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Question #1f1cc

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Answer 1 · 2018-01-30T17:30:17

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

Explanation:

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

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

${ln e}^{- 2 x} = ln (e - 1)$ $lne^(-2x)=ln(e-1)$

$- 2 x \cdot ln e = ln (e - 1)$ $-2x*lne=ln(e-1)$

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

$- 2 x = ln (e - 1)$ $-2x=ln(e-1)$

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

Answer 2 · 2018-01-30T17:33:06

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

Explanation:

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

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

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

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