Google
Câu hỏi: Cho các số $a, b>0$ thỏa mãn ${{\log }_{a}}b=2$. Giá trị của ${{\log }_{{{a}^{2}}}}{{b}^{2}}+{{b}^{{{\log }_{a}}2}}$ bằng
A. $4$.
B. $12$.
C. $6$.
D. $10$.
A. $4$.
B. $12$.
C. $6$.
D. $10$.
Ta có: ${{\log }_{a}}b=2\Leftrightarrow b={{a}^{2}}$
${{\log }_{{{a}^{2}}}}{{b}^{2}}+{{b}^{{{\log }_{a}}2}}={{\log }_{{{a}^{2}}}}{{\left( {{a}^{2}} \right)}^{2}}+{{\left( {{a}^{2}} \right)}^{{{\log }_{a}}2}}={{\log }_{{{a}^{2}}}}{{a}^{4}}+{{\left( {{a}^{{{\log }_{a}}2}} \right)}^{2}}=4.\dfrac{1}{2}.{{\log }_{a}}a+{{\left( 2 \right)}^{2}}=2+4=6$.
${{\log }_{{{a}^{2}}}}{{b}^{2}}+{{b}^{{{\log }_{a}}2}}={{\log }_{{{a}^{2}}}}{{\left( {{a}^{2}} \right)}^{2}}+{{\left( {{a}^{2}} \right)}^{{{\log }_{a}}2}}={{\log }_{{{a}^{2}}}}{{a}^{4}}+{{\left( {{a}^{{{\log }_{a}}2}} \right)}^{2}}=4.\dfrac{1}{2}.{{\log }_{a}}a+{{\left( 2 \right)}^{2}}=2+4=6$.
Đáp án C.