Okay, so I need to find the domain of the function f(x) = (2 - x) / log(2 - log(x - 2)). Hmm, the domain is all the real numbers x where the function is defined. That means I need to make sure that the denominator isn't zero, the logarithm arguments are positive, and so on. Let me break this down step by step.

First, let's look at the denominator: log(2 - log(x - 2)). Since logarithm is only defined for positive arguments, the expression inside the logarithm, which is 2 - log(x - 2), must be greater than 0. So, 2 - log(x - 2) > 0. Let me note that down.

Also, the argument of the inner logarithm is x - 2. Since log is defined only for positive numbers, x - 2 must be greater than 0. So, x - 2 > 0, which simplifies to x > 2. That's the first condition.

So far, x must be greater than 2. But we also have the condition on the outer logarithm. Let's handle that. The outer logarithm's argument is 2 - log(x - 2), which must be greater than 0. So:

2 - log(x - 2) > 0  
=> log(x - 2) < 2

Since log here is base 10, this inequality translates to:

x - 2 < 10^2  
x - 2 < 100  
x < 102

Therefore, combining this with the previous condition x > 2, we have x ∈ (2, 102). But wait, there's more. We also need to ensure that the denominator log(2 - log(x - 2)) is not zero because division by zero is undefined. So, log(2 - log(x - 2)) ≠ 0.

When does log(2 - log(x - 2)) equal zero? That happens when the argument is 1, since log(1) = 0. Therefore:

2 - log(x - 2) = 1  
=> log(x - 2) = 1  
=> x - 2 = 10^1  
=> x - 2 = 10  
=> x = 12

So, x cannot be 12. Therefore, we need to exclude x = 12 from our domain. So, the domain so far is (2, 12) ∪ (12, 102).

But let me check if there are any other restrictions. The numerator is (2 - x), which is a polynomial and defined for all real numbers, so no issues there. The denominator's logarithm is already handled. So, the only restrictions are x > 2, x < 102, and x ≠ 12.

Wait, but let me verify the step where I converted log(x - 2) < 2 to x - 2 < 100. Since log base 10 is a strictly increasing function, yes, that's correct. So, if log(a) < b, then a < 10^b. So, that step is valid.

Another thing to check: the argument inside the inner logarithm, x - 2, must be positive, so x > 2, which we already have. So, that's covered.

But let me also check the expression 2 - log(x - 2). We already set that greater than 0, leading to log(x - 2) < 2, which gives x < 102. So, that's correct.

Now, the denominator cannot be zero, which we addressed by excluding x = 12. So, putting it all together, the domain is (2, 12) ∪ (12, 102). But let me double-check if there's any other point where the denominator might be undefined or the expression undefined.

Wait, let's consider the inner logarithm again. The inner log is log(x - 2), which is defined for x > 2. Then, the outer log is log(2 - log(x - 2)), which requires that 2 - log(x - 2) > 0, which we handled by x - 2 < 100, so x < 102. Also, the denominator cannot be zero, which we dealt with.

Is there a possibility that the denominator could be negative? Because if the denominator is negative, the function f(x) would be negative, but the problem is only about the domain, so negative values are allowed as long as the denominator isn't zero. So, the denominator can be positive or negative, as long as it's not zero. So, the only restrictions are x > 2, x < 102, and x ≠ 12.

Therefore, the domain is all real numbers between 2 and 102, excluding 12. So, in interval notation, that's (2, 12) ∪ (12, 102).

But let me test a value in each interval to make sure.

First, take x = 3, which is in (2, 12). Then, log(3 - 2) = log(1) = 0. Then, 2 - 0 = 2, so log(2) is defined and not zero. So, denominator is log(2) ≈ 0.3010, which is fine. So, x=3 is allowed.

Next, x = 11. Then, log(11 - 2) = log(9) ≈ 0.9542. Then, 2 - 0.9542 ≈ 1.0458. Then, log(1.0458) ≈ 0.0194, which is not zero. So, denominator is defined and non-zero. So, x=11 is allowed.

Now, x=12. Let's check. Then, log(12 - 2) = log(10) = 1. Then, 2 - 1 = 1. Then, log(1) = 0. So, denominator is zero, which is undefined. Hence, x=12 is excluded.

Take x=100, which is in (12, 102). log(100 - 2) = log(98) ≈ 1.9912. Then, 2 - 1.9912 ≈ 0.0088. Then, log(0.0088) ≈ -2.052, which is defined and non-zero. So, denominator is defined. So, x=100 is allowed.

Take x=102. Then, log(102 - 2) = log(100) = 2. Then, 2 - 2 = 0. Then, log(0) is undefined. So, x=102 is excluded. So, that's why x must be less than 102.

Similarly, x approaching 2 from the right: log(x-2) approaches log(0+) which is -infinity. Then, 2 - (-infinity) is infinity. Then, log(infinity) is infinity. So, denominator approaches infinity, which is fine. So, x approaching 2 is allowed, but x=2 is excluded because x must be greater than 2.

So, all the test cases check out. Therefore, the domain is (2, 12) union (12, 102).

But let me check if there's any other point where the denominator could be zero. We already found x=12. Let's solve 2 - log(x - 2) = 10^k for some k? Wait, no. The denominator is log(2 - log(x - 2)) = 0. So, 2 - log(x - 2) = 10^0 = 1. So, that's the only solution. So, x=12 is the only point where the denominator is zero. Therefore, that's the only exclusion.

Therefore, the domain is x ∈ (2, 12) ∪ (12, 102). So, in interval notation, that's two intervals: from 2 to 12, not including 12, and from 12 to 102, not including 12 and 102.

I think that's the final answer. Let me just recap:

1. Inner logarithm requires x - 2 > 0 ⇒ x > 2.
2. Outer logarithm requires 2 - log(x - 2) > 0 ⇒ log(x - 2) < 2 ⇒ x - 2 < 100 ⇒ x < 102.
3. Denominator cannot be zero: log(2 - log(x - 2)) ≠ 0 ⇒ 2 - log(x - 2) ≠ 1 ⇒ log(x - 2) ≠ 1 ⇒ x ≠ 10 + 2 = 12.

Therefore, combining all these conditions, x must be in (2, 12) ∪ (12, 102). Yep, that seems right.
Therefore, the domain of the function $f(x)$ is $\boxed{(2, 12) \cup (12, 102)}$.
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