Question

Решите неравенство:
log4 (x+3) + log4 (x+15) ≤ 3

Answer 1

$log_4(x+3)+log_4(x+15) \leq 3\\ \left \{ {{x+3\ \textgreater \ 0} \atop {x+15\ \textgreater \ 0}} \right. \left \{ {{x\ \textgreater \ -3} \atop {x\ \textgreater \ -15}} \right.$
x∈(-3;+∞)
$log_4(x^2+18x+45) \leq 3\\x^2+18x+45 \leq 64\\x^2+18x-19 \leq 0\\x^2+18x-19=0\\a+b+c=0\\x_1=1\\x_2=-19\\(x+19)(x-1) \leq 0$
x∈[-19;1]
Ответ :(-3;1]

Answer 2

Вот так,но не уверена