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

Question

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

Comments

Что во втором логарифме в скобках

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X+2

Answer 1

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)" alt=" 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)" align="absmiddle" class="latex-formula">

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)" alt="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)" align="absmiddle" class="latex-formula">