Question

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

• A. ${\log }_{10}\left(2-x\right)$
• B. $\frac{1}{2}{\log }_{10}\left(\frac{1+x}{1-x}\right)$
• C. $\frac{1}{2}{\log }_{10}\left(2x-1\right)$
• D. $\frac{1}{4}\log \left(\frac{2x}{2-x}\right)$

Answer 1

Answer: (B)

Let $y=f(x)$ $\Rightarrow x=f^{-1}(y)$
Then, $y=\frac{10^x-10^{-x}}{10^x+10^{-x}}$
$\Rightarrow \left(10^x+10^{-x}\right)y=\left(10^x-10^{-x}\right)$
$\Rightarrow \left(10^{2x}+1\right)y=\left(10^{2x}-1\right)$
$\Rightarrow 10^{2x}(y-1)=-(y+1)$
$\Rightarrow 10^{2x}=\frac{y+1}{1-y}$
$\Rightarrow 2x={\log }_{10}\left(\frac{1+y}{1-y}\right)$
$\Rightarrow x=\frac{1}{2}{\log }_{10}\left(\frac{1+y}{1-y}\right)$
$\Rightarrow f^{-1}(y)=\frac{1}{2}{\log }_{10}\left(\frac{1+y}{1-y}\right)$
Hence, the required inverse function is
$\frac{1}{2}{\log }_{10}\left(\frac{1+x}{1-x}\right)$