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The inverse function of $f(\mathrm{x})=\frac{8^{2 \mathrm{x}}-8^{-2 \mathrm{x}}}{8^{2 \mathrm{x}}+8^{-2 \mathrm{x}}}, \mathrm{x} \in(-1,1),$ is
$\frac{1}{4}\left(\log _{8} e\right) \log _{e}\left(\frac{1-x}{1+x}\right)$
$\frac{1}{4} \log _{e}\left(\frac{1-x}{1+x}\right)$
$\frac{1}{4}\left(\log _{8} e\right) \log _{e}\left(\frac{1+x}{1-x}\right)$
$\frac{1}{4} \log _{e}\left(\frac{1+x}{1-x}\right)$
Solution
$f(\mathrm{x})=\mathrm{y}=\frac{8^{4 \mathrm{x}}-1}{8^{4 \mathrm{x}}+1}=1-\frac{2}{8^{4 \mathrm{x}}+1}$
so, $8^{4 x}+1=\frac{2}{1-y} \Rightarrow 8^{4 x}=\frac{1+y}{1-y}$
$\Rightarrow \mathrm{x}=\ln \left(\frac{1+\mathrm{y}}{1-\mathrm{y}}\right) \times \frac{1}{4(\mathrm{n} 8}=f^{-1}(\mathrm{y})$
Hence, $f^{-1}(\mathrm{x})=\frac{1}{4} \log _{8} \mathrm{e}l\mathrm{n}\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)$
