Answer
$\left\{2\right\}$
Work Step by Step
Since $\log_b y=x$ implies $b^x=y$, the given equation, $
\log_x 16=4
$, implies
\begin{align*}\require{cancel}
x^4&=16
.\end{align*}
Taking the fourth root of both sides, the equation above is equivalent to
\begin{align*}\require{cancel}
x&=\pm\sqrt{16}
\\&=
\pm2
.\end{align*}
If $x=-2,$ the expression $\log_x16$ becomes $\log_{-2}16$. This is undefined since in $\log_bx$, $b$ and $x$ are positive numbers with $b\ne1$.
Hence, the solution set of the equation $
\log_x 16=4
$ is $
\left\{2\right\}
$.
