jmc

We begin by evaluating the RHS of the equation. Since $27=3^3$, we know that $27^{\frac{1}{3}}=3$ and ${\log }_{27}3=\frac{1}{3}$. This allows us to simplify the original equation as ${\log }_{x}4=\frac{1}{3}$. Writing this equation in exponential form, we get $x^{\frac{1}{3}}=4$, which gives us the solution $x=4^3=\boxed{64}$.