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