Solve for x log base 2 of x=0

Rewrite ${\displaystyle {\log }_2\left(x\right)=0}$ 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 ${\displaystyle b^y=x}$.