#### Answer

$x=\{ 1,100 \}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\log x^2=(\log x)^2
,$ use the properties of logarithms. Then use concepts of solving quadratic equations to solve the values of $x.$ Finally, do checking of the solutions with the original equation.
$\bf{\text{Solution Details:}}$
Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent
\begin{array}{l}\require{cancel}
2\log x=(\log x)^2
.\end{array}
Using the properties of equality and factoring the $GCF,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
0=(\log x)^2-2\log x
\\\\
(\log x)^2-2\log x=0
\\\\
\log x(\log x-2)=0
.\end{array}
Equating each factor to $0$ (Zero Product Property), the solutions are
\begin{array}{l}\require{cancel}
\log x=0
\\\\\text{OR}\\\\
\log x-2=0
.\end{array}
Since $\log_by=x$ is equivalent to $y=b^x$, the equations above, in exponential form, are equivalent to
\begin{array}{l}\require{cancel}
\log x=0
\\\\
\log_{10} x=0
\\\\
x=10^0
\\\\
x=1
\\\\\text{OR}\\\\
\log x-2=0
\\\\
\log_{10} x=2
\\\\
x=10^2
\\\\
x=100
.\end{array}
Upon checking, both $
x=\{ 1,100 \}
,$ satisfy the original equation.