### Instruction: For what values of $x$ is $\frac{\log{(3-x)}}{\sqrt{x-1}}$ defined?### Response: To find the values of $x$ for which the expression is defined, I need to consider the domains of the functions involved: logarithm, square root, and division.
For the logarithm function, the argument $(3-x)$ must be positive, otherwise the function is undefined.
So I need to solve the inequality $3-x > 0$, which gives me $x < 3$.
For the square root function, the radicand $(x-1)$ must be non-negative, otherwise the function is undefined.
So I need to solve the inequality $x-1 \geq 0$, which gives me $x \geq 1$.
For the division function, the denominator $\sqrt{x-1}$ must not be zero, otherwise the function is undefined.
So I need to exclude the value $x=1$ from the domain.
Putting these conditions together, I get that the expression is defined only when $1 < x < 3$.
This is the intersection of the domains of the logarithm and the square root functions, excluding the endpoint $x=1$.