Solve for x log base 3 of x-1=2

Rewrite ${\displaystyle {\log }_3(x-1)=2}$ in exponential form using the definition of a logarithm. If ${\displaystyle x}$ and ${\displaystyle b}$ are positive real numbers and ${\displaystyle b}$${\displaystyle \ne }$${\displaystyle 1}$, then ${\displaystyle {\log }_b\left(x\right)=y}$ is equivalent to ${\displaystyle b^y=x}$.