Answer
$\left\{\dfrac{1}{9}\right\}$
Work Step by Step
Since $\log_b y=x$ implies $y=b^x$, the given equation, $
\log_x\dfrac{1}{81}=2
$, implies
\begin{align*}\require{cancel}
\dfrac{1}{81}&=x^2
\\\\
x^2&=\dfrac{1}{81}
.\end{align*}
Taking the square root of both sides (Square Root Property), and solving the variable, then
\begin{align*}\require{cancel}
x&=\pm\sqrt{\dfrac{1}{81}}
\\\\
x&=\pm\sqrt{\left(\dfrac{1}{9}\right)^2}
\\\\
x&=\pm\dfrac{1}{9}
.\end{align*}
If $x=-\dfrac{1}{9}$, the term $\log_x \dfrac{1}{81}$ of the original equation becomes $\log_{-1/9} \dfrac{1}{81}$. This is undefined since in $\log_b x$, $x$ and $b$ should be positive numbers with $b\ne1$.
Hence, the solution set of the equation $
\log_x\dfrac{1}{81}=2
$ is $
\left\{\dfrac{1}{9}\right\}
$.
