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Câu hỏi: Cho ${{\log }_{2}}5=a; {{\log }_{5}}3=b.$ Tính ${{\log }_{5}}24$ theo $a$ và $b$.
A. ${{\log }_{5}}24=\dfrac{3+ab}{a}.$
B. ${{\log }_{5}}24=\dfrac{a+3b}{a}.$
C. ${{\log }_{5}}24=\dfrac{a+b}{3ab}.$
D. ${{\log }_{5}}24=\dfrac{3a+b}{b}.$
Ta có ${{\log }_{5}}24={{\log }_{5}}\left( {{3.2}^{3}} \right)={{\log }_{5}}3+3{{\log }_{5}}2$ $={{\log }_{5}}3+\dfrac{3}{{{\log }_{2}}5}=b+\dfrac{3}{a}=\dfrac{ab+3}{a}$.
A. ${{\log }_{5}}24=\dfrac{3+ab}{a}.$
B. ${{\log }_{5}}24=\dfrac{a+3b}{a}.$
C. ${{\log }_{5}}24=\dfrac{a+b}{3ab}.$
D. ${{\log }_{5}}24=\dfrac{3a+b}{b}.$
Ta có ${{\log }_{5}}24={{\log }_{5}}\left( {{3.2}^{3}} \right)={{\log }_{5}}3+3{{\log }_{5}}2$ $={{\log }_{5}}3+\dfrac{3}{{{\log }_{2}}5}=b+\dfrac{3}{a}=\dfrac{ab+3}{a}$.
Đáp án A.