6・16 Let $a, b$ be positive numbers, and satisfy $a^{b}=b^{a}$ and $b=9a$, then the value of $a$ is
(A) 9.
(B) $\frac{1}{9}$.
(C) $\sqrt[9]{9}$.
(D) $\sqrt[3]{9}$.
(E) $\sqrt[4]{3}$.
(36th American High School Mathematics Examination, 1985)
[Solution 1] Given $a^{9 a}=(9 a)^{a}$, then $a^{9 a}=9^{a} \cdot a^{a}$, which means $a^{8 a}=9^{a}$, or $a^{8 a}=\left(9^{\frac{1}{8}}\right)^{8 a}$.

Thus, $a=9^{\frac{1}{8}}=\sqrt[8]{9}=\sqrt[4]{3}$.
Therefore, the answer is $(E)$.
[Solution 2] Given $a^{9 a}=(9 a)^{a}$, taking the logarithm on both sides, we have
$$
9 a \lg a=a(\lg 9+\lg a),
$$

which simplifies to $8 \lg a=\lg 9$, or $\lg a=\lg 9 \frac{1}{8}$, then
$$
a=9^{\frac{1}{8}}=3^{\frac{1}{4}}=\sqrt[4]{3} \text {. }
$$

Therefore, the answer is $(E)$.