6.62 For all positive numbers $x$ not equal to 1, the algebraic expression $\frac{1}{\log _{3} x}+\frac{1}{\log _{4} x}+\frac{1}{\log _{5} x}$ equals
(A) $\frac{1}{\log _{60} x}$.
(B) $\frac{1}{\log _{x} 60}$.
(C) $\frac{1}{\left(\log _{3} x\right)\left(\log _{4} x\right)\left(\log _{5} x\right)}$.
(D) $\frac{12}{\left(\log _{3} x\right)+\left(\log _{4} x\right)+\left(\log _{5} x\right)}$.
(E) $\frac{\log _{2} x}{\left(\log _{3} x\right)\left(\log _{5} x\right)}+\frac{\log _{3} x}{\left(\log _{2} x\right)\left(\log _{5} x\right)}+\frac{\log _{5} x}{\left(\log _{2} x\right)\left(\log _{3} x\right)}$.
(29th American High School Mathematics Examination, 1978)
[Solution] By $\log _{b} a=\frac{1}{\log _{a} b}$, then
$$
\frac{1}{\log _{3} x}+\frac{1}{\log _{4} x}+\frac{1}{\log _{5} x}=\log _{x} 3+\log _{x} 4+\log _{x} 5=\log _{x} 60=\frac{1}{\log _{60} x} \text {. }
$$
Therefore, the answer is $(A)$.