To define the logarithm $\log_b a$, three conditions must be met: 1) The argument $a > 0$, 2) The base $b > 0$, and 3) The base $b \neq 1$. For $f(x) = \log_{(x-1)}(5-x)$, we require: (i) $5-x > 0 \Rightarrow x 0 \Rightarrow x > 1$. (iii) $x-1 \neq 1 \Rightarrow x \neq 2$. Combining these, $x$ must be in the interval $(1, 5)$ but cannot be $2$. Thus, the domain is $(1, 2) \cup (2, 5)$. Option 1 is wrong because it includes $x=2$, which makes the base 1. Option 3 is wrong because logs are undefined for zero or negative numbers. Option 4 is wrong because it unnecessarily excludes the interval $(1, 2)$.