#### Answer

$x=\{ -2,2 \}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\log_2 \sqrt{2x^2}=\dfrac{3}{2}
,$ convert to the equivalent exponential form. Then square both sides and use the properties of equality to isolate $x^2.$ Then take the square root of both sides. Finally, do checking of the solutions with the original equation.
$\bf{\text{Solution Details:}}$
Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{2x^2}=2^{\frac{3}{2}}
.\end{array}
Squaring both sides results to
\begin{array}{l}\require{cancel}
(\sqrt{2x^2})^2=\left( 2^{\frac{3}{2}} \right)^2
\\\\
2x^2=2^3
\\\\
2x^2=8
\\\\
x^2=\dfrac{8}{2}
\\\\
x^2=4
.\end{array}
Taking the square root of both sides results to
\begin{array}{l}\require{cancel}
x=\pm\sqrt{4}
\\\\
x=\pm2
.\end{array}
Upon checking, both $
x=\{ -2,2 \}
,$ satisfy the original equation.