Biết ${{\log }_{2}}3=a , {{\log }_{3}}5=b$. Khi đó ${{\log }_{15}}12$ bằng

Ta có: ${{\log }_{15}}12=\frac{{{\log }_{3}}12}{{{\log }_{3}}15}=\frac{1+2{{\log }_{3}}2}{1+{{\log }_{3}}5}=\frac{1+\frac{2}{{{\log }_{2}}3}}{1+{{\log }_{3}}5}=\frac{1+\frac{2}{a}}{1+b}=\frac{a+2}{a\left( b+1 \right)}$.