$8 \cdot 39$ If $|x-8 y|+(4 y-1)^{2}=0$, then the value of $\log _{2} y^{x}$ is
(A) 0.
(B) $\frac{1}{4}$.
(C) $\frac{1}{16}$.
(D) -4.
(E) $\sqrt{2}$.
(China Beijing Junior High School Mathematics Competition, 1982)
[Solution] From the given, we know that $x-8y=0$, and $4y-1=0$ which gives $x=2, y=\frac{1}{4}$. Therefore, $\log _{2} y^{x}=\log _{2}\left(\frac{1}{4}\right)^{2}=-4$.

So the answer is $(D)$.