### Instruction: For what values of $x$ is $\frac{\log{(3-x)}}{\sqrt{x-1}}$ defined?### Response: I know that a fraction is defined only when the denominator is not zero, so I need to avoid any values of $x$ that make $\sqrt{x-1}$ equal to zero.
To do that, I can solve the equation $\sqrt{x-1} = 0$ by squaring both sides and getting $x-1 = 0$, which implies that $x = 1$.
So $x = 1$ is one value that I need to exclude from the domain of the function, because it makes the denominator zero.
I also know that a logarithm is defined only when the argument is positive, so I need to avoid any values of $x$ that make $3-x \leq 0$.
To do that, I can solve the inequality $3-x \leq 0$ by adding $x$ to both sides and getting $3 \leq x$.
So any value of $x$ that is greater than or equal to $3$ is another value that I need to exclude from the domain of the function, because it makes the argument of the logarithm non-positive.
Finally, I also know that a square root is defined only when the radicand is non-negative, so I need to avoid any values of $x$ that make $x-1 < 0$.
To do that, I can solve the inequality $x-1 < 0$ by adding $1$ to both sides and getting $x < 1$.
So any value of $x$ that is less than $1$ is another value that I need to exclude from the domain of the function, because it makes the radicand of the square root negative.