The inverse function of f(x) = frac{8^{2x} - | vedclass

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(C) Let $y = f(x) = \frac{8^{2x} - 8^{-2x}}{8^{2x} + 8^{-2x}}$.Multiply numerator and denominator by $8^{2x}$:$y = \frac{8^{4x} - 1}{8^{4x} + 1}$.Now,

Vedclass · Accepted Answer

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

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(C) Let $y = f(x) = \frac{8^{2x} - 8^{-2x}}{8^{2x} + 8^{-2x}}$.Multiply numerator and denominator by $8^{2x}$:$y = \frac{8^{4x} - 1}{8^{4x} + 1}$.Now,solve for $x$ in terms of $y$:$y(8^{4x} + 1) = 8^{4x} - 1$$y \cdot 8^{4x} + y = 8^{4x} - 1$$1 + y = 8^{4x}(1 - y)$$8^{4x} = \frac{1 + y}{1 - y}$.Taking $\log_{8}$ on both sides:$4x = \log_{8} \left(\frac{1 + y}{1 - y}\right)$.Using the change of base formula $\log_{8} A = \frac{\ln A}{\ln 8} = (\log_{8} e) \ln A$:$4x = (\log_{8} e) \ln \left(\frac{1 + y}{1 - y}\right)$.$x = \frac{1}{4} (\log_{8} e) \ln \left(\frac{1 + y}{1 - y}\right)$.Replacing $y$ with $x$,the inverse function is $f^{-1}(x) = \frac{1}{4} (\log_{8} e) \ln \left(\frac{1 + x}{1 - x}\right)$.

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

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