The inverse function of f(x)=82x-8-2x82x+8-2x,x∈(-1,1), is

y=82x-8-2x82x+8-2x⇒1+y1-y=82x8-2x⇒84x=1+y1-y⇒4x=log81+y1-y⇒x=14log81+y1-y∴f-1(x)=14log81+x1-x=14log8e loge1+x1-x

The inverse function of f(x)=82x-8-2x82x+8-2x,x∈(-1,1), is

y=82x-8-2x82x+8-2x⇒1+y1-y=82x8-2x⇒84x=1+y1-y⇒4x=log81+y1-y⇒x=14log81+y1-y∴f-1(x)=14log81+x1-x=14log8e loge1+x1-x

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

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$\frac{1}{4} \log_{e} (\frac{1 + x}{1 - x})$

$\frac{1}{4} \log_{e} (\frac{1 - x}{1 + x})$

$\frac{1}{4} (\log_{8} e) \log_{e} (\frac{1 - x}{1 + x})$

$\frac{1}{4} (\log_{8} e) \log_{e} (\frac{1 + x}{1 - x})$

answer is D.

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Detailed Solution

$\begin{array}{l} y = \frac{8^{2 x} - 8^{- 2 x}}{8^{2 x} + 8^{- 2 x}} \\ \Rightarrow \frac{1 + y}{1 - y} = \frac{8^{2 x}}{8^{- 2 x}} \Rightarrow 8^{4 x} = (\frac{1 + y}{1 - y}) \\ \Rightarrow 4 x = \log_{8} (\frac{1 + y}{1 - y}) \Rightarrow x = \frac{1}{4} \log_{8} (\frac{1 + y}{1 - y}) \\ ∴ f^{- 1} (x) = \frac{1}{4} \log_{8} (\frac{1 + x}{1 - x}) = \frac{1}{4} (\log_{8} e) \log_{e} (\frac{1 + x}{1 - x}) \end{array}$