# Question #1f1cc

Jan 30, 2018

$x = - \ln \frac{e - 1}{2}$

#### Explanation:

$e - {e}^{- 2 x} = 1$

${e}^{- 2 x} = e - 1$

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

$- 2 x \cdot \ln e = \ln \left(e - 1\right)$

$- 2 x \cdot 1 = \ln \left(e - 1\right)$

$- 2 x = \ln \left(e - 1\right)$

$x = - \ln \frac{e - 1}{2}$

Jan 30, 2018

$x = \ln \left[\frac{1}{\sqrt{e - 1}}\right]$

#### Explanation:

$e - {e}^{- 2 x}$ can be written as $e - {\left[{e}^{-} x\right]}^{2}$ so the expression becomes $- e - {\left[{e}^{-} x\right]}^{2} = 1$, tidying up both sides gives ${\left[{e}^{-} x\right]}^{2} = \left[e - 1\right]$, and then taking the sqare root of both sides,

${e}^{-} x = \sqrt{e - 1}$, and therefore ${e}^{x} = \frac{1}{\sqrt{e - 1}}$.

Taking logs of both sides, $\ln {e}^{x} = \ln \left[\frac{1}{\sqrt{e - 1}}\right]$ $b u t \ln \left[{e}^{x}\right] = x$ and so $x = \ln \left[\frac{1}{\sqrt{e - 1}}\right]$. Hope this helps.