### 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 expression undefined.
One condition is that the denominator cannot be zero, since that would make the fraction undefined.
So I need to find the value of $x$ that makes $\sqrt{x-1}=0$, which is equivalent to $x-1=0$ or $x=1$.
This means that $x=1$ is an excluded value for the expression.
Another condition is that the radicand of the square root cannot be negative, since that would make the expression imaginary.
So I need to find the values of $x$ that make $x-1<0$, which is equivalent to $x<1$.
This means that any value of $x$ that is less than $1$ is also an excluded value for the expression.
A third condition is that the argument of the logarithm cannot be zero or negative, since that would make the expression undefined or imaginary.
So I need to find the values of $x$ that make $3-x\leq 0$, which is equivalent to $x\geq 3$.
This means that any value of $x$ that is greater than or equal to $3$ is also an excluded value for the expression.
Therefore, the only values of $x$ that make the expression defined are those that satisfy $1<x<3$.