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Câu hỏi: Cho $a>0$ và đặt ${{\log }_{2}}a=x$. Tính ${{\log }_{8}}\left( 4{{a}^{3}} \right)$ theo $x$.
A. ${{\log }_{8}}\left( 4{{a}^{3}} \right)=3x+2$.
B. ${{\log }_{8}}\left( 4{{a}^{3}} \right)=x+\dfrac{2}{3}$.
C. ${{\log }_{8}}\left( 4{{a}^{3}} \right)=9x+6$.
D. ${{\log }_{8}}\left( 4{{a}^{3}} \right)=-\dfrac{3x+2}{3}$.
A. ${{\log }_{8}}\left( 4{{a}^{3}} \right)=3x+2$.
B. ${{\log }_{8}}\left( 4{{a}^{3}} \right)=x+\dfrac{2}{3}$.
C. ${{\log }_{8}}\left( 4{{a}^{3}} \right)=9x+6$.
D. ${{\log }_{8}}\left( 4{{a}^{3}} \right)=-\dfrac{3x+2}{3}$.
Ta có :
$\begin{aligned}
& {{\log }_{8}}\left( 4{{a}^{3}} \right)={{\log }_{{{2}^{3}}}}\left( 4{{a}^{3}} \right)=\dfrac{1}{3}{{\log }_{2}}\left( {{2}^{2}}.{{a}^{3}} \right)=\dfrac{1}{3}\left( {{\log }_{2}}{{2}^{2}}+{{\log }_{2}}{{a}^{3}} \right) \\
& =\dfrac{1}{3}\left( 2+3{{\log }_{2}}a \right)=\dfrac{2}{3}+{{\log }_{2}}a=x+\dfrac{2}{3} \\
\end{aligned}$
$\begin{aligned}
& {{\log }_{8}}\left( 4{{a}^{3}} \right)={{\log }_{{{2}^{3}}}}\left( 4{{a}^{3}} \right)=\dfrac{1}{3}{{\log }_{2}}\left( {{2}^{2}}.{{a}^{3}} \right)=\dfrac{1}{3}\left( {{\log }_{2}}{{2}^{2}}+{{\log }_{2}}{{a}^{3}} \right) \\
& =\dfrac{1}{3}\left( 2+3{{\log }_{2}}a \right)=\dfrac{2}{3}+{{\log }_{2}}a=x+\dfrac{2}{3} \\
\end{aligned}$
Đáp án B.