# How do you solve  ln sqrt(x + 2) = 1?

Sep 14, 2015

$x = {e}^{2} - 2$

#### Explanation:

$\ln \sqrt{x + 2} = 1 \implies$ If: ${\log}_{a} \left(x\right) = y \Leftrightarrow x = {a}^{y}$, then:

$\sqrt{x + 2} = {e}^{1} = e \implies$square both sides:

$x + 2 = {e}^{2} \implies$subtract 2 from both sides:

$x = {e}^{2} - 2 \implies$

All roots must be checked in the original equation to verify that they work and are not "Extraneous" roots that were introduced during the squaring process, so:

$\ln \sqrt{x + 2} = 1 \implies$ at: $x = {e}^{2} - 2$
$= \ln \sqrt{{e}^{2} - 2 + 2}$
$= \ln \sqrt{{e}^{2}}$
$= \ln \left(e\right)$
$= 1$
$1 = 1 \implies$hence verified: $x = {e}^{2} - 2$, is a valid root.