Question

 Помогите решить 
log по основанию 2 (x-4) < 2
log по основанию 3 (x+) < 1

Answer 1

$log_2(x-4) < 2\ left { {{x-4>0} atop {log_2(x-4) < 2}} right.<=> left { {{x>4} atop {x-4<2^2}} right. <=>\ <=> left { {{x>4} atop {x-4<4}} right. <=> left { {{x>4} atop {x<8}} right. (4; 8)$

$log_3(x+2) < 1\ left { {{x+2>0} atop {log_3(x+2) < 1}} right.<=> left { {{x>-2} atop {x+2<2^1}} right. <=>\ <=> left { {{x>-2} atop {x+2<2}} right. <=> left { {{x>-2} atop {x<0}} right. (-2; 0)$