The equation 3lnx=lb3x is solved by the sqaure root of 3.
Thanks
Note that $\displaystyle \displaystyle \begin{align*} x = 0 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} x = -\sqrt{3} \end{align*}$ are NOT acceptable solutions, because $\displaystyle \displaystyle \begin{align*} \ln{x} \end{align*}$ is only defined for $\displaystyle \displaystyle \begin{align*} x > 0 \end{align*}$.