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Câu hỏi: Cho ${{\log }_{a}}x=2, {{\log }_{b}}x=3$ với a, b là các số thực lớn hơn 1. Tính $P={{\log }_{\dfrac{a}{{{b}^{2}}}}}x$.
A. 5.
B. $-6$.
C. $\dfrac{1}{6}$.
D. $\dfrac{-1}{6}$.
${{\log }_{a}}x=2, {{\log }_{b}}x=3$ $\Rightarrow x\ne 1$
Do đó $P={{\log }_{\dfrac{a}{{{b}^{2}}}}}x=\dfrac{1}{{{\log }_{x}}\dfrac{a}{{{b}^{2}}}}=\dfrac{1}{{{\log }_{x}}a-{{\log }_{x}}{{b}^{2}}}=\dfrac{1}{{{\log }_{x}}a-2{{\log }_{x}}b}=\dfrac{1}{\dfrac{1}{{{\log }_{a}}x}-2\dfrac{1}{{{\log }_{b}}x}}=-6$.
A. 5.
B. $-6$.
C. $\dfrac{1}{6}$.
D. $\dfrac{-1}{6}$.
${{\log }_{a}}x=2, {{\log }_{b}}x=3$ $\Rightarrow x\ne 1$
Do đó $P={{\log }_{\dfrac{a}{{{b}^{2}}}}}x=\dfrac{1}{{{\log }_{x}}\dfrac{a}{{{b}^{2}}}}=\dfrac{1}{{{\log }_{x}}a-{{\log }_{x}}{{b}^{2}}}=\dfrac{1}{{{\log }_{x}}a-2{{\log }_{x}}b}=\dfrac{1}{\dfrac{1}{{{\log }_{a}}x}-2\dfrac{1}{{{\log }_{b}}x}}=-6$.
Đáp án B.