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

[latex] 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)[/latex] [latex]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)[/latex]