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Mathematics Question on Functions

The inverse function of f(x)=82x−8−2x82x+8−2x,xϵ(−1,1),f\left(x\right) = \frac{8^{2x}-8^{-2x}}{8^{2x} + 8^{-2x}}, x\epsilon \left(-1, 1\right), is __________.

A

14(log8 e)loge(1−x1+x)\frac{1}{4}\left(log_{8} \,e\right) log_{e} \left(\frac{1-x}{1+x}\right)

B

14(log8 e)loge(1+x1−x)\frac{1}{4}\left(log_{8} \,e\right) log_{e} \left(\frac{1+x}{1-x}\right)

C

14loge(1+x1−x)\frac{1}{4} log_{e} \left(\frac{1+x}{1-x}\right)

D

14loge(1−x1+x)\frac{1}{4} log_{e} \left(\frac{1- x}{1+ x}\right)

Answer

14(log8 e)loge(1+x1−x)\frac{1}{4}\left(log_{8} \,e\right) log_{e} \left(\frac{1+x}{1-x}\right)

Explanation

Solution

f(x)=y=84x−184x+1=1−284x+1f \left(x\right)=y=\frac{8^{4x}-1}{8^{4x}+1}=1-\frac{2}{8^{4x}+1}
so, 84x+1=21−y⇒84x=1+y1−y8^{4x}+1=\frac{2}{1-y} \Rightarrow 8^{4x}=\frac{1+y}{1-y}
⇒x=ℓn(1+y1−y)×14ℓn8=f−1(y)\Rightarrow x=\ell n\left(\frac{1+y}{1-y}\right)\times \frac{1}{4\ell n8}=f ^{-1}\left(y\right)
Hence, f−1(x)=14logg eℓn(1+x1−x)f ^{-1}\left(x\right)=\frac{1}{4}log_{g}\,e\ell n\left(\frac{1+x}{1-x}\right)