Question

The inverse of $f\left(x\right)=\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$ is

• A. $\frac{1}{6}{\log }_{10}\left(\frac{1+x}{1-x}\right)$
• B. $\frac{1}{6}{\log }_{10}\left(\frac{x}{1-x}\right)$
• C. $\frac{1}{6}{\log }_e\left(\frac{1+x}{1-x}\right)$
• D. $\frac{1}{6}{\log }_e\left(\frac{1-x}{1+x}\right)$

Answer 1

The correct option is C $\frac{1}{6}{\log }_e\left(\frac{1+x}{1-x}\right)$

$y=\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$

$\Rightarrow y=\frac{e^{6x}-1}{e^{6x}+1}$

$\Rightarrow \frac{y+1}{y-1}=\frac{e^{6x}-1+e^{6x}+1}{e^{6x}-1-e^{6x}-1}$

$\Rightarrow \frac{y+1}{y-1}=\frac{2e^{6x}}{-2}$

$\Rightarrow \frac{1+y}{1-y}=e^{6x}$

Taking ${\log }_e$ we get ${\log }_e\left(\frac{1+y}{1-y}\right)=6x$

$x=\frac{1}{6}\left({\log }_e\left(\frac{1+y}{1-y}\right)\right)$

Hence $f^{-1}$$=\frac{1}{6}\left({\log }_e\left(\frac{1+x}{1-x}\right)\right)$