6.53 rounded to the thousandth place, $\lg 2$ is $0.301, \lg 3$ is $0.477 . \log _{5} 10$ the best approximation is
(A) $\frac{8}{7}$.
(B) $\frac{9}{7}$.
(C) $\frac{10}{7}$.
(D) $\frac{11}{7}$.
(E) $\frac{12}{7}$.
(30th American High School Mathematics Examination, 1979)
[Solution] Since $\log _{b} a=\frac{1}{\log _{a} b}$, then
$$
\log _{5} 10=\frac{1}{\lg 5}=\frac{1}{\lg 10-\lg 2} \approx \frac{1}{0.699} \approx \frac{10}{7} .
$$

Therefore, the answer is $(C)$.