Question

The inverse of the function $y=\frac{{10}^x-{10}^{-x}}{{10}^x+{10}^{-x}}$ is

• A. $\left(\frac{1}{2}\right){\log }_{10}\left(\frac{x+1}{x-1}\right)$
• B. $\left(\frac{1}{2}\right){\log }_{10}\left(\frac{2+x}{2-x}\right)$
• C. $\left(\frac{1}{2}\right){\log }_{10}\left(\frac{1-x}{1+x}\right)$
• D. None of these

Answer 1

The correct option is A

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

Explanation for the correct option:

Compute the required value:

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

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

Hence, the correct option is $\mathrm{Option}\left(\mathrm{A}\right)$