### 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 function undefined, or invalid.
The first thing that comes to mind is the denominator: I can't have $\sqrt{x-1}$ be zero or negative, since that would either make the function undefined or involve imaginary numbers.
So, I need $x-1$ to be positive, which means $x > 1$.
That's one condition, but I also need to consider the numerator: $\log{(3-x)}$.
I know that the logarithm function is only defined for positive arguments, so I need $3-x$ to be positive as well.
This means $x < 3$.
So, combining these two conditions, I have $1 < x < 3$ as the range of values that make the function defined.
I can check this by plugging in some values in this interval, such as $x=2$, and see that the function gives a valid output.
Therefore, the answer is $1 < x < 3$.