Answer
$x\in\{1,100\}$
Work Step by Step
In order for the equation to be defined,
$x\gt 0\qquad (*).$
On the LHS, apply$ \quad\log_{a}M^{p}=p\cdot\log_{a}M$
$ 2\log x=(\log x)^{2}\qquad$ ... let $t=\log x$
$2t=t^{2}$
$0=t^{2}(t-2)$
Solutions for t: $t=0,t=2.$
Bring back x
$\left[\begin{array}{lll}
\log x=0 & ... & \log x=2\\
x=10^{0} & & x=10^{2}\\
x=1 & & x=100
\end{array}\right]$
Both satisfy (*), so both are valid solutions.
