Bài 104217

$\Leftrightarrow\frac{2}{9}.9^{\log_2x}=6^{\log_2x}-4^{\log_2x}$

Chia cả 2 vế cho $ 6^{\log_2x}$:

$\frac{2}{9}(\frac{3}{2})^{\log_2x}=1-(\frac{2}{3})^{\log_2x}$

Đặt $t=(\frac{3}{2})^{\log_2x}>0$

$PT\Leftrightarrow\frac{2}{9}t=1-\frac{1}{t}\Leftrightarrow\frac{2}{9}t^2-t+1=0\Leftrightarrow\left[\begin{array}{I} t=3\\ t=\frac{3}{2}\end{array}\right.$

$\Leftrightarrow\left[\begin{array}{I}\log_2x=\log_{\frac{3}{2}} 3\\\log_2x=1\end{array}\right.\Leftrightarrow\left[\begin{array}{I} x=2^{\log_{\frac{3}{2}}3}\\ x=2\end{array}\right.$