If `f: Rvec(-1,1)` defined by `f(x)=(10^x-10^(-x))/(10^x+10^(-x))` is invertible, find `f^(-1)`

Explanation

`f(x)=frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}`


Let `f(x)=y`


`Rightarrow frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}=y`


`Rightarrow frac{10^{2 x}-1}{10^{2 x}-1}=y`


`Rightarrow 10^{2 pi}-1=y(10^{2 x}+1)`


`Rightarrow 10^{2 pi}-10^{2 x_{x}}=y+1`


`Rightarrow 10^{2 x}(1-y)=y+1`


`Rightarrow 10^{2 pi}=frac{y+1}{1-y}`


...