jmc

We know ${\log }_{4}3=x$ so $4^x=3$. We also know ${\log }_{2}27=kx$ so $2^{kx}=27$. We need to combine these equations, but note that $27=3^3$, so cube the first equation: $(4^x{)}^3=3^3=27$, so $4^{3x}=27=2^{kx}$. But $4=2^2$, so we can substitute to get the same base: $(2^2{)}^{3x}=2^{kx}$, so $2^{6x}=2^{kx}$. Then $6x=kx$ and $\boxed{k=6}$.