Question

Помогите с решением!
log_2(x+4)>=log_(4x+16)(8)

Answer 1

$log_{2}(x+4) ge log_{(4x+16)}8 \ \ ODZ: left{ begin{gathered} x + 4 > 0 \ 4x+16 > 0 \ 4x + 16 ne 1 end{gathered} right. ; left{ begin{gathered} x > -4 \ x ne -dfrac{15}{4} \ end{gathered} right. ; x in (-4;-dfrac{15}{4})cup ( -dfrac{15}{4}; +infty)$

$log_{2}(x+4) ge dfrac{1}{log_{8}(4x+16)} \ \ log_{2}(x+4) ge dfrac{1}{frac{1}{3}log_{2}(4x+16)} \ \ log_{2}(x+4) ge dfrac{3}{log_{2}(4(x+4))} \ \ log_{2}(x+4) ge dfrac{3}{log_{2}(4) + log_{2}(x+4)} \ \ log_{2}(x+4) ge dfrac{3}{2 + log_{2}(x+4)} \ \ log_{2}(x+4) = t \ \ t ge dfrac{3}{2+t} \ \ dfrac{t^{2}+2t-3}{2+t} ge 0 \ \ dfrac{(t-1)(t+3)}{2+t} ge 0 (1)$

$left[ begin{gathered} -3 le t < -2 \ t ge 1 \ end{gathered} right.$

$left[ begin{gathered} -3 le log_{2}(x+4) < -2 \ log_{2}(x+4) ge 1 \ end{gathered} right.$

$left[ begin{gathered} left{ begin{gathered} log_{2}(x+4) ge -3 \ log_{2}(x+4) < -2 \ end{gathered} right. \ log_{2}(x+4) ge 1 \ end{gathered} right.$

$left[ begin{gathered} left{ begin{gathered} log_{2}(x+4) ge log_{2}2^{-3} \ log_{2}(x+4) < log_{2}2^{-2} \ end{gathered} right. \ log_{2}(x+4) ge log_{2}2^{1} \ end{gathered} right.$

$left[ begin{gathered} left{ begin{gathered} x+4 ge 2^{-3} \ x+4 < 2^{-2} \ end{gathered} right. \ x+4 ge 2 \ end{gathered} right.$

$left[ begin{gathered} left{ begin{gathered} x ge -dfrac{31}{8} \ x < -dfrac{15}{4} \ end{gathered} right. \ x ge -2 \ end{gathered} right.$

$x in [-dfrac{31}{8}; -dfrac{15}{4}) cup [-2;+infty)$

С учётом ОДЗ (2):

$x in [-dfrac{31}{8}; -dfrac{15}{4}) cup [-2;+infty)$

Ответ: x ∈ [-31/8; -15/4)∪ [-2; +∞)