Question

$f:R\to (-1,1),f\left(x\right)=\frac{{10}^x-{10}^{-x}}{{10}^x+{10}^{-x}}$. Find $f^{-1}$, if it exists.

Answer 1

$f\left(x\right)=\frac{{10}^x-{10}^{-x}}{{10}^x+{10}^{-x}}$

$y=\frac{{10}^x-1/{10}^x}{{10}^x+1/10x}=\frac{{10}^{2x}-1}{{10}^{2x}+1}$

Let $u={10}^{2x}$

$y=\frac{u-1}{u+1}$

$y(u+1)=u-1$

$uy+y=u-1$

$uy-u=-1-y$

$u(y-1)=-1-y$

$u=\frac{1+y}{1-y}$

${10}^{2x}=\frac{1+y}{1-y}$

Appying $lo{g}_{10}$

${\left(\frac{1}{2}\right)}^{2x}={\left(\frac{1}{2}\right)}^{lo{g}_{10}}\left(\frac{1+y}{1-y}\right)$

$\therefore f^{-1}\left(x\right)=\frac{1}{2}lo{g}_{10}\left(\frac{1+y}{1-y}\right)$.