ĐK :$x>0$
$(*) \Leftrightarrow
{2}{.}{{9}^{{lo}{{g}_{2}}{x - 1}}}{ =
}{{x}^{{lo}{{g}_{x}}{6}{.lo}{{g}_{2}}
{x}}}{ - }{{2}^{{2lo}{{g}_{2}}{x}}}$
$\Leftrightarrow\frac{2}{9}.9^{\log_2x}=6^{\log_2x}-4^{\log_2x}$
Chia cả 2 vế
cho $ 6^{\log_2x}$:
$\frac{2}{9}(\frac{3}{2})^{\log_2x}=1-(\frac{2}{3})^{\log_2x}$
Đặt
$t=(\frac{3}{2})^{\log_2x}>0$
$PT\Leftrightarrow\frac{2}{9}t=1-\frac{1}{t}\Leftrightarrow\frac{2}{9}t^2-t+1=0\Leftrightarrow\left[\begin{array}{I}
t=3\\ t=\frac{3}{2}\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{I}\log_2x=\log_{\frac{3}{2}}
3\\\log_2x=1\end{array}\right.\Leftrightarrow\left[\begin{array}{I}
x=2^{\log_{\frac{3}{2}}3}\\ x=2\end{array}\right.$