Question

решите неравенство:
($\frac{1}{6}$)^(x+1) ≥3

Answer 1

Ответ:

$(-\infty;-log_618]$

Объяснение:

$(\frac{1}{6})^{x+1}\geq 3\\\\(\frac{1}{6})^x*\frac{1}{6}\geq 3\\\\(\frac{1}{6})^x\geq 3:\frac{1}{6}\\\\(\frac{1}{6})^x\geq 3*6\\\\(\frac{1}{6})^x\geq 18\\\\0 < \frac{1}{6} < 1\\\\x\leq log_{\frac{1}{6}}18\\\\x\leq \log_{6^{-1}}18\\\\x\leq-log_618\\\\x\in(-\infty;-log_618]$