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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

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