Question

If $x=\frac{e^y-e^{-y}}{e^y+e^{-y}}$, then $y$ is equal to

• A. $\frac{1}{2}{\log }_e\frac{1+x}{1-x}$
• B. $\frac{1}{2}{\log }_e\frac{1-x}{1+x}$
• C. ${\log }_e\frac{1+x}{1-x}$
• D. $2{\log }_e\frac{1+x}{1-x}$

Answer 1

The correct option is A $\frac{1}{2}{\log }_e\frac{1+x}{1-x}$

From given $x$, we get $1+x=\frac{2e^y}{e^y+e^{-y}}$ and $1-x=\frac{2e^{-y}}{e^y+e^{-y}}$

By dividing them both we get $\frac{1+x}{1-x}=e^{2y}$, apply log on both sides, we get $y=\frac{1}{2}{\log }_e\frac{1+x}{1-x}$