| log3 (x) | больше (1/2)        (по основанию 3)

[latex]|log_{3}x|>\frac{1}{2}[/latex]                              ОДЗ: x>0   [latex]log_{3}x>\frac{1}{2}[/latex]               и              [latex]log_{3}x<-\frac{1}{2}[/latex]   [latex]x>3^\frac{1}{2}[/latex]                                       [latex]x<3^{-\frac{1}{2}}[/latex]   [latex]x>\sqrt{3}[/latex]                                               [latex]x<\frac{1}{\sqrt{3}}[/latex]                                                                                 [latex]x<\frac{\sqrt{3}}{3}[/latex]   Учитывая ОДЗ получим ответ:  [latex]x\in(0;\frac{\sqrt{3}}{3})\cup(\sqrt{3};+\infty)[/latex]