Answer
$(5,\infty)$
or $\{x|x \gt 5\}$
Work Step by Step
First, we see that if we want $\displaystyle \frac{1}{x-5}$ to be defined,
$x-5\neq 0\quad \Rightarrow x\neq 5.$
Next,
the argument of a logarithmic function must be greater than zero.
So, for g to be defined, we must have
$\displaystyle \frac{1}{x-5} \gt 0 \quad $... The LHS is positive when the denominator is positive,
$x-5 \gt 0 $
$\mathrm{x} \gt 5$
The domain of f is $(5,\infty)$
or $\{x|x \gt 5\}$
