A) \[\frac{1}{4}{{\log }_{e}}\left( \frac{1-x}{1+x} \right)\]
B) \[\frac{1}{4}(lo{{g}_{8}}e)lo{{g}_{e}}\left( \frac{1+x}{1-x} \right)\]
C) \[\frac{1}{4}{{\log }_{e}}\left( \frac{1+x}{1-x} \right)\]
D) \[\frac{1}{4}(lo{{g}_{8}}e)lo{{g}_{e}}\left( \frac{1-x}{1+x} \right)\]
Correct Answer: B
Solution :
[b] \[y=\frac{{{8}^{2x}}-{{8}^{-2x}}}{{{8}^{2x}}+{{8}^{-2x}}}\] \[\Rightarrow y=\frac{{{x}^{4x}}-1}{{{8}^{4x}}+1}\] \[\Rightarrow {{8}^{4x}}\cdot y+y={{8}^{8x}}-1\] \[\Rightarrow 1+y={{8}^{4x}}(1-y)\] \[\Rightarrow {{8}^{4x}}=\frac{1+y}{1-y}\] \[\Rightarrow 4x={{\log }_{8}}\left( \frac{1+y}{1-y} \right)\] \[\therefore {{f}^{-1}}(x)=\frac{1}{4}{{\log }_{8}}\left( \frac{1+x}{1-x} \right)=\frac{1}{4}{{\log }_{8}}e{{\log }_{e}}\left( \frac{1+x}{1-x} \right)\]
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