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
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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