Question

If x=$\frac{a^y-a^{-y}}{a^y+a^{-y}}$
To show $y=\frac{1}{2}{\log }_a\left(\frac{1+x}{1-x}\right)$

Answer 1

We have,

$x=\frac{a^y-a^{-y}}{a^y+a^{-y}}$ ……. (1)

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

$x{a}^{2y}+x=a^{2y}-1$

$1+x=a^{2y}-x{a}^{2y}$

$1+x=(1-x)a^{2y}$

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

${\log }_a\left(\frac{1+x}{1-x}\right)=2y$

$y=\frac{1}{2}{\log }_a\left(\frac{1+x}{1-x}\right)$

Hence, proved.