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Câu hỏi: Nếu ${{\log }_{8}}a+{{\log }_{4}}{{b}^{2}}=5$ và ${{\log }_{4}}{{a}^{2}}+{{\log }_{8}}b=7$ thì giá trị của ${{\log }_{2}}(ab)$ bằng bao nhiêu?
A. 9
B. 18
C. 1
D. 3
A. 9
B. 18
C. 1
D. 3
$\left\{ \begin{aligned}
& {{\log }_{8}}a+{{\log }_{4}}{{b}^{2}}=5 \\
& {{\log }_{4}}{{a}^{2}}+{{\log }_{8}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{3}{{\log }_{2}}2+\dfrac{1}{2}{{\log }_{2}}{{b}^{2}}=5 \\
& \dfrac{1}{2}{{\log }_{2}}{{a}^{2}}+\dfrac{1}{3}{{\log }_{2}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{2}}\left( {{a}^{\dfrac{1}{3}}}b \right)={{\log }_{2}}{{2}^{5}} \\
& {{\log }_{2}}\left( a{{b}^{\dfrac{1}{3}}} \right)={{\log }_{2}}{{2}^{7}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{\dfrac{1}{3}}}b={{2}^{5}} \\
& a{{b}^{\dfrac{1}{3}}}={{2}^{7}} \\
\end{aligned} \right.$
Suy ra ${{\left( ab \right)}^{\dfrac{3}{4}}}={{2}^{12}}\Leftrightarrow ab={{\left( {{2}^{12}} \right)}^{\dfrac{3}{4}}}={{2}^{9}}\Rightarrow {{\log }_{2}}(ab)={{\log }_{2}}{{2}^{9}}=9$.
& {{\log }_{8}}a+{{\log }_{4}}{{b}^{2}}=5 \\
& {{\log }_{4}}{{a}^{2}}+{{\log }_{8}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{3}{{\log }_{2}}2+\dfrac{1}{2}{{\log }_{2}}{{b}^{2}}=5 \\
& \dfrac{1}{2}{{\log }_{2}}{{a}^{2}}+\dfrac{1}{3}{{\log }_{2}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{2}}\left( {{a}^{\dfrac{1}{3}}}b \right)={{\log }_{2}}{{2}^{5}} \\
& {{\log }_{2}}\left( a{{b}^{\dfrac{1}{3}}} \right)={{\log }_{2}}{{2}^{7}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{\dfrac{1}{3}}}b={{2}^{5}} \\
& a{{b}^{\dfrac{1}{3}}}={{2}^{7}} \\
\end{aligned} \right.$
Suy ra ${{\left( ab \right)}^{\dfrac{3}{4}}}={{2}^{12}}\Leftrightarrow ab={{\left( {{2}^{12}} \right)}^{\dfrac{3}{4}}}={{2}^{9}}\Rightarrow {{\log }_{2}}(ab)={{\log }_{2}}{{2}^{9}}=9$.
Đáp án A.