Question

Решите логарифмические неравенства ДАЮ 65 БАЛЛОВ:
1)$log_{0,5} (2-5x) \leq -2$
2)$log_2 (4-3x) \leq -3$
3)$lod_4 (3-4x) \geq -1$
4)$log_{0,2} (15-2x) \geq -2$
5)$log_{0,8} (3-5x) \geq 0$

Answer 1

Ответ:

Пошаговое объяснение:

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Answer 2

$1)log_{0,5}(2-5x)\leq -2\\x<\frac{2}{5} \\2-5x\geq 0,5^{-2}<=>2-5x\geq 4<=>-5x\geq 2=>x\leq -\frac{2}{5}\\2)log_2(4-3x) \leq-3\\x<\frac{4}{3}\\4-3x\leq 2^{-3}<=>4-3x\leq \frac{1}{8}<=>-3x\leq\frac{31}{8} =>x\geq \frac{31}{24} \\\frac{31}{24}\leq x <\frac{4}{3} \\3)log_4(3-4x )\geq -1\\x<\frac{3}{4} \\3-4x\geq 4^{-1}<=>3-4x\geq \frac{1}{4} <=>-4x\geq -\frac{11}{4} =>x\leq \frac{11}{16}\\4)log_{0,2}(15-2x)\geq-2\\x<\frac{15}{2}\\12-2x\leq 0,2^{-2}<=>15-2x\leq 25<=>-2x\leq 10=>x\geq -5\\-5\leq x<\frac{15}{2}$

$5)log_{0,8}(3-5x)\geq 0\\x<\frac{3}{5} \\3-5x\leq 0,8^0<=>3-5x\leq 1<=>-5x\leq -2=>x\geq \frac{2}{5}\\\frac{2}{5} \leq x<\frac{3}{5}$