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Câu hỏi: Cho ${{\log }_{3}}5=a$. Tính ${{\log }_{243}}1125$ theo $a$.
A. $\dfrac{3+2a}{5}$.
B. $\dfrac{3a}{5}$.
C. $\dfrac{2+3a}{4}$.
D. $\dfrac{2+3a}{5}$.
A. $\dfrac{3+2a}{5}$.
B. $\dfrac{3a}{5}$.
C. $\dfrac{2+3a}{4}$.
D. $\dfrac{2+3a}{5}$.
Ta có ${{\log }_{243}}1125={{\log }_{{{3}^{5}}}}\left( {{3}^{2}}{{.5}^{3}} \right)=\dfrac{1}{5}\left( {{\log }_{3}}{{3}^{2}}+{{\log }_{_{3}}}{{5}^{3}} \right)=\dfrac{2+3a}{5}$.
Đáp án D.
