PT đã cho $<=>{{4}^{1+\lg x}}-{{6}^{\lg x}}={{2.3}^{\lg {{x}^{2}}+2}}<=>{{4.4}^{\lg x}}-{{6}^{\lg x}}={{18.9}^{\lg x}}<=>4{{[{{(\frac{2}{3})}^{\lg x}}\text{ }\!\!]\!\!\text{ }}^{2}}-{{(\frac{2}{3})}^{\lg x}}-18=0$

Đặt $t={{(\frac{2}{3})}^{\lg x}}>0=>4{{t}^{2}}-t-18=0<=>t=\frac{9}{4}$(vì t>0)

$=>{{(\frac{2}{3})}^{\operatorname{lgx}}}=\frac{9}{4}={{(\frac{2}{3})}^{-2}}<=>\lg x=-2<=>x=\frac{1}{100}(TM)$