Question

Решить неравенство log1/2 (8-2x)≤-4

Answer 1

$\displaystyle\bf\\ODZ:\\\\8-2x > 0\\\\-2x > -8\\\\\boxed{x < 4}\\\\\\\log_{\dfrac{1}{2} } (8-2x)\leq -4\\\\\\0 < \frac{1}{2} < 1 \ \ \ \Rightarrow \ \ \ 8-2x\geq \Big(\frac{1}{2} \Big)^{-4}\\\\\\8-2x\geq2^{4} \\\\\\8-2x\geq 16\\\\\\-2x\geq 8\\\\\\x\leq -4\\\\\\x\in\Big(-\infty \ ; \ -4\Big]$