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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.

Find the domain of the function f(x) = 1 / log_3(2 - x).

Learn how to find the domain of the function f(x) = 1 / log_3(2 - x). This problem involves understanding the properties of logarithmic functions, ensuring the argument is positive, and excluding specific values that make the denominator zero. Suitable for high school students in Grades 9-12.

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