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 $x \in \boxed{(2,12) \cup (12,102)}$
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  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.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: (2,12) \cup (12,102)
extra_info.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.
extra_info.split: test