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Câu hỏi: Cho $a>0$, $b>0$ và $ab\ne 1$ thỏa mãn $3\ln a+7\ln b=0$, khi đó ${{\log }_{\sqrt{ab}}}b\sqrt[3]{a}$ bằng
A. $3$.
B. $-3$.
C. $\dfrac{1}{3}$.
D. $-\dfrac{1}{3}$.
A. $3$.
B. $-3$.
C. $\dfrac{1}{3}$.
D. $-\dfrac{1}{3}$.
Ta có: $3\ln a+7\ln b=0\Leftrightarrow {{a}^{3}}{{b}^{7}}=1\Leftrightarrow {{a}^{3}}=\dfrac{1}{{{b}^{7}}}\Leftrightarrow a=\dfrac{1}{{{b}^{\dfrac{7}{3}}}}={{b}^{-\dfrac{7}{3}}}$.
Khi đó ${{\log }_{\sqrt{ab}}}b\sqrt[3]{a}={{\log }_{\sqrt{{{b}^{-\dfrac{7}{3}}}.b}}}b.{{b}^{-\dfrac{7}{9}}}={{\log }_{{{b}^{-\dfrac{2}{3}}}}}{{b}^{\dfrac{2}{9}}}=-\dfrac{3}{2}.\dfrac{2}{9}{{\log }_{b}}b=-\dfrac{1}{3}$.
Khi đó ${{\log }_{\sqrt{ab}}}b\sqrt[3]{a}={{\log }_{\sqrt{{{b}^{-\dfrac{7}{3}}}.b}}}b.{{b}^{-\dfrac{7}{9}}}={{\log }_{{{b}^{-\dfrac{2}{3}}}}}{{b}^{\dfrac{2}{9}}}=-\dfrac{3}{2}.\dfrac{2}{9}{{\log }_{b}}b=-\dfrac{1}{3}$.
Đáp án D.