Question

log1/4 (x²+6x)≥-2
1/25× 9 loga √14+0,5
допоможіть будь ласка

Answer 1

$a)\\ \(\log_{\frac{1}{4}}(x^2+6x) \geq -2\)\(\frac{1}{4}^{-2} \leq x^2 + 6x\)\(4^2 \leq x^2 + 6x\)\(16 \leq x^2 + 6x\)\(x^2 + 6x - 16 \geq 0\)b) \(\frac{1}{25} \times 9 \log_a(\sqrt{14} + 0.5)\)\(\frac{1}{25} \times 9 \log_a(\sqrt{14} + 0.5)\)\(= \frac{9}{25} \log_a(\sqrt{14} + 0.5)\)$