6.24 $x=\frac{1}{\log _{\frac{1}{2}} \frac{1}{3}}+\frac{1}{\log _{\frac{1}{5}} \frac{1}{3}}$ is in the interval
(A) $(-2,-1)$.
(B) $(1,2)$.
(C) $(-3,-2)$.
(D) $(2,3)$.
(China High School Mathematics League, 1983)
［Solution］ $\because x=\frac{\lg \frac{1}{2}}{\lg \frac{1}{3}}+\frac{\lg \frac{1}{5}}{\lg \frac{1}{3}}=\frac{\lg 10}{\lg 3}$,
and
$$
\begin{array}{l} 
\frac{\lg 9}{\lg 3}<\frac{\lg 10}{\lg 3}<\frac{\lg 27}{\lg 3}, \text{ i.e., } 2<x<3, \\
\therefore \quad x \in(2,3)
\end{array}
$$

Therefore, the answer is $(D)$.