Question

Решите неравенство:
а). $log_{0,5} frac{x}{3} geq -2$
б). $log_{0,6} (2x-1)< log_{0,6}x$

Answer 1

$log_{0,5} frac{x}{3} geq -2, \  left { {{frac{x}{3}>0,} atop {frac{x}{3}leq 0,5^{-2};}} right.  left { {x>0,} atop {xleq 12;}} right. \ 0<x leq 12; \  xin(0;12],$

$log_{0,6} (2x-1)<log_{0,6}x, \ left { {{2x-1>0,} atop {x>0;}} right. left { {{x>0,5,} atop {x>0;}} right. x>0,5; \ 2x-1>x, \ left { {{x>0,5,} atop {x>1;}} right. x>1; \ xin(1;+infty).$