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Simplify: $\tan h\left(\ln \frac{e}{x}\right)$

$\tanh \left(\ln \frac{e}{x}\right)=$

$=\frac{e^{\ln \frac{e}{x}}-e^{-\ln \frac{e}{x}}}{e^{\ln \frac{e}{x}}+e^{-\ln \frac{e}{x}}}$

$=\frac{\frac{e}{x}-\frac{x}{e}}{\frac{e}{x}+\frac{x}{e}}$

$=\frac{\left(\frac{e}{x}-\frac{x}{e}\right) e x}{\left(\frac{e}{x}+\frac{x}{e}\right) e x}$

$=\frac{e^{2}-x^{2}}{e^{2}+x^{2}}$

Simplify $\cosh \left(\frac{1}{2} \ln (x+1)-\ln 2\right) \quad$ and find its value in a rational form

$\cosh \left(\frac{1}{2} \ln (x+1)-\ln (2)\right)$

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

$=\cosh (\ln \sqrt{x+1}-\ln (2))=\cosh \left(\ln \left(\frac{\sqrt{x+1}}{2}\right)\right)$

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

$=\frac{1}{2}\left[e^{\ln \left(\frac{\sqrt{x+1}}{2}\right)}-e^{-\ln \left(\frac{\sqrt{x+1}}{2}\right)}\right]$

$=\frac{1}{2}\left[\frac{\sqrt{x+1}}{2}+\frac{2}{\sqrt{x+1}}\right]$

$=\frac{1}{2}\left[\frac{\sqrt{x+1} \cdot \sqrt{x+1}}{2 \sqrt{x+1}}+\frac{2 \cdot 2}{\sqrt{x+1} \cdot 2}\right]$

$=\frac{1}{2} \cdot \frac{x+1+4}{2 \sqrt{x+1}}=\frac{x+5}{4 \sqrt{x+1}}$

Determine whether each of the following statements is true or false, and justify your answer.
$\cos h x=\frac{5}{4}$ only if $x=\ln 2$

$\cosh x=\frac{5}{4}$

$\frac{e^{x}+e^{-x}}{2}=\frac{5}{4}$

$4\left(e^{x}+e^{-x}\right)=2 \cdot 5 \rightarrow 4 e^{x}+4 e^{-x}=10$

$\div 2$

$2 e^{x}+2 e^{-x}=5 \quad \cdot e^{x}$

$2 e^{2 x}+2=5 e^{x}$

$\rightarrow 2 e^{2 x}-5 e^{x}+2=0$

let $y=e^{x}$

$2 y^{2}-5 y+2=0$

$(2 y-1)(y-2)=0$

$2 y-1=0$
$2 y=1$
$y=1 / 2$
$e^{x}=1 / 2$

$x=\ln (1 / 2)=-\ln (2)$

$y-2=0$
$y=2$
$e^{x}=2$
$x=\ln (2)$

The statements is false