Nếu log 8 a+log 4 b^2=5 và

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} \left\{\begin{array}{l}{\log }_{8}a+{\log }_4{b}^2=5\\ {\log }_4{a}^2+{\log }_{8}b=7\end{array}\right.\iff \left\{\begin{array}{l}\frac{1}{3}{\log }_{2}2+\frac{1}{2}{\log }_2{b}^2=5\\ \frac{1}{2}{\log }_2{a}^2+\frac{1}{3}{\log }_{2}b=7\end{array}\right.\iff \left\{\begin{array}{l}{\log }_2\left(a^{\frac{1}{3}}b\right)={\log }_2{2}^5\\ {\log }_2\left(a{b}^{\frac{1}{3}}\right)={\log }_2{2}^7\end{array}\right. \\ \iff \left\{\begin{array}{l}{a}^{\frac{1}{3}}b=2^5\\ a{b}^{\frac{1}{3}}=2^7\end{array}\right. \end{gathered}$$

Suy ra ${\left(ab\right)}^{\frac{3}{4}}=2^{12}\iff ab={\left(2^{12}\right)}^{\frac{3}{4}}=2^9\Rightarrow {\log }_2\left(ab\right)={\log }_2{2}^9=9$.