$8 \cdot 24$ satisfies the equation $\left(\log _{3} x\right)\left(\log _{x} 5\right)=\log _{3} 5$ for the positive number $x$
(A) only 3 and 5.
(B) only 3, 5, and 15.
(C) only numbers of the form $5^{n} \cdot 3^{m}$, where $n$ and $m$ are positive integers.
(D) all positive numbers $x \neq 1$.
(E) none of these.
(26th American High School Mathematics Examination, 1975)
[Solution] If $x>0$ and $x \neq 1$, let $a=\log _{3} x, \quad b=\log _{x} 5$ and $c=\log _{3} 5$, then $x=3^{a}, \quad 5=x^{b}$ and $5=3^{c}$.

Thus $3^{a b}=3^{c}$ or $a b=c$.

That is, when $x \neq 1$, $a b=c$.
Therefore, the answer is $(D)$.