${\log }_9\left(3{\log }_2(1+{\log }_3(1+2{\log }_{2}x))\right)=\frac{1}{2}$ ,  then $x=$

${\log }_9\left(3{\log }_2(1+{\log }_3(1+2{\log }_{2}x))\right)=\frac{1}{2}$

${\log }_9\left(3{\log }_2(1+{\log }_3(1+2{\log }_{2}x))\right)=\frac{1}{2}$

${\log }_2(1+{\log }_3(1+2{\log }_{2}x))=1$

$1+{\log }_3(1+2{\log }_{2}x)=2$

${\log }_3(1+2{\log }_{2}x)=1$

$1+2{\log }_{2}x=3$

$2{\log }_{2}x=2$

${\log }_{2}x=1$

$x=2$