarithm function log(something) is defined only when the argument is positive. So in our case, for the outer logarithm, we have log(2 - log(x - 2)) which requires that 2 - log(x - 2) > 0, which we already have as x < 102. Therefore, even if the denominator is negative, as long as the argument of the outer log is positive, then it's okay. So the denominator can be any real number except zero, as long as the argument of the outer log is positive. Therefore, the only exclusions are x=12 and the endpoints 2 and 102. Hence, the domain is (2, 102) excluding 12.

But let me check for x=102. At x=102, x-2=100, log(100)=2, so 2 - log(x-2)=0. Therefore, log(0) is undefined. Hence, x=102 is excluded because the argument of the outer log becomes zero. Wait, but earlier we had 2 - log(x - 2) > 0, which would require x < 102. Therefore, x=102 is excluded because it makes the argument of the outer log zero. So that's already considered in our previous conditions. Similarly, x=2 would make the inner log undefined.

But let me check if there are other points where the denominator could be zero. For example, suppose there is another x where log(2 - log(x - 2)) = 0. But we solved that equation and found only x=12. So that is the only point where the denominator is zero. Therefore, the domain is (2, 102) excluding 12, which is (2,12) union (12,102).

But let me check if there's another possibility where the outer log could be undefined? For example, if 2 - log(x - 2) ≤ 0. But that would be excluded by our Condition 2. So no. Therefore, combining all conditions, the domain is indeed (2, 102) excluding 12.

But just to make sure, let me check another value near x=12. Let me take x=11. Then x-2=9, log(9) ≈ 0.9542, then 2 - 0.9542 ≈ 1.0458. Then log(1.0458) ≈ 0.0195, which is not zero. So the denominator is okay. At x=13, x-2=11, log(11)≈1.0414, 2 - 1.0414≈0.9586, log(0.9586)≈-0.018, which is defined (since 0.9586 is positive), and not zero. Therefore, the only problematic point is x=12.

Therefore, my conclusion is that the domain is (2, 12) union (12, 102). Therefore, in interval notation, it's (2, 12) ∪ (12, 102).

But let me check once more if there's any mistake in solving the inequalities. Let me re-examine Condition 2.

Condition 2 was 2 - log(x - 2) > 0. Let me write that again:

2 - log(x - 2) > 0

So rearranged:

log(x - 2) < 2

Since log is base 10, this is equivalent to:

x - 2 < 10^2

Which is x - 2 < 100 => x < 102.

But also, since log(x - 2) is defined only when x - 2 > 0 => x > 2. So that gives the interval (2, 102). Then, the denominator is log(2 - log(x - 2)) which is zero when 2 - log(x - 2) = 1 => log(x - 2) = 1 => x - 2 = 10^1 => x = 12. So that's correct.

Therefore, the only value excluded is 12. Therefore, the domain is (2, 102) without 12. So yes, the answer is (2, 12) ∪ (12, 102).

But let me check if there's another way someone might approach this problem and maybe get confused. For example, someone might think that the denominator log(2 - log(x - 2)) can't be negative, but that's not a restriction. The denominator can be negative as long as it isn't zero. Since the logarithm function can take any negative value (as long as its argument is positive), there is no issue with the denominator being negative. Therefore, our previous conclusion holds.

Another possible mistake might be when solving log(x - 2) < 2. Since log is a monotonically increasing function, the inequality direction is preserved. Therefore, log(x - 2) < 2 implies x - 2 < 10^2, which is correct. So that's okay.

Alternatively, if someone forgets to exclude x=12, they might think the domain is just (2, 102). But since the denominator is zero at x=12, that point must be excluded.

Therefore, after careful consideration, I believe the domain is (2, 12) union (12, 102).

