### 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 what conditions must be met for the expression to be defined.
I know that the logarithm function is only defined for positive arguments, so $3-x$ has to be positive.
That means $x$ has to be less than $3$, or $x < 3$.
I also know that the square root function is only defined for non-negative arguments, so $x-1$ has to be non-negative.
That means $x$ has to be greater than or equal to $1$, or $x \geq 1$.
Finally, I know that the fraction is undefined when the denominator is zero, so $x-1$ cannot be zero.
That means $x$ cannot be equal to $1$, or $x \neq 1$.
To summarize, the expression is defined when $x < 3$, $x \geq 1$, and $x \neq 1$.
I can write this as a compound inequality: $1 < x < 3$.