$8 \cdot 9$ a real number $x$, such that $64^{x-1}$ divided by $4^{x-1}$ equals $256^{2 x}$, then $x$ equals
(A) $-\frac{2}{3}$.
(B) $-\frac{1}{3}$.
(C) 0 .
(D) $\frac{1}{4}$.
(E) $\frac{3}{8}$.
(19th American High School Mathematics Examination, 1978)
[Solution] From the given $\frac{64^{x-1}}{4^{x-1}}=256^{2 x}$, we have
$$
\begin{array}{l}
\frac{64^{x-1}}{4^{x-1}}=\left(\frac{64}{4}\right)^{x-1}=16^{x-1}, \\
\text { and } 256^{2 x}=16^{2 \cdot 2 x}=16^{4 x}, \\
\therefore \quad 4 x=x-1, \text { which gives } \quad x=-\frac{1}{3}. \\
\end{array}
$$

Therefore, the answer is $(B)$.