### Instruction: For what values of $x$ is $\frac{\log{(3-x)}}{\sqrt{x-1}}$ defined?### Response: To find the values of $x$ that make the expression defined, I need to consider two things: the domain of the logarithm function and the domain of the square root function.
The logarithm function is only defined for positive arguments, so I need to have $3-x > 0$.
This means that $x < 3$.
The square root function is only defined for non-negative arguments, so I need to have $x-1 \geq 0$.
This means that $x \geq 1$.
Combining these two inequalities, I get that $1 \leq x < 3$.
This is the interval of values of $x$ that make the expression defined.