AHSME 1968 Problem 2

The real value of $x$ such that $64^{x-1}$ divided by $4^{x-1}$ equals $256^{2x}$ is: $\text{(A) } -\frac{2}{3} \text{(B) } -\frac{1}{3} \text{(C) } 0 \text{(D) } \frac{1}{4} \text{(E) } \frac{3}{8}$