Question

ПОМОГИТЕ ПОЖ!!! Решите неравенство: log^2 по основанию 1/6 x>4

Answer 1

$log^2_{frac{1}{6}}x>4, \ x>0, \ log^2_{frac{1}{6}}x-4>0, \ (log_{frac{1}{6}}x-2)(log_{frac{1}{6}}x+2)>0,$
$left [ {{ left { {{log_{frac{1}{6}}x-2<0,} atop {log_{frac{1}{6}}x+2<0;}} right. } atop {left { {{log_{frac{1}{6}}x-2>0,} atop {log_{frac{1}{6}}x+2>0;}} right.}} right. left [ {{ left { {{log_{frac{1}{6}}x<2,} atop {log_{frac{1}{6}}x<-2;}} right. } atop {left { {{log_{frac{1}{6}}x>2,} atop {log_{frac{1}{6}}x>-2;}} right.}} right. left [ {{log_{frac{1}{6}}x<-2,} atop {log_{frac{1}{6}}x>2;}} right.$ 
$left [ {{x>(frac{1}{6})^{-2},} atop {x<(frac{1}{6})^{2};}} right. left [ {{x>36,} atop {x<frac{1}{36};}} right. \ xin(0;frac{1}{36})cup(36;+infty)$