<Question>
What is the domain of the function $f(x) = \frac{2-x}{\log(2-\log(x-2))}$, where $\log$ is the base $10$ logarithm function? Express your answer in interval notation.
</Question>

<think>
<Plan_1>
Okay, so I need to find the domain of the function f(x) = (2 - x) / log(10)(2 - log10(x - 2)). Hmm, the domain of a function includes all the real numbers x for which the function is defined. Since this function has a logarithm in the denominator, I need to make sure that the argument of the logarithm is positive, and also that the denominator isn't zero. Plus, the inner logarithm log10(x - 2) requires that x - 2 be positive. Let me break this down step by step.

First, let's look at the innermost part: log10(x - 2). The logarithm is defined when its argument is positive.
So, x - 2 > 0, which means x > 2. That's the first condition.

Next, the argument of the outer logarithm is 2 - log10(x - 2). For the logarithm to be defined, this expression must be positive.
So, 2 - log10(x - 2) > 0. Let me solve that inequality.

<prompt_1.1> Solve the inequality 2 - log10(x - 2) > 0 as per the above plan. </prompt_1.1>
-----

But wait, let me check with an example. Suppose x = 3.

<prompt_1.2> Verify the domain by checking specific values of x, such as x = 3 and x = 111. </prompt_1.2>

Wait, but what about x approaching 2 from above?

<prompt_1.3> Analyze the behavior of the function as x approaches the boundaries of the domain. </prompt_1.3>
-----

Let me just check if there's any other possible source of error. For example, could the denominator ever be zero within the domain?

<prompt_1.4> Check if the denominator can be zero within the domain. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's rewrite this as log10(x - 2) < 2. Because if I move the log term to the other side, it becomes log10(x - 2) < 2. Now, solving log10(x - 2) < 2. Converting this logarithmic inequality to exponential form: x - 2 < 10^2, which is x - 2 < 100. So, x < 102. But I also have to remember that x - 2 > 0 from the inner logarithm, so combining these, x is in (2, 102).
</execution_1.1>

<execution_1.2>
Then log10(3 - 2) = log10(1) = 0. So 2 - 0 = 2, which is positive. So denominator is log10(2) ≈ 0.3010, so f(3) = (2 - 3)/log10(2) = -1 / 0.3010 ≈ -3.32, which is defined.
</execution_1.2>

<execution_1.3>
If x approaches 2 from above, say x = 2.1, then x - 2 = 0.1, so log10(0.1) = -1. Then 2 - (-1) = 3, which is positive. So the outer log is defined. The denominator is log10(3) ≈ 0.477, so the function is defined. 

If x approaches 100, then x - 2 = 98, log10(98) ≈ 1.991, so 2 - 1.991 ≈ 0.009, which is positive but very small. So log10(0.009) ≈ -2.045, which is negative. So the denominator approaches a negative number, but since it's not zero, the function is still defined.
</execution_1.3>

<execution_1.4>
Suppose log10(2 - log10(x - 2)) = 0. Then, 2 - log10(x - 2) = 1, since log10(1) = 0. So, 2 - log10(x - 2) = 1 => log10(x - 2) = 1 => x - 2 = 10^1 = 10 => x = 12. Is x = 12 in the domain? Let's check:

For x = 12, check the inner log: log10(12 - 2) = log10(10) = 1. Then, 2 - 1 = 1, which is positive, so the outer log is defined. Then, log10(1) = 0, so denominator is 0. Therefore, x = 12 is excluded from the domain, which matches our earlier result. So, we need to exclude x = 12 from the interval.
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, but hold on.
Wait, the original function is log(2 - log(x - 2)), and the base is 10. So, the steps I took should be correct. Let me confirm:

<prompt_2.1> Verify the steps taken to solve the inequality and confirm the domain. </prompt_2.1>
-----

Based on execution_1.2:
What if x approaches 102 from below?

<prompt_2.2> Analyze the behavior of the function as x approaches the upper boundary of the domain. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
1. Inner log: x - 2 > 0 => x > 2.

