Find the Inverse y = log of x+1

Rewrite the equation as ${\displaystyle \log (y+1)=x}$.

Rewrite ${\displaystyle \log (y+1)=x}$ in exponential form using the definition of a logarithm. If ${\displaystyle x}$ and ${\displaystyle b}$ are positive real numbers and ${\displaystyle b\ne 1}$, then ${\displaystyle {\log }_b\left(x\right)=y}$ is equivalent to ${\displaystyle b^y=x}$.

Solve for ${\displaystyle y}$.

To verify the inverse, check if ${\displaystyle f^{-1}\left(f\left(x\right)\right)=x}$ and ${\displaystyle f\left(f^{-1}\left(x\right)\right)=x}$.

Evaluate ${\displaystyle f^{-1}\left(f\left(x\right)\right)}$.

Evaluate ${\displaystyle f\left(f^{-1}\left(x\right)\right)}$.

Since ${\displaystyle f^{-1}\left(f\left(x\right)\right)=x}$ and ${\displaystyle f\left(f^{-1}\left(x\right)\right)=x}$, then ${\displaystyle f^{-1}\left(x\right)={10}^x-1}$ is the inverse of ${\displaystyle f\left(x\right)=\log (x+1)}$.