Câu hỏi: Đặt $m={{\log }_{6}}2$, $n={{\log }_{6}}5$. Khi đó, ${{\log }_{3}}5$ bằng A. $\dfrac{n}{m-1}$. B. $\dfrac{n}{m+1}$. C. $\dfrac{n}{1-m}$. D. $\dfrac{m}{n}$.
Ta có ${{\log }_{3}}5=\dfrac{{{\log }_{6}}5}{{{\log }_{6}}3}=\dfrac{{{\log }_{6}}5}{1-{{\log }_{6}}2}=\dfrac{n}{1-m}$.