2. Outer log: 2 - log10(x - 2) > 0 => log10(x - 2) < 2 => x - 2 < 100 => x < 102. So x must be between 2 and 102. But wait, is that all? Wait, the denominator can't be zero, so we have to make sure that log10(2 - log10(x - 2)) ≠ 0. Because if the denominator is zero, the function is undefined. So when does log10(2 - log10(x - 2)) = 0? That would be when 2 - log10(x - 2) = 1, because log10(1) = 0. So solving 2 - log10(x - 2) = 1 => log10(x - 2) = 1 => x - 2 = 10^1 = 10 => x = 12. Therefore, x = 12 must be excluded from the domain.
</execution_2.1>

<execution_2.2>
Let's say x = 101. Then x - 2 = 99, log10(99) ≈ 1.9956, so 2 - 1.9956 ≈ 0.0044. Then log10(0.0044) ≈ -2.356, so denominator is a negative number, but not zero. So the function is defined, just negative.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So, combining all the conditions:

<prompt_3.1> Combine all conditions to find the domain of the function. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
1. x - 2 > 0 => x > 2.

2. 2 - log10(x - 2) > 0 => x < 102.

3. x ≠ 12 (from the denominator being zero).

Therefore, the domain is all x in (2, 102) except x = 12. So in interval notation, that would be (2, 12) union (12, 102).
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Wait, but let me confirm again. The denominator is log10(2 - log10(x - 2)). The denominator cannot be zero, so 2 - log10(x - 2) ≠ 1, which as we saw leads to x = 12. So x = 12 must be excluded.
-----

Based on execution_1.4:
Then, the domain is (2, 12) union (12, 102).
-----

Based on execution_3.1:
But let me check another value in (2,12). Let's take x = 10.

<prompt_4.1> Verify the domain by checking specific values of x in the intervals (2, 12) and (12, 102). </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Then x - 2 = 8, log10(8) ≈ 0.9031, so 2 - 0.9031 ≈ 1.0969. Then log10(1.0969) ≈ 0.0403, so denominator is positive. So the function is defined. If x = 12, as before, denominator is zero. If x approaches 12 from below, say x = 11.9, then x - 2 = 9.9, log10(9.9) ≈ 0.9956, 2 - 0.9956 ≈ 1.0044, log10(1.0044) ≈ 0.0019, so denominator is a small positive number, so function is defined. From above, x = 12.1: x - 2 = 10.1, log10(10.1) ≈ 1.0043, 2 - 1.0043 ≈ 0.9957, log10(0.9957) ≈ -0.0021, so denominator is negative, but not zero.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
So, yes, x = 12 is the only point in (2, 102) where the denominator is zero, so we have to exclude it.
Therefore, the domain is (2, 12) union (12, 102).

Let me just check if there are any other points where the denominator could be zero.

<prompt_5.1> Confirm the domain and check for other points where the denominator could be zero. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Suppose 2 - log10(x - 2) = 10^k, where k is some number. Wait, the denominator is log10(2 - log10(x - 2)), so the denominator is zero when 2 - log10(x - 2) = 1, as we saw. So only when 2 - log10(x - 2) = 1. So that equation gives x = 12 as the only solution. Therefore, only x = 12 needs to be excluded.

Therefore, the domain is (2, 12) union (12, 102).
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
But let me check the endpoints again.

<prompt_6.1> Perform a final verification of the domain and conclude the solution. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
If x = 2, the inner log is undefined. If x = 102, the outer log's argument is zero, which is undefined. So yes, open intervals at both ends. And x = 12 is excluded because denominator is zero. So the final answer should be (2, 12) ∪ (12, 102).
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
**Final Answer**
The domain of the function is \boxed{(2, 12) \cup (12, 102)}.
</Plan_7>

<Execution_7>
No prompt to execute.
</Execution_7>

<Final_answer>
Thus, the domain of $f(x)$ is the union of two intervals: $(2, 12)$ and $(12, 102)$, which can be written in interval notation as $\boxed{(2, 12) \cup (12, 102)}$.
</Final_answer>
</think>