inequação logaritmica com módulo

[latex]\\\mathrm{|log_3(x)|=\left\{\begin{matrix} \mathrm{log_3(x),se\ x\geq 1\ (i)}\\ \mathrm{-log_3(x),se\ x<1\ (ii)} \end{matrix}\right.}\\\\ \mathrm{Par x\geq 1:log_3(x)-log_3(x)-3<0\to -3<0,\ o\ que\ \acute{e}\ verdade\ para\ todo\ x\in\mathbb{R}\ (iii)}\\\\ \mathrm{Assim:(i)\ \cap \ (iii)\to x\geq 1\ \cap \ \mathbb{R}\ \therefore\ S_1=[1,+\infty[}\\\\ \mathrm{Para\ x<1:-log_3(x)-log_3(x)-3<0\to log_3\left ( x^2 \right )>-3\to |x|>\frac{1}{3\sqrt{3}}\ \therefore\ \cancel{\mathrm{x<-\frac{1}{3\sqrt{3}}}}\ \vee\ x>\frac{1}{3\sqrt{3}}}\\\\ \mathrm{Mas\ note\ que\ devido\ ao\ logaritmando\ x\ deve\ ser\ maior\ que\ zero\ \therefore\ x>\frac{1}{3\sqrt{3}}\ (iv)}\\\\ \mathrm{Assim:(ii)\ \cap \ (iv)\to x<1\ \cap \ x>\frac{1}{3\sqrt{3}}\ \therefore\ S_2=\left [ \frac{1}{3\sqrt{3}},1 \right ]}\\\\ \mathrm{S=S_1\ \cup \ S_2\to S=[1,+\infty[\ \cup \ \left [ \frac{1}{3\sqrt{3}},1 \right ]\ \therefore\ S=\left [ \frac{1}{3\sqrt{3}},+\infty \right [}[/latex]

O seu gabarito está errado.