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Câu hỏi: Biết ${{\log }_{a}}b=-2$, tính ${{\log }_{b}}{{a}^{2}}{{b}^{3}}$.
A. ${{\log }_{b}}{{a}^{2}}{{b}^{3}}=2$.
B. ${{\log }_{b}}{{a}^{2}}{{b}^{3}}=6$.
C. ${{\log }_{b}}{{a}^{2}}{{b}^{3}}=4$.
D. ${{\log }_{b}}{{a}^{2}}{{b}^{3}}=7$.
A. ${{\log }_{b}}{{a}^{2}}{{b}^{3}}=2$.
B. ${{\log }_{b}}{{a}^{2}}{{b}^{3}}=6$.
C. ${{\log }_{b}}{{a}^{2}}{{b}^{3}}=4$.
D. ${{\log }_{b}}{{a}^{2}}{{b}^{3}}=7$.
Ta có ${{\log }_{b}}{{a}^{2}}{{b}^{3}}=\dfrac{{{\log }_{a}}{{a}^{2}}{{b}^{3}}}{{{\log }_{a}}b}=\dfrac{{{\log }_{a}}{{a}^{2}}+{{\log }_{a}}{{b}^{3}}}{{{\log }_{a}}b}=\dfrac{2+3{{\log }_{a}}b}{{{\log }_{a}}b}=\dfrac{2+3.\left( -2 \right)}{-2}=2$.
Đáp án A.