$6 \cdot 60$ If $x=\left(\log _{8} 2\right)^{\log _{2} 8}$, then $\log _{3} x$ equals
(A) -3 .
(B) $-\frac{1}{3}$.
(C) $\frac{1}{3}$.
(D) 3 .
(E) 9 .
(17th American High School Mathematics Examination, 1966)
[Solution] Let $x=\left(\log _{8} 2\right)^{\log _{2} 8}=\left(\log _{8} 2\right)^{3}=\left(\frac{1}{\log _{2} 8}\right)^{3}=\left(\frac{1}{3}\right)^{3}$, $\therefore \quad \log _{3} x=\log _{3}\left(\frac{1}{3}\right)^{3}=3 \log _{3} 3^{-1}=-3$. Therefore, the answer is $(A)$.