### Instruction: For what values of $x$ is $\frac{\log{(3-x)}}{\sqrt{x-1}}$ defined?### Response: To find the values of $x$ that make this expression defined, I need to consider what restrictions are imposed by the logarithm and the square root functions.
The logarithm function $\log{y}$ is only defined for positive values of $y$.
That means that $3-x$ must be positive, or $x<3$.
The square root function $\sqrt{y}$ is only defined for non-negative values of $y$.
That means that $x-1$ must be non-negative, or $x\geq 1$.
Therefore, the expression is defined only when both conditions are satisfied, or $1\leq x<3$.