jmc

We begin by solving for $x$ from the first equation ${\log }_6(4x)=2$. Expressing this in exponential form, we find that $4x=6^2$, giving us $x=\frac{6^2}{4}=9$. After plugging this value of $x$ into ${\log }_{x}27$, we end up with the expression ${\log }_{9}27$. Since $27=(9)(3)=(9^1)(9^{\frac{1}{2}})=9^{\frac{3}{2}}$, we see that ${\log }_{9}27=\boxed{\frac{3}{2}}$.