Question

Let $\frac{e^x-e^{-x}}{e^x+e^{-x}}=ln\sqrt{\frac{1+x}{1-x}},$ then find $x$

Answer 1

Consider $f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}=1-\frac{2}{e^{2x}+1}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{4e^{2x}}{(e^{2x}+1{)}^2}>0,\forall x\in R$
$\Rightarrow f\left(x\right)$ is an increasing function.
$\Rightarrow$ Domain : R, Range :$(-1,1)$
For $f:R\to (-1,1)$,
$f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}},f^{-x}:(-1,1)\to R$
$\Rightarrow x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
$e^y=\sqrt{\frac{1+x}{}1-x}\Rightarrow f^{\prime }\left(x\right)=\ln \sqrt{\frac{1+x}{1-x}}$.
Hence, given equation is equivalent to $f\left(x\right)=f^{\prime }\left(x\right)$.
$\iff f\left(x\right)=x$ (as f is an increasing function)
$\Rightarrow \ln \sqrt{\frac{1+x}{1-x}}=x\Rightarrow \frac{1+x}{1-x}=e^{2x}$
Now, draw the graph of $y=\frac{1+x}{1-x}$ and $y=e^{2x}$. They intersect each other at $x=0$.