$13 \cdot 43$ Among $2^{\frac{1}{2}}, 3^{\frac{1}{3}}, 8^{\frac{1}{8}}, 9^{\frac{1}{9}}$, the largest and the second largest numbers in order are
(A) $3^{\frac{1}{3}}, 2^{\frac{1}{2}}$.
(B) $3^{\frac{1}{3}}, 8^{\frac{1}{8}}$.
(C) $3^{\frac{1}{3}}, 9^{\frac{1}{9}}$.
(D) $8^{\frac{1}{8}}, 9^{\frac{1}{9}}$.
(E) None of these.
(23rd American High School Mathematics Examination, 1972)
[Solution] First note that $\left(2^{\frac{1}{2}}\right)^{6}=8\left(9^{\frac{1}{9}}\right)^{18}=81$, so $2^{\frac{1}{2}}>9^{\frac{1}{9}}$. And $2^{\frac{1}{2}}>8^{\frac{1}{8}}=2^{\frac{3}{8}}$.
Therefore, $3^{\frac{1}{3}}, 2^{\frac{1}{2}}$ are the largest and second largest numbers among these, respectively.
Hence, the answer is $(A)$.