Question

please solve the 11 th question​

Answer 1

Question :

If $\sf x = \dfrac{e {}^{y} - e {}^{ - y} }{e {}^{y} + e {}^{ - y} }$

Prove that :

$\sf y = \frac{1}{2} \ log( \frac{1 + x}{1 - x} )$

Solution :

$\sf x = \dfrac{e {}^{y} - e {}^{ - y} }{e {}^{ {}^{y} } + e {}^{ - y} } </p><p>$ $\implies \sf x(e {}^{y} + e {}^{ - y} ) = e {}^{y} - e {}^{ - y}$

Let $\bf\:e{}^{y}=t$

$\implies \sf x(t +t {}^{ - 1} ) = (t - t {}^{ - 1} )$

$\implies \sf x(t + \frac{1}{t} ) = (t - \frac{1}{t} )$

$\implies \sf x( \frac{t {}^{2} + 1 }{ \cancel{t}} ) = ( \frac{t {}^{2} - 1}{ \cancel{t}} )$

$\implies \sf x(t {}^{2} + 1) = (t {}^{2} - 1)$

$\implies \sf xt {}^{2} + x = t {}^{2} - 1$

$\implies \sf 1 + x = t {}^{2} - xt {}^{2}$

$\implies \sf (1 + x) = t {}^{2} (1 - x)$

$\implies \sf t {}^{2} = \frac{1 + x}{1 - x}$

$\implies \sf t = \sqrt{ \frac{1 + x}{1 - x} }$

$\implies \sf e {}^{y} = \sqrt{ \frac{1 + x}{1 - x} }$

Take log on both sides

$\implies \sf \log_{e}(e {}^{y} ) = \log_{e}( \sqrt{ \frac{1 + x}{1 - x} } )$

We know that $\bf\:\log(x{}^{n})=n\log\:x$

$\implies \sf y \ log_{e}(e) = \frac{1}{2} log_{e}( \frac{1 + x}{1 - x} )$

$\implies \bf y = \ log_{e}( \dfrac{1 + x}{1 - x} )$

Hence proved !