Question

Inverse of function $f\left(x\right)=\frac{{10}^x-{10}^{-x}}{{10}^x+{10}^{-x}}$ is

• A. ${\log }_{10}(2-x)$
• B. $\frac{1}{2}{\log }_{10}\left(\frac{1+x}{1-x}\right)$
• C. $\frac{1}{2}{\log }_{10}(2x-1)$
• D. $\frac{1}{2}{\log }_{10}\left(\frac{2x}{2-x}\right)$

Answer 1

The correct option is B $\frac{1}{2}{\log }_{10}\left(\frac{1+x}{1-x}\right)$

Let $f\left(x\right)=y$, then

$\frac{{10}^x-{10}^{-x}}{{10}^x+{10}^{-x}}=y$

$\Rightarrow \frac{{10}^{2x}-1}{{10}^{2x}+1}=y$

$\Rightarrow {10}^{2x}=\frac{1+y}{1-y}$ .... By Componendo and dividendo

$\Rightarrow x=\frac{1}{2}{\log }_{10}\left(\frac{1+y}{1-y}\right)$

$\Rightarrow f^{-1}\left(y\right)=\frac{1}{2}{\log }_{10}\left(\frac{1+y}{1-y}\right)$

$\therefore f^{-1}\left(x\right)=\frac{1}{2}{\log }_{10}\left(\frac{1+x}{1-x}\right)$.