8-38 If $(8 y-1)^{2}+|x-16 y|=0$, then the value of $\log _{2} y^{x}$ is
(A) -6.
(B) 0.
(C) $\frac{1}{64}$.
(D) $\frac{1}{8}$.
(Shanghai Junior High School Mathematics Competition, 1983)
[Solution] From the given conditions and $(8 y-1)^{2} \geqslant 0,|x-16 y| \geqslant 0$, we have $\left\{\begin{array}{l}8 y-1=0, \\ x-16 y=0 .\end{array}\right.$ Solving these, we get $\left\{\begin{array}{l}x=2, \\ y=\frac{1}{8} .\end{array}\right.$ $\therefore \quad \log _{2} y^{x}=2 \log _{2} \frac{1}{8}=-6$. Therefore, the answer is $(A)$.