### Instruction: For what values of $x$ is $\frac{\log{(3-x)}}{\sqrt{x-1}}$ defined?### Response: To answer this question, I need to think about the conditions that make the expression undefined.
One condition is that I cannot take the logarithm of a negative number or zero, so I need $3-x > 0$ or equivalently $x < 3$.
Another condition is that I cannot take the square root of a negative number, so I need $x-1 \geq 0$ or equivalently $x \geq 1$.
The third condition is that I cannot divide by zero, so I need $\sqrt{x-1} \neq 0$ or equivalently $x \neq 1$.
Combining these three conditions, I get that $1 < x < 3$ is the interval where the expression is defined.
Alternatively, I could write this as the set of values $\{x \mid 1 < x < 3\}$.