Question

Решить логарифмическое неравенство

Answer 1

$log_2(x+1)^2*log_{\frac{1}{3}}x^2 - 4log_2(x+1) + 4 log_3(-x)+4 \leq 0\\-4log_2(x+1)*log_3(-x) - 4log_2(x+1) + 4log_3(-x)+4\leq 0\\log_3(-x) = a, log_2(x+1)=b\\-4ab - 4b + 4a + 4 \leq 0\\ab + b - a - 1 \geq 0\\b(a+1)-(a+1)\geq 0\\(a+1)(b-1)\geq 0\\(log_3(-x)+1)(log_2(x+1)-1)\geq 0\\(log_3(-x)-log_3\frac{1}{3} )(log_2(x+1)-log_22)\geq 0\\(-x-\frac{1}{3})(x+1-2)\geq 0\\(x+\frac{1}{3})(x-1)\leq 0\\x\in [-\frac{1}{3}; 1]\\ ODZ: x\in (-1; 0)\\Answer: x \in [-\frac{1}{3};0)$