Математика: Срешением! log_2(x+4) =log_(4x+16)(8)

С учётом ОДЗ (2):ответ: x ∈ [-31/8; -15/4)∪ [-2; +∞)...

Срешением! log_2(x+4)> =log_(4x+16)(8)

Ответ:

snizhanadob

16.08.2020 22:56

$\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.$