$6 \cdot 31$ If $\lg 2=a, \lg 3=b$, then $\log _{5} 12$ equals
(A) $\frac{a+b}{1+a}$.
(B) $\frac{2 a+b}{1+a}$.
(C) $\frac{a+2 b}{1+a}$.
(D) $\frac{2 a+b}{1-a}$.
(E) $\frac{a+2 b}{1-a}$.
(12th American High School Mathematics Examination, 1961)
［Solution］By the change of base formula, we have
$$
\log _{5} 12=\frac{\lg 12}{\lg 5}=\frac{\lg 3+2 \lg 2}{\lg 10-\lg 2}=\frac{2 a+b}{1-a} .
$$

Therefore, the correct choice is $(D)$.