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

$\implies x = {e}^{4} - 1$
$\ln \sqrt{x + 1} = 2$
$\implies {\log}_{e} \sqrt{x + 1} = 2$
$\implies \sqrt{x + 1} = {e}^{2}$
$\implies \left(x + 1\right) = {e}^{4}$
$\implies x = {e}^{4} - 1$