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  • The inverse function of <img alt="f(x)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}},x\epsilon (-1,1)," src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3D%5Cfrac%7B8%5E%7B2x%7D-8%5E%7B-2x%7D%7D%7B8%5E%7B2x%7D&plus;8%5E%7B-2x%7D%7D%2Cx%5Cepsil

The inverse function of f(x)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}},x\epsilon (-1,1), is
Option: 1 \frac{1}{4}(\log _{8}e)\log _{e}\left ( \frac{1-x}{1+x} \right )
Option: 2 \frac{1}{4}(\log _{8}e)\log _{e}\left ( \frac{1+x}{1-x} \right )
Option: 3 \frac{1}{4}\log _{e}\left ( \frac{1+x}{1-x} \right )
Option: 4 \frac{1}{4}\log _{e}\left ( \frac{1-x}{1+x} \right )
 

Answers (1)

best_answer

\\f(x)=\frac{8^{2 x}-8^{-2 x}}{8^{2 x}+8^{-2 x}} \\ f(x)=\frac{8^{4 x}-1}{8^{4 x}+1} \text{put }8^{4 x}=t \\ y=\frac{t-1}{t+1}\\ {y t+y=t-1} \\ {\frac{y+1}{1-y}=t} \\ {\frac{y+1}{1-y}=8^{4 x}}\\ ln\left(\frac{y+1}{1-y}\right)=4 x \ln 8 \quad

\\x=\frac{1}{4 \ln 8} \ln \left(\frac{y+1}{1-y}\right)\\\\f^{-1}(x)=\frac{1}{4 \ln 8} \ln \left(\frac{x+1}{1-x}\right)\\

Correct Option (2)

Posted by

Kuldeep Maurya

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