Question

If $x=\tan h^{-1}\left(y\right)$ 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 $2x$

We have,

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

We lnow that,

$\tan h^{-1}\left(x\right)=\frac{1}{2}{\log }_e\left(\frac{1+x}{1-x}\right)$

then,

$x=\tan h^{-1}\left(y\right)$

$x=\frac{1}{2}{\log }_e\left(\frac{1+y}{1-y}\right)$

$2x={\log }_e\left(\frac{1+y}{1-y}\right)$

and

${\log }_e(\frac{1+y}{1-y}=2x)$

Hence, this is the answer