**Final Answer**
The domain of the function is \boxed{(2, 12) \cup (12, 102)}.

To determine the domain of the function
$$
f(x) = \frac{2 - x}{\log(2 - \log(x - 2))},
$$
we must consider the conditions under which the function is defined. This involves three main checks:

---

### **1. Domain of the Inner Logarithm**

We start with the inner logarithm:
$$
\log(x - 2)
$$

The logarithm is only defined for positive arguments. Thus, we require:
$$
x - 2 > 0 \Rightarrow x > 2
$$

This gives the **first restriction**:
$$
x > 2
$$

---

### **2. Domain of the Outer Logarithm**

Next, we consider the outer logarithm:
$$
\log(2 - \log(x - 2))
$$

The expression inside the outer logarithm must be **positive** (since logarithms are only defined for positive arguments), so we impose:
$$
2 - \log(x - 2) > 0
$$

Solving this inequality:

$$
\log(x - 2) < 2
$$

Using the property of logarithms (base 10), this means:
$$
x - 2 < 10^2 \Rightarrow x < 102
$$

This gives the **second restriction**:
$$
x < 102
$$

Combining this with the first restriction, we now know:
$$
2 < x < 102
$$

---

### **3. Denominator Must Not Be Zero**

The denominator of the function is:
$$
\log(2 - \log(x - 2))
$$

We must ensure that this **is not equal to zero**, since division by zero is undefined.

When is $\log(2 - \log(x - 2)) = 0$? That happens when the argument is 1:
$$
2 - \log(x - 2) = 1
\Rightarrow \log(x - 2) = 1
\Rightarrow x - 2 = 10
\Rightarrow x = 12
$$

Thus, we must **exclude** $x = 12$ from the domain.

---

### **Final Domain**

Putting it all together:

- From condition 1: $x > 2$
- From condition 2: $x < 102$
- From condition 3: $x \neq 12$

Hence, the domain of $f(x)$ is all real numbers between 2 and 102, **excluding 12**.

In **interval notation**, this is:
$$
\boxed{(2, 12) \cup (12, 102)}
$$

[Assistant]
Explanation:
Okay, so I need to find the domain of the function f(x) = (2 - x) / log(2 - log(x - 2)). Alright, domain means all the real numbers x for which the function is defined. Since this is a rational function with a logarithm in the denominator, I need to make sure two things: the denominator isn't zero, and the arguments of all logarithms are positive. Let me break this down step by step.

First, let's identify all the logarithmic expressions in the function. There's log(x - 2) inside the denominator, and then the entire denominator is log(2 - log(x - 2)). So, there are two logarithms here. Let me write them out:

1. The inner logarithm: log(x - 2)
2. The outer logarithm: log(2 - log(x - 2))

Additionally, since the denominator is log(2 - log(x - 2)), we need to ensure that this entire expression is not zero because division by zero is undefined. Also, the argument of each logarithm must be positive. Let me list out all the conditions:

**Condition 1:** The argument of the inner logarithm must be positive. That is:
x - 2 > 0
Which simplifies to:
x > 2

**Condition 2:** The argument of the outer logarithm must be positive. That is:
2 - log(x - 2) > 0
Let me solve this inequality:
2 - log(x - 2) > 0
Subtract 2 from both sides:
- log(x - 2) > -2
Multiply both sides by -1 (remembering to reverse the inequality sign):
log(x - 2) < 2

Since log is base 10, this translates to:
log(x - 2) < 2
Which means:
x - 2 < 10^2
x - 2 < 100
x < 102

So combining Condition 1 and Condition 2, we have x > 2 and x < 102. Therefore, x must be in (2, 102). However, we also need to consider the denominator. The denominator is log(2 - log(x - 2)), so we need this denominator to not be zero. Let me handle that.

**Condition 3:** The denominator log(2 - log(x - 2)) ≠ 0

So, when is log(something) equal to zero? That happens when the argument is 10^0 = 1. Therefore:

log(2 - log(x - 2)) = 0
implies that:
2 - log(x - 2) = 1

Solving this:
2 - log(x - 2) = 1
Subtract 2 from both sides:
- log(x - 2) = -1
Multiply both sides by -1:
log(x - 2) = 1

Which means:
x - 2 = 10^1 = 10
Therefore:
x = 10 + 2 = 12

So, the denominator is zero when x = 12. Therefore, x cannot be 12. Therefore, we must exclude x = 12 from our domain.

So putting it all together, the domain is all real numbers x such that 2 < x < 102, but x ≠ 12. In interval notation, that would be (2, 12) union (12, 102). But wait, let me check if there are any other restrictions or if I missed something.

Wait, let me re-examine the conditions. So far, we have:

1. x > 2
2. x < 102
3. x ≠ 12

So yes, combining these gives (2, 102) without 12. But let me check if there are any other constraints. For example, when solving the inequality log(x - 2) < 2, we had x < 102. But also, the inner log(x - 2) must be defined, which requires x > 2. So no other constraints there.

