Question

The inverse function of $f\left(x\right)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}},x\in (-1,1)$, is

• A. $\frac{1}{4}\left({\log }_{8}e\right){\log }_e\left(\frac{1-x}{1+x}\right)$
• B. $\frac{1}{4}\left({\log }_{8}e\right){\log }_e\left(\frac{1+x}{1-x}\right)$
• C. $\frac{1}{4}{\log }_e\left(\frac{1+x}{1-x}\right)$
• D. $\frac{1}{4}{\log }_e\left(\frac{1-x}{1+x}\right)$

Answer 1

The correct option is B $\frac{1}{4}\left({\log }_{8}e\right){\log }_e\left(\frac{1+x}{1-x}\right)$

$f\left(x\right)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}}=\frac{8^{4x}-1}{8^{4x}+1}$
Let $f\left(x\right)=y$
$\Rightarrow y=\frac{8^{4x}-1}{8^{4x}+1}$
$f\left(x\right)=y\Rightarrow x=f^{-1}\left(y\right)$
Applying componendo - dividendo, we get
$\frac{y+1}{y-1}=\frac{2\times 8^{4x}}{-2}$
$\Rightarrow \frac{y+1}{y-1}=-8^{4x}\Rightarrow 8^{4x}=\frac{1+y}{1-y}$
$\Rightarrow x=\frac{1}{4}{\log }_8\left(\frac{1+y}{1-y}\right)$
$\Rightarrow f^{-1}\left(y\right)=\frac{1}{4}{\log }_8\left(\frac{1+y}{1-y}\right)$
$\Rightarrow f^{-1}\left(x\right)=\frac{1}{4}{\log }_8\left(\frac{1+x}{1-x}\right)$
$\Rightarrow f^{-1}\left(x\right)=\frac{1}{4}\left({\log }_{8}e\right)\left({\log }_e\frac{1+x}{1-x}\right)$