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Câu hỏi: Giả sử $a$, $b$ là các số thực dương tùy ý, ${{\log }_{4}}\left( {{a}^{6}}{{b}^{2}} \right)$ bằng
A. $12{{\log }_{2}}a-4{{\log }_{2}}b$.
B. $12{{\log }_{2}}a+4{{\log }_{2}}b$.
C. $3{{\log }_{2}}a+{{\log }_{2}}b$.
D. $3{{\log }_{2}}a-{{\log }_{2}}b$.
A. $12{{\log }_{2}}a-4{{\log }_{2}}b$.
B. $12{{\log }_{2}}a+4{{\log }_{2}}b$.
C. $3{{\log }_{2}}a+{{\log }_{2}}b$.
D. $3{{\log }_{2}}a-{{\log }_{2}}b$.
Ta có
${{\log }_{4}}\left( {{a}^{6}}{{b}^{2}} \right)=\dfrac{1}{2}{{\log }_{2}}\left( {{a}^{6}}{{b}^{2}} \right)=\dfrac{1}{2}\left( {{\log }_{2}}{{a}^{6}}+{{\log }_{2}}{{b}^{2}} \right)=\dfrac{1}{2}\left( 6{{\log }_{2}}a+2{{\log }_{2}}b \right)=3{{\log }_{2}}a+{{\log }_{2}}b$.
${{\log }_{4}}\left( {{a}^{6}}{{b}^{2}} \right)=\dfrac{1}{2}{{\log }_{2}}\left( {{a}^{6}}{{b}^{2}} \right)=\dfrac{1}{2}\left( {{\log }_{2}}{{a}^{6}}+{{\log }_{2}}{{b}^{2}} \right)=\dfrac{1}{2}\left( 6{{\log }_{2}}a+2{{\log }_{2}}b \right)=3{{\log }_{2}}a+{{\log }_{2}}b$.
Đáp án C.