## College Algebra (11th Edition)

$x=\{ -2,2 \}$
$\bf{\text{Solution Outline:}}$ 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.