The inverse function of
Question:

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

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

  2. (2) $\frac{1}{4}\left(\log _{8} e\right) \log _{e}\left(\frac{1-x}{1+x}\right)$

  3. (3) $\frac{1}{4} \log _{e}\left(\frac{1-x}{1+x}\right)$

  4. (4) $\frac{1}{4}\left(\log _{8} e\right) \log _{e}\left(\frac{1+x}{1-x}\right)$


Correct Option: 1

Solution:

$y=\frac{8^{2 x}-8^{-2 x}}{8^{2 x}+8^{-2 x}}$

$\frac{1+y}{1-y}=\frac{8^{2 x}}{8^{-2 x}} \Rightarrow 8^{4 x}=\frac{1+y}{1-y}$

$\Rightarrow \quad 4 x=\log _{8}\left(\frac{1+y}{1-y}\right)$

$\Rightarrow \quad x=\frac{1}{4} \log _{8}\left(\frac{1+y}{1-y}\right)$

$\therefore \quad f^{-1}(x)=\frac{1}{4} \log _{8}\left(\frac{1+x}{1-x}\right)$

 

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