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Câu hỏi: Cho ${{\log }_{a}}b=2,{{\log }_{b}}c=3$. Tính ${{\log }_{c}}a$.
A. ${{\log }_{c}}a=\dfrac{2}{3}$.
B. ${{\log }_{c}}a=6$.
C. ${{\log }_{c}}a=\dfrac{3}{2}$.
D. ${{\log }_{c}}a=\dfrac{1}{6}$.
Ta có ${{\log }_{c}}a={{\log }_{c}}b\cdot {{\log }_{b}}a=\dfrac{1}{{{\log }_{a}}b\cdot {{\log }_{b}}c}=\dfrac{1}{2.3}=\dfrac{1}{6}.$
A. ${{\log }_{c}}a=\dfrac{2}{3}$.
B. ${{\log }_{c}}a=6$.
C. ${{\log }_{c}}a=\dfrac{3}{2}$.
D. ${{\log }_{c}}a=\dfrac{1}{6}$.
Ta có ${{\log }_{c}}a={{\log }_{c}}b\cdot {{\log }_{b}}a=\dfrac{1}{{{\log }_{a}}b\cdot {{\log }_{b}}c}=\dfrac{1}{2.3}=\dfrac{1}{6}.$
Đáp án D.