Question

If x= e^y - e^(-y)/ e^y+ e^(-y) , show that, y= 1/2 { log (1+x/1-x) with base e}

Answer 1

$x= \frac{ e^{y}- e^{-y} }{ e^{y} + e^{-y} }$
Multiplying both side by $( e^{y} + e^{-y} )$
[tex]x e^{y} +x e^{-y} = e^{y} - e^{-y} [/tex]
Rearranging
$x e^{y} - e^{y} = -xe^{-y} - e^{-y}$
Multiplying by -1
$e^{y}-xe^{y}=xe^{-y}+e^{-y}$
$e^{y}(1-x)=e^{-y}(1+x)$
$\frac{ e^{y} }{e^{-y}} = \frac{1+x}{1-x}$
$e^{2y}= \frac{1+x}{1-x}$
Taking log of base e on both sides
$2y log_{e} e=log_{e}( \frac{1+x}{1-x} )$
But $log_{e}e=1$
∴ $y = \frac{1}{2} log_{e} (\frac{1+x}{1-x} )$