Answer: Option D
$$\eqalign{
& {\log _5}12 = {\log _5}\left( {3 \times 4} \right) \cr
& = {\log _5}3 + {\log _5}4 \cr
& = {\log _5}3 + 2{\log _5}2 \cr
& = \frac{{{{\log }_{10}}3}}{{{{\log }_{10}}5}} + \frac{{2{{\log }_{10}}2}}{{{{\log }_{10}}5}} \cr} $$
$$ = \frac{{{{\log }_{10}}3}}{{{{\log }_{10}}10 - {{\log }_{10}}2}}$$   $$ + \frac{{2{{\log }_{10}}2}}{{{{\log }_{10}}10 - {{\log }_{10}}2}}$$
$$\eqalign{
& = \frac{b}{{1 - a}} + \frac{{2a}}{{1 - a}} \cr
& = \frac{{2a + b}}{{1 - a}} \cr} $$