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

f(x)=82x−8−2x82x+8−2x× 82x82x=84x−184x+1Let f−1(x)=y⇒ x=f(y)⇒ x=84y−184y+1⇒ x84y+x=84y−1⇒ x84y−84y=−x−1⇒ 84y(x−1)=−x−1⇒ 84y=1+x1−x⇒ 4y=log8(1+x1−x)⇒ y=14log8(1+x1−x)∴ f−1(x)=14 log8(1+x1−x)

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

f(x)=82x−8−2x82x+8−2x× 82x82x=84x−184x+1Let f−1(x)=y⇒ x=f(y)⇒ x=84y−184y+1⇒ x84y+x=84y−1⇒ x84y−84y=−x−1⇒ 84y(x−1)=−x−1⇒ 84y=1+x1−x⇒ 4y=log8(1+x1−x)⇒ y=14log8(1+x1−x)∴ f−1(x)=14 log8(1+x1−x)

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$T h e i n v e r s e f u n c t i o n o f f (x) = \frac{8^{2 x} - 8^{- 2 x}}{8^{2 x} + 8^{- 2 x}}, x \in (- 1, 1), i s$

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

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

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

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

answer is D.

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Detailed Solution

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