1. Решите неравенство: \( \log _{2 x} 0,25 \leqslant \log _{2} 32 x-1 \)

Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{0.25}{2x}=\frac{\left(32x-1\right)}{2}\) - step1: Find the domain: \(\frac{0.25}{2x}=\frac{\left(32x-1\right)}{2},x\neq 0\) - step2: Remove the parentheses: \(\frac{0.25}{2x}=\frac{32x-1}{2}\) - step3: Divide the terms: \(\frac{1}{8x}=\frac{32x-1}{2}\) - step4: Cross multiply: \(2=8x\left(32x-1\right)\) - step5: Rewrite the expression: \(2=2\times 4x\left(32x-1\right)\) - step6: Evaluate: \(1=4x\left(32x-1\right)\) - step7: Swap the sides: \(4x\left(32x-1\right)=1\) - step8: Expand the expression: \(128x^{2}-4x=1\) - step9: Move the expression to the left side: \(128x^{2}-4x-1=0\) - step10: Solve using the quadratic formula: \(x=\frac{4\pm \sqrt{\left(-4\right)^{2}-4\times 128\left(-1\right)}}{2\times 128}\) - step11: Simplify the expression: \(x=\frac{4\pm \sqrt{\left(-4\right)^{2}-4\times 128\left(-1\right)}}{256}\) - step12: Simplify the expression: \(x=\frac{4\pm \sqrt{528}}{256}\) - step13: Simplify the expression: \(x=\frac{4\pm 4\sqrt{33}}{256}\) - step14: Separate into possible cases: \(\begin{align}&x=\frac{4+4\sqrt{33}}{256}\\&x=\frac{4-4\sqrt{33}}{256}\end{align}\) - step15: Simplify the expression: \(\begin{align}&x=\frac{1+\sqrt{33}}{64}\\&x=\frac{4-4\sqrt{33}}{256}\end{align}\) - step16: Simplify the expression: \(\begin{align}&x=\frac{1+\sqrt{33}}{64}\\&x=\frac{1-\sqrt{33}}{64}\end{align}\) - step17: Check if the solution is in the defined range: \(\begin{align}&x=\frac{1+\sqrt{33}}{64}\\&x=\frac{1-\sqrt{33}}{64}\end{align},x\neq 0\) - step18: Find the intersection: \(\begin{align}&x=\frac{1+\sqrt{33}}{64}\\&x=\frac{1-\sqrt{33}}{64}\end{align}\) - step19: Rewrite: \(x_{1}=\frac{1-\sqrt{33}}{64},x_{2}=\frac{1+\sqrt{33}}{64}\) Solve the equation \( \frac{\text{\log}(0.25)}{\text{\log}(2)} = \frac{\text{\log}(32x-1)}{\text{\log}(2)} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{0.25}{2}=\frac{\left(32x-1\right)}{2}\) - step1: Remove the parentheses: \(\frac{0.25}{2}=\frac{32x-1}{2}\) - step2: Divide the terms: \(\frac{1}{8}=\frac{32x-1}{2}\) - step3: Swap the sides: \(\frac{32x-1}{2}=\frac{1}{8}\) - step4: Multiply both sides of the equation by \(2:\) \(\frac{32x-1}{2}\times 2=\frac{1}{8}\times 2\) - step5: Multiply the terms: \(32x-1=\frac{2}{8}\) - step6: Evaluate: \(32x-1=\frac{1}{4}\) - step7: Move the constant to the right side: \(32x=\frac{1}{4}+1\) - step8: Add the numbers: \(32x=\frac{5}{4}\) - step9: Multiply by the reciprocal: \(32x\times \frac{1}{32}=\frac{5}{4}\times \frac{1}{32}\) - step10: Multiply: \(x=\frac{5}{128}\) Solve the equation \( 2x > 0 \). Solve the inequality by following steps: - step0: Solve for \(x\): \(2x>0\) - step1: Rewrite the expression: \(x>0\) Solve the equation \( 32x-1=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(32x-1=0\) - step1: Move the constant to the right side: \(32x=0+1\) - step2: Remove 0: \(32x=1\) - step3: Divide both sides: \(\frac{32x}{32}=\frac{1}{32}\) - step4: Divide the numbers: \(x=\frac{1}{32}\) Давайте решим неравенство \( \log_{2x} 0,25 \leqslant \log_{2} (32x-1) \) шаг за шагом. ### Шаг 1: Преобразование логарифмов Сначала преобразуем логарифмы. Мы знаем, что \( 0.25 = \frac{1}{4} = 2^{-2} \), поэтому: \[ \log_{2}(0.25) = -2 \] Теперь у нас есть: \[ \log_{2x}(0.25) = \frac{\log_{2}(0.25)}{\log_{2}(2x)} = \frac{-2}{\log_{2}(2) + \log_{2}(x)} = \frac{-2}{1 + \log_{2}(x)} \] Теперь неравенство выглядит так: \[ \frac{-2}{1 + \log_{2}(x)} \leqslant \log_{2}(32x-1) \] ### Шаг 2: Решение уравнения Теперь решим уравнение \( 32x - 1 = 0 \): \[ 32x - 1 = 0 \implies 32x = 1 \implies x = \frac{1}{32} \] ### Шаг 3: Условия для логарифмов Логарифм определен только для положительных значений, поэтому: 1. \( 2x > 0 \implies x > 0 \) 2. \( 32x - 1 > 0 \implies x > \frac{1}{32} \) ### Шаг 4: Объединение условий Теперь у нас есть два условия: 1. \( x > 0 \) 2. \( x > \frac{1}{32} \) Наиболее строгим условием является \( x > \frac{1}{32} \). ### Шаг 5: Итоговое решение Таким образом, решение неравенства \( \log_{2x} 0,25 \leqslant \log_{2} (32x-1) \) будет: \[ x > \frac{1}{32} \] Это означает, что все значения \( x \), которые больше \( \frac{1}{32} \), удовлетворяют данному неравенству.