㏒1/3 (2-3х)/х ≥-1 помогите

log1/3[(2-3x)/x] >=-1 .................................. ОДЗ: (2-3x)/x >0 __-____(0)_____+____(2/3)_____- x e (0; 2/3) ..................................... log1/3(2-3x) - log1/3(x) >=-1 log1/3(2-3x) >= log1/3(x)-1 log1/3(2-3x) >= log1/3(x) - log1/3(1/3) log1/3(2-3x) >= log1/3 ( x/ 1/3) log1/3(2-3x) >= log1/3(3x) 2-3x<= 3x -3x-3x<=-2 -6x<=-2 6x>=2 x>=1/3 С учетом ОДЗ: x e [1/3; 2/3)