$6 \cdot 59$ If $\log _{M} N=\log _{N} M$, and $M \neq N, \quad M N>0$, also $M \neq 1$, $N \neq 1$, then $M N$ equals
(A) $\frac{1}{2}$.
(B) 1 .
(C) 2 .
(D) 10 .
(E) a number greater than 2 and less than 10.
(17th American High School Mathematics Examination, 1966)
[Solution] By $\log _{M} N=\frac{1}{\log _{N} M}$, hence from the given condition, $\left(\log _{N} M\right)^{2}=1$, that is,
$$
\log _{N} M= \pm 1 .
$$

If $\log _{N} M=1$, then $M=N$, which contradicts the given condition;
If $\log _{N} M=-1$, then $M=N^{-1}$, i.e., $M N=1$. Therefore, the answer is $(B)$.