To find the domain of the function $f(x) = \frac{1}{\log_3(2 - x)}$, we need to ensure that the denominator is valid. Specifically:

1. Logarithm Validity: For $\log_3(2 - x)$ to be defined, the argument $2 - x$ must be positive: $2 - x > 0 \quad \Rightarrow \quad x < 2.$

2. Non-zero Denominator: The logarithm cannot be zero because it is in the denominator. This means $\log_3(2 - x) \neq 0$. Since $\log_3(1) = 0$, we want to exclude values of $x$ such that $2 - x = 1$: $2 - x = 1 \quad \Rightarrow \quad x = 1.$

Thus, the domain of $f(x)$ is $(-\infty, 1) \cup (1, 2)$, excluding $x = 1$, where the denominator would be zero.

Summary of Domain:

$x \in (-\infty, 1) \cup (1, 2).$

Would you like more details on any part of the process?

Here are five related questions to consider:

1. What is the domain of $f(x) = \log_5(4x - 7)$?
2. How does the base of a logarithm affect its properties?
3. Can logarithmic functions have negative arguments?
4. What would the range of this function be?
5. How would a vertical shift affect the domain of $f(x)$?

Tip: Always ensure the argument of a logarithmic function is positive before solving for the domain.