If $${\log _{10}}2 = a$$  and $${\log _{10}}3 = b,$$   then $${\log _5}12$$ = ?

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