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

Jun 16, 2018

THe solution is $= {e}^{2} - 2$

#### Explanation:

Solve the equation as follows :

$\ln \sqrt{x + 2} = 1$

$\implies$, $\ln {\left(x + 2\right)}^{\frac{1}{2}} = 1$

$\implies$, $\frac{1}{2} \ln \left(x + 2\right) = 1$

$\implies$, $\ln \left(x + 2\right) = 2$

$\implies$, $x + 2 = {e}^{2}$

$\implies$, $x = {e}^{2} - 2 = 5.389$

Jun 16, 2018

$x = {e}^{2} - 2 \approx 5.3890560989$

#### Explanation:

Remember that $\ln \left(a\right) = c$ means ${\log}_{e} \left(a\right) = c$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXXXXXXXX}} \Rightarrow {e}^{c} = a$

So
$\textcolor{w h i t e}{\text{XXX}} \ln \left(\sqrt{x + 2}\right) = 1$
means
$\textcolor{w h i t e}{\text{XXX}} {e}^{1} = \sqrt{x + 2}$

and therefore
$\textcolor{w h i t e}{\text{XXX}} {e}^{2} = x + 2$
and
$\textcolor{w h i t e}{\text{XXX}} x + 2 \left(\textcolor{m a \ge n t a}{- 2}\right) = {e}^{2} \textcolor{m a \ge n t a}{- 2}$
or
$\textcolor{w h i t e}{\text{XXX}} x = {e}^{2} - 2$

...an approximation to this value can be determined using a calculator.