Question

The inverse of $\frac{7^x-7^{-x}}{7^x+7^{-x}}$ is

• A. $\left(\frac{1}{2}\right){\log }_7\left(\frac{1+x}{1-x}\right)$
• B. $\left(\frac{1}{2}\right){\log }_7\left(\frac{1-x}{1+x}\right)$
• C. ${\log }_{\frac{1}{2}}\left(\frac{1-x}{1+x}\right)$
• D. ${\log }_e\left(\frac{1+x}{1-x}\right)$

Answer 1

The correct option is A

$\left(\frac{1}{2}\right){\log }_7\left(\frac{1+x}{1-x}\right)$

Explanation for correct the option:

Compute the required value:

$f\left(x\right)=\frac{7^x-7^{-x}}{7^x+7^{-x}}$

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

$$\begin{gathered} \Rightarrow y=\frac{7^x-7^{-x}}{7^x+7^{-x}} \\ \Rightarrow y=\frac{7^{2x}-1}{7^{2x}+1} \\ \Rightarrow \frac{y+1}{y-1}=7^{2x}\left[\because componendoanddividendo\right] \\ \Rightarrow \log \left(\frac{y+1}{y-1}\right)=2x\log \left(7\right) \\ \Rightarrow \left(\frac{1}{2}\right){\log }_7\left(\frac{y+1}{y-1}\right)=x \\ \therefore f^{-1}\left(x\right)=\left(\frac{1}{2}\right){\log }_7\left(\frac{x+1}{x-1}\right) \end{gathered}$$

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