# Question #984b4

Apr 14, 2017

$x = \sqrt{\frac{1}{e - 1}}$

#### Explanation:

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

By Log Property: $r \ln x = \ln {x}^{r}$,

$R i g h t a r r o w \ln \left(1 + {x}^{2}\right) = 1 + \ln {x}^{2}$

By raising $e$ to both sides,

$R i g h t a r r o w {e}^{\ln \left(1 + {x}^{2}\right)} = {e}^{1 + \ln {x}^{2}} = {e}^{1} \cdot {e}^{\ln {x}^{2}}$

By Inverse Property: ${e}^{\ln x} = x$,

$R i g h t a r r o w 1 + {x}^{2} = e {x}^{2}$

By subtracting ${x}^{2}$ from both sides,

$R i g h t a r r o w 1 = e {x}^{2} - {x}^{2} = \left(e - 1\right) {x}^{2}$

By dividing both sides by $\left(e - 1\right)$,

$R i g h t a r r o w \frac{1}{e - 1} = {x}^{2}$

By taking the square-root of both sides,

$\pm \sqrt{\frac{1}{e - 1}} = x$

Since the domain of $\ln x$ is $x > 0$, we have

$x = \sqrt{\frac{1}{e - 1}}$

I hope that this was clear.