Cho ${{\log }_{2}}6=a.$ Khi đó ${{\log }_{3}}18$ tính theo $a$ là:

Ta có ${{\log }_{2}}6=a\Leftrightarrow {{\log }_{2}}\left( 2.3 \right)=a\Leftrightarrow 1+{{\log }_{2}}3=a\Leftrightarrow {{\log }_{2}}3=a-1.$
Khi đó ${{\log }_{3}}18={{\log }_{3}}\left( {{2.3}^{2}} \right)={{\log }_{3}}2+2=\dfrac{1}{a-1}+2=\dfrac{2a-1}{a-1}.$