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Câu hỏi: Cho các số $a,b>0,a\ne 1$ thõa mãn ${{\log }_{ab}}\dfrac{a}{b}=\dfrac{1}{3}.$ Giá trị của ${{\log }_{{{a}^{3}}}}\left( a{{b}^{6}} \right)$ bằng
A. $\dfrac{8}{3}$.
B. $\dfrac{13}{4}$.
C. $\dfrac{8}{9}$.
D. $\dfrac{4}{3}$.
A. $\dfrac{8}{3}$.
B. $\dfrac{13}{4}$.
C. $\dfrac{8}{9}$.
D. $\dfrac{4}{3}$.
Ta có: ${{\log }_{ab}}\dfrac{a}{b}={{\log }_{ab}}a-{{\log }_{ab}}b=\dfrac{1}{1+{{\log }_{a}}b}-\dfrac{1}{{{\log }_{b}}a+1}=\dfrac{1}{3}$
Đặt ${{\log }_{a}}b=t\Rightarrow \dfrac{1}{1+t}-\dfrac{1}{\dfrac{1}{t}+1}=\dfrac{1}{1+t}-\dfrac{t}{1+t}=\dfrac{1-t}{1+t}=\dfrac{1}{3}\Rightarrow t=\dfrac{1}{2}$
Nên: ${{\log }_{{{a}^{3}}}}\left( a{{b}^{6}} \right)={{\log }_{{{a}^{3}}}}a+{{\log }_{{{a}^{3}}}}{{b}^{6}}=\dfrac{1}{3}+2{{\log }_{a}}b=\dfrac{1}{3}+1=\dfrac{4}{3}$.
Đặt ${{\log }_{a}}b=t\Rightarrow \dfrac{1}{1+t}-\dfrac{1}{\dfrac{1}{t}+1}=\dfrac{1}{1+t}-\dfrac{t}{1+t}=\dfrac{1-t}{1+t}=\dfrac{1}{3}\Rightarrow t=\dfrac{1}{2}$
Nên: ${{\log }_{{{a}^{3}}}}\left( a{{b}^{6}} \right)={{\log }_{{{a}^{3}}}}a+{{\log }_{{{a}^{3}}}}{{b}^{6}}=\dfrac{1}{3}+2{{\log }_{a}}b=\dfrac{1}{3}+1=\dfrac{4}{3}$.
Đáp án D.
