Question

lf $x={\tanh }^{-1}y$, then ${\log }_e\left(\frac{1+y}{1-y}\right)=$

• A. $x$
• B. $4x$
• C. $2x$
• D. $3x$

Answer 1

The correct option is C $2x$

Given $x=tan{h}^{-1}y$
$\Rightarrow tanhx=y=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{sinhx}{coshx}$
$y+1=\frac{2e^x}{e^x+e^{-x}}$ - 1
$1-y=1+\frac{e^{-x}-e^x}{e^x+e^{-x}}=\frac{2e^{-x}}{e^x+e^{-x}}$ - 2
divide 1 by 2,
$\frac{y+1}{1-y}=e^{2x}$
$ln\left(\frac{1+y}{1-y}\right)=ln{e}^{2x}=2x$