Ta có: ĐK x > 0 Khi đó$4{{\log }_{2}}2x-{{x}^{{{\log }_{2}}6}}={{2.3}^{{{\log }_{2}}4{{x}^{2}}}}\Leftrightarrow {{4}^{1+{{\log }_{2}}x}}-{{6}^{{{\log }_{2}}x}}={{2.3}^{2{{\log }_{2}}2x}}$

$\Leftrightarrow {{4.4}^{{{\log }_{2}}x}}-{{6}^{{{\log }_{2}}x}}-{{18.9}^{{{\log }_{2}}x}}=0\Leftrightarrow 4.{{\left( \frac{4}{9} \right)}^{^{{{\log }_{2}}x}}}-{{\left( \frac{2}{3} \right)}^{{{\log }_{2}}x}}-18=0\xrightarrow{t={{\left( \frac{2}{3} \right)}^{{{\log }_{2}}x}}>0}4{{t}^{2}}-2t-18=0\Rightarrow t=\frac{9}{4}$ $\Rightarrow {{\log }_{2}}x=-2\Leftrightarrow x=\frac{1}{4}$.