Answer
$x=\pm \sqrt{e^5}$ or $x\approx \pm12.182$
Work Step by Step
$\ln x^2=5$ (Take the exponent to both sides)
$e^{\ln x^2}=e^5$ (Use the property $e^{\ln a}$)
$x^2=e^5$
$x=\pm \sqrt{e^5}$
$x=\pm 12.18249....$
$x\approx 12.182$
Thus, $x=\pm \sqrt{e^5}$ or $x\approx \pm12.182$.
