2. The value of $x=\frac{1}{\log _{\frac{1}{2}} \frac{1}{3}}+\frac{1}{\log _{\frac{1}{5}} \frac{1}{3}}$ belongs to the interval
(A) $(-2,-1)$;
(B) $(1,2)$;
(C) $(-3,-2)$;
(D) $(2,3)$.
2. (D)
$$
\begin{array}{cc}
\because & x=\log _{\frac{1}{3}} \frac{1}{2}+\log _{\frac{1}{3}} \frac{1}{5}=\log _{\frac{1}{3}} \frac{1}{10}=\log _{3} 10, \\
\therefore & 2=\log _{3} 9<\log _{3} 10<\log _{3} 27=3 . \\
\therefore & x \in(2,3) .
\end{array}
$$