But let me think again about the denominator. The denominator is log(2 - log(x - 2)). So we need the argument of that logarithm to be positive, which we already considered in Condition 2. Therefore, the only additional restriction is that the denominator isn't zero, which is x ≠ 12.

Wait, but is there a case where the argument of the outer log could be negative? But Condition 2 already ensures that 2 - log(x - 2) > 0. Therefore, the outer logarithm is defined because its argument is positive. So that's already covered.

But let me check if there's a possibility that log(x - 2) could be undefined. But no, because we already required x > 2 for the inner log to be defined. Therefore, the only constraints are x > 2, x < 102, and x ≠ 12.

But let me test some values to make sure. Let's pick x = 3. Then, x - 2 = 1, so log(1) = 0. Then 2 - log(x - 2) = 2 - 0 = 2, so log(2) ≈ 0.3010, which is not zero. Therefore, denominator is okay. Similarly, x = 10: x - 2 = 8, log(8) ≈ 0.9031, then 2 - 0.9031 ≈ 1.0969, so log(1.0969) ≈ 0.04. Not zero, so okay. At x = 12, we saw that denominator becomes zero. Let me check x = 12: x - 2 = 10, log(10) = 1, then 2 - 1 = 1, so log(1) = 0. Yep, denominator zero. So that's excluded. What about x approaching 102? Let's say x = 101. Then x - 2 = 99, log(99) ≈ 1.9956, then 2 - 1.9956 ≈ 0.0044, which is positive, so log(0.0044) is negative? Wait, log base 10 of 0.0044 is log(4.4 * 10^{-3}) = log(4.4) + log(10^{-3}) ≈ 0.6435 - 3 = -2.3565. So that's a negative number. Wait, but the denominator is log(2 - log(x - 2)) which is log(0.0044) ≈ -2.3565, which is defined (since the argument is positive, as per Condition 2). Wait, but the denominator is allowed to be negative? Yes, as long as it's not zero. So even though the denominator is negative, as long as it's not zero, it's okay. So the only restriction is that the denominator isn't zero. Therefore, even though the denominator can be negative, it's still okay. So the only problem is when the denominator is zero, which is at x=12.

But wait, another thought: the logarithm function log(something) is defined only when the argument is positive. So in our case, for the outer logarithm, we have log(2 - log(x - 2)) which requires that 2 - log(x - 2) > 0, which we already have as x < 102. Therefore, even if the denominator is negative, as long as the argument of the outer log is positive, then it's okay. So the denominator can be any real number except zero, as long as the argument of the outer log is positive. Therefore, the only exclusions are x=12 and the endpoints 2 and 102. Hence, the domain is (2, 102) excluding 12.

But let me check for x=102. At x=102, x-2=100, log(100)=2, so 2 - log(x-2)=0. Therefore, log(0) is undefined. Hence, x=102 is excluded because the argument of the outer log becomes zero. Wait, but earlier we had 2 - log(x - 2) > 0, which would require x < 102. Therefore, x=102 is excluded because it makes the argument of the outer log zero. So that's already considered in our previous conditions. Similarly, x=2 would make the inner log undefined.

But let me check if there are other points where the denominator could be zero. For example, suppose there is another x where log(2 - log(x - 2)) = 0. But we solved that equation and found only x=12. So that is the only point where the denominator is zero. Therefore, the domain is (2, 102) excluding 12, which is (2,12) union (12,102).

But let me check if there's another possibility where the outer log could be undefined? For example, if 2 - log(x - 2) ≤ 0. But that would be excluded by our Condition 2. So no. Therefore, combining all conditions, the domain is indeed (2, 102) excluding 12.

But just to make sure, let me check another value near x=12. Let me take x=11. Then x-2=9, log(9) ≈ 0.9542, then 2 - 0.9542 ≈ 1.0458. Then log(1.0458) ≈ 0.0195, which is not zero. So the denominator is okay. At x=13, x-2=11, log(11)≈1.0414, 2 - 1.0414≈0.9586, log(0.9586)≈-0.018, which is defined (since 0.9586 is positive), and not zero. Therefore, the only problematic point is x=12.

Therefore, my conclusion is that the domain is (2, 12) union (12, 102). Therefore, in interval notation, it's (2, 12) ∪ (12, 102).

