Question

3 с проверкой можно? можно 1 задание с log

Answer 1

Ответ:

Пошаговое объяснение:

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Answer 2

1.

$log_{4} (3x - 5) = -2\\ log_{4} (3x - 5) = -2, x > \frac{5}{3} \\3x - 5 = 4^{-2} \\3x - 5 = \frac{1}{4^{2} } \\3x - 5 = \frac{1}{16 } \\3x = \frac{1}{16} +5\\3x = \frac{81}{16} \\x = \frac{27}{16} , x > \frac{5}{3} \\x = \frac{27}{16} = 1\frac{11}{16}$

2.

$25^{7x-2} \leq \frac{1}{25} \\5^{2(7x-2)} \leq 5^{-2}\\5^{14x-4} \leq 5^{-2}\\14x - 4 \leq -2\\14x \leq -2 +4\\14x \leq 2\\x \leq \frac{2}{14} \\x \leq \frac{1}{7}$

3.

$\sqrt{\frac{6}{5x-4} } = \frac{1}{7} \\\frac{6}{5x-4} =\frac{1}{49} \\294 = 5x - 4\\-5x = -4-294\\-5x = -298\\x = \frac{298}{5}$

Проверка:

$\sqrt{\frac{6}{5* \frac{298}{5}-4 } } = \frac{1}{7}\\\sqrt{\frac{6}{298-4} } = \frac{1}{7} \\\sqrt{\frac{6}{294} } = \frac{1}{7} \\\sqrt{\frac{1}{49} } = \frac{1}{7} \\{\frac{1}{7} } = \frac{1}{7}$