8-19 If $\log _{7}\left[\log _{3}\left(\log _{2} x\right)\right]=0$, then $x^{-\frac{1}{2}}$ equals
(A) $\frac{1}{3}$.
(B) $\frac{1}{2 \sqrt{3}}$.
(C) $\frac{1}{3 \sqrt{3}}$.
(D) $\frac{1}{\sqrt{42}}$.
(E) None of the above.
(34th American High School Mathematics Examination, 1983)
[Solution] To find the value of $x^{-\frac{1}{2}}$, we first solve for $x$ from the original equation. From $\log _{7}\left[\log _{3}\left(\log _{2} x\right)\right]=0$, we have $\log _{3}\left(\log _{2} x\right)=1$, which means $\log _{2} x=3$, so $x=2^{3}=8$.
Therefore,
$$
x^{-\frac{1}{2}}=\frac{1}{\sqrt{8}}=\frac{1}{2 \sqrt{2}} \text {. }
$$

Hence, the answer is $(E)$.