Đặt ${{\log }_{2}}3=a$. Khi đó ${{\log }_{12}}18$ bằng

Ta có: ${{\log }_{12}}18=\dfrac{{{\log }_{2}}18}{{{\log }_{2}}12}=\dfrac{{{\log }_{2}}\left( {{2.3}^{2}} \right)}{{{\log }_{2}}\left( {{2}^{2}}.3 \right)}=\dfrac{1+2{{\log }_{2}}3}{2+{{\log }_{2}}3}=\dfrac{1+2a}{2+a}$