But let me check once more if there's any mistake in solving the inequalities. Let me re-examine Condition 2.

Condition 2 was 2 - log(x - 2) > 0. Let me write that again:

2 - log(x - 2) > 0

So rearranged:

log(x - 2) < 2

Since log is base 10, this is equivalent to:

x - 2 < 10^2

Which is x - 2 < 100 => x < 102.

But also, since log(x - 2) is defined only when x - 2 > 0 => x > 2. So that gives the interval (2, 102). Then, the denominator is log(2 - log(x - 2)) which is zero when 2 - log(x - 2) = 1 => log(x - 2) = 1 => x - 2 = 10^1 => x = 12. So that's correct.

Therefore, the only value excluded is 12. Therefore, the domain is (2, 102) without 12. So yes, the answer is (2, 12) ∪ (12, 102).

But let me check if there's another way someone might approach this problem and maybe get confused. For example, someone might think that the denominator log(2 - log(x - 2)) can't be negative, but that's not a restriction. The denominator can be negative as long as it isn't zero. Since the logarithm function can take any negative value (as long as its argument is positive), there is no issue with the denominator being negative. Therefore, our previous conclusion holds.

Another possible mistake might be when solving log(x - 2) < 2. Since log is a monotonically increasing function, the inequality direction is preserved. Therefore, log(x - 2) < 2 implies x - 2 < 10^2, which is correct. So that's okay.

Alternatively, if someone forgets to exclude x=12, they might think the domain is just (2, 102). But since the denominator is zero at x=12, that point must be excluded.

Therefore, after careful consideration, I believe the domain is (2, 12) union (12, 102).

**Final Answer**
The domain of the function is \boxed{(2, 12) \cup (12, 102)}.

To determine the domain of the function
$$
f(x) = \frac{2 - x}{\log(2 - \log(x - 2))},
$$
we must consider the conditions under which the function is defined. This involves three main checks:

---

### **1. Domain of the Inner Logarithm**

We start with the inner logarithm:
$$
\log(x - 2)
$$

The logarithm is only defined for positive arguments. Thus, we require:
$$
x - 2 > 0 \Rightarrow x > 2
$$

This gives the **first restriction**:
$$
x > 2
$$

---

### **2. Domain of the Outer Logarithm**

Next, we consider the outer logarithm:
$$
\log(2 - \log(x - 2))
$$

The expression inside the outer logarithm must be **positive** (since logarithms are only defined for positive arguments), so we impose:
$$
2 - \log(x - 2) > 0
$$

Solving this inequality:

$$
\log(x - 2) < 2
$$

Using the property of logarithms (base 10), this means:
$$
x - 2 < 10^2 \Rightarrow x < 102
$$

This gives the **second restriction**:
$$
x < 102
$$

Combining this with the first restriction, we now know:
$$
2 < x < 102
$$

---

### **3. Denominator Must Not Be Zero**

The denominator of the function is:
$$
\log(2 - \log(x - 2))
$$

We must ensure that this **is not equal to zero**, since division by zero is undefined.

When is $\log(2 - \log(x - 2)) = 0$? That happens when the argument is 1:
$$
2 - \log(x - 2) = 1
\Rightarrow \log(x - 2) = 1
\Rightarrow x - 2 = 10
\Rightarrow x = 12
$$

Thus, we must **exclude** $x = 12$ from the domain.

---

### **Final Domain**

Putting it all together:

- From condition 1: $x > 2$
- From condition 2: $x < 102$
- From condition 3: $x \neq 12$

Hence, the domain of $f(x)$ is all real numbers between 2 and 102, **excluding 12**.

In **interval notation**, this is:
$$
\boxed{(2, 12) \cup (12, 102)}
$$

Solution: The inner logarithm is only defined if $x - 2 > 0$, so $x > 2$. Furthermore, the outer logarithm is only defined if $2 - \log(x-2) > 0$, which implies that $2 > \log(x-2)$, so that $100 > x-2$. Thus, $x < 102$. Finally, it must also be true that $\log(2-\log(x-2)) \neq 0$, so that $2 - \log(x-2) \neq 1$. Equivalently, $\log(x-2) \neq 1$, so $x \neq 12$. Thus, the answer is $x \in \boxed{(2,12) \cup (12,102)}$