Question

LG(5x-7)>2; log0,1(5-3x)<-2 решение простейших логарифмических неравенств

Answer 1

$lg(5x-7) textgreater 2 \ \lg(5x-7) textgreater 2cdot 1 \ \ 1=lg10 \ \lg(5x-7) textgreater 2cdot lg10 \ \lg(5x-7) textgreater 2cdot lg10 \ \ lg(5x-7) textgreater lg10^2 \ \ left { {{5x-7 textgreater 100} atop {5x-7 textgreater 0}} right. \ \ 5x textgreater 107 \ \ x textgreater 21,4$

$log0,1(5-3x) textless -2 \ \ log0,1(5-3x) textless -2cdot log{0,1}0,1 \ \ log0,1(5-3x) textless log{0,1}(0,1)^{-2} \ \ left { {{5-3x textgreater 0} atop {5-3x textgreater 100}} right. \ \ 5-3x textgreater 100 \ \ -3x textgreater 95 \ \ x textless -31 frac{2}{3}$