Question

The inversion function of $f\left(x\right)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}},x\in (-1,1)$ is

• A. $\frac{1}{4}{\log }_8\left(e\right){\log }_e\left(\frac{1-x}{1+x}\right)$
• B. $\frac{1}{4}{\log }_8\left(e\right){\log }_e\left(\frac{1+x}{1-x}\right)$
• C. $\frac{1}{4}{\log }_e\left(\frac{1+x}{1-x}\right)$
• D. $\frac{1}{4}{\log }_e\left(\frac{1-x}{1+x}\right)$

Answer 1

The correct option is B

$\frac{1}{4}{\log }_8\left(e\right){\log }_e\left(\frac{1+x}{1-x}\right)$

Step 1: Simplify the equation in terms of $y$.

Given: $f\left(x\right)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}},x\in (-1,1)$

Assume that, $f\left(x\right)=y$ and so, $x=f^{-1}\left(y\right)$

Simplify the given equation to get $x$ in terms of $y$.

$$\begin{gathered} y=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}} \\ \Rightarrow y=\frac{8^{4x}-1}{8^{4x}+1} \\ \Rightarrow y\left(8^{4x}+1\right)=8^{4x}-1 \\ \Rightarrow 8^{4x}y+y=8^{4x}-1 \\ \Rightarrow 8^{4x}y-8^{4x}=-y-1 \\ \Rightarrow 8^{4x}(y-1)=-1(y+1) \\ \Rightarrow 8^{4x}=\frac{y+1}{1-y} \end{gathered}$$

Take $\log$ on both sides.

$$\begin{gathered} {\log }_e\left(8^{4x}\right)={\log }_e\left(\frac{y+1}{1-y}\right) \\ \Rightarrow \frac{{\log }_8\left(8^{4x}\right)}{{\log }_8\left(e\right)}={\log }_e\left(\frac{y+1}{1-y}\right)\left[\because {\log }_a\left(b\right)=\frac{{\log }_c\left(b\right)}{{\log }_c\left(a\right)}\right] \\ \Rightarrow 4x={\log }_8\left(e\right)·{\log }_e\left(\frac{y+1}{1-y}\right)[\because {\log }_a(a)=1] \\ \Rightarrow x=\frac{1}{4}{\log }_8\left(e\right)·{\log }_e\left(\frac{y+1}{1-y}\right) \end{gathered}$$

Since, $x=f^{-1}\left(y\right)$.

So, $f^{-1}\left(y\right)=\frac{1}{4}{\log }_8\left(e\right)·{\log }_e\left(\frac{y+1}{1-y}\right)$

Step 2: Replace $y$ and $x$.

Replace $y$ by $x$.

$f^{-1}\left(x\right)=\frac{1}{4}{\log }_8\left(e\right)·{\log }_e\left(\frac{x+1}{1-x}\right)$

Therefore, the inverse of the given function is $f^{-1}\left(x\right)=\frac{1}{4}{\log }_8\left(e\right)·{\log }_e\left(\frac{x+1}{1-x}\right)$.

Hence, Option (B) is the correct answer.