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Question Number 196683 by MATHEMATICSAM last updated on 29/Aug/23

If y = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} then show that

x = \frac{1}{2} \log_{e} (\frac{1 + y}{1 - y}) .

Answered by Frix last updated on 29/Aug/23

y = \frac{e^{2 x} - 1}{e^{2 x} + 1}

e^{2 x} = \frac{1 + y}{1 - y}

2 x = \ln \frac{1 + y}{1 - y}

x = \frac{1}{2} \ln \frac{1 + y}{1 - y}

Answered by som(math1967) last updated on 29/Aug/23

\frac{1}{y} = \frac{e^{x} + \frac{1}{e^{x}}}{e^{x} - \frac{1}{e^{x}}}

\Rightarrow \frac{1 + y}{1 - y} = \frac{2 e^{x}}{\frac{2}{e^{x}}}

\Rightarrow \frac{1 + y}{1 - y} = e^{2 x}

\Rightarrow {l o g}_{e} (\frac{1 + y}{1 - y}) = {l o g}_{e} e^{2 x}

∴ 2 x = {l o g}_{e} (\frac{1 + y}{1 - y})

x = \frac{1}{2} {l o g}_{e} (\frac{1 + y}{1 - y})