Question

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

Answer 1

Answer: 0

Let $f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
$\Rightarrow y=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
$\Rightarrow y=\frac{e^{2x}-1}{e^{2x}+1}$
$\Rightarrow e^{2x}y+y=e^{2x}-1$
$e^{2x}\left(y-1\right)=-1-y$
$e^{2x}=\frac{1+y}{1-y}$
$e^x=\sqrt{\frac{1+y}{1-y}}$
$x=\ln \sqrt{\frac{1+y}{1-y}}$
$f^{-1}\left(y\right)=\ln \sqrt{\frac{1+y}{1-y}}$
$f^{-1}\left(x\right)=\ln \sqrt{\frac{1+x}{1-x}}$
Now, given equation is
$\frac{e^x-e^{-x}}{e^x+e^{-x}}=\ln \sqrt{\frac{1+x}{1-x}}$
$\Rightarrow f\left(x\right)=f^{-1}\left(x\right)$
The solution of $f\left(x\right)=f^{-1}\left(x\right)$ is always obtained on $y=x$ line
$\Rightarrow \ln \sqrt{\frac{1+x}{1-x}}=x$
$\Rightarrow e^{2x}=\frac{1+x}{1-x}$

Only one solution $x=0$ .