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1. \( \left(2\right. \) балла) \( 2^{2-\mathrm{x}}=32 \) 2. \( \left(2\right. \) балла) \( 81^{\mathrm{x}-4}=\frac{1}{3} \) 3. \( \left(2\right. \) балла) \( \left(\frac{1}{4}\right) 4 \mathrm{x}-13=\frac{1}{64} \) 4. \( \left(3\right. \) балла) \( 5^{\mathrm{x}+2}-5^{\mathrm{x}}=120 \) 5. (3 балла \( \left(4^{\mathrm{x}}-6 \cdot 2^{\mathrm{x}}+8=0\right. \) 6. (3 балла) \( 9^{\mathrm{x}}+3 \cdot 3^{\mathrm{x}}-18=0 \)

Ask by Norton Bowman. in Russia
Feb 03,2025

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Answer

**1. \(2^{2 - x} = 32\)** - \(x = -3\) **2. \(81^{x - 4} = \frac{1}{3}\)** - \(x = 3.75\) **3. \(\left(\frac{1}{4}\right) \cdot 4x - 13 = \frac{1}{64}\)** - \(x \approx 13.015625\) **4. \(5^{x + 2} - 5^{x} = 120\)** - \(x = 1\) **5. \(4^{x} - 6 \cdot 2^{x} + 8 = 0\)** - \(x = 1\) или \(x = 2\) **6. \(9^{x} + 3 \cdot 3^{x} - 18 = 0\)** - \(x = 1\)

Solution

Конечно, вот решения заданных уравнений: --- **1. \(2^{2 - x} = 32\) (2 балла)** **Решение:** \[ 32 = 2^5 \] \[ 2^{2 - x} = 2^5 \Rightarrow 2 - x = 5 \Rightarrow x = 2 - 5 = -3 \] **Ответ:** \(x = -3\) --- **2. \(81^{x - 4} = \frac{1}{3}\) (2 балла)** **Решение:** \[ 81 = 3^4 \] \[ (3^4)^{x - 4} = 3^{-1} \Rightarrow 3^{4(x - 4)} = 3^{-1} \] \[ 4(x - 4) = -1 \Rightarrow 4x - 16 = -1 \Rightarrow 4x = 15 \Rightarrow x = \frac{15}{4} = 3.75 \] **Ответ:** \(x = \frac{15}{4}\) или \(x = 3.75\) --- **3. \(\left(\frac{1}{4}\right) \cdot 4x - 13 = \frac{1}{64}\) (2 балла)** **Решение:** \[ \left(\frac{1}{4}\right) \cdot 4x = x \] \[ x - 13 = \frac{1}{64} \Rightarrow x = 13 + \frac{1}{64} = \frac{832}{64} + \frac{1}{64} = \frac{833}{64} \approx 13.015625 \] **Ответ:** \(x = \frac{833}{64}\) или \(x \approx 13.015625\) --- **4. \(5^{x + 2} - 5^{x} = 120\) (3 балла)** **Решение:** \[ 5^{x + 2} = 5^x \cdot 25 \] \[ 5^x \cdot 25 - 5^x = 120 \Rightarrow 5^x (25 - 1) = 120 \Rightarrow 5^x \cdot 24 = 120 \] \[ 5^x = \frac{120}{24} = 5 \Rightarrow 5^x = 5^1 \Rightarrow x = 1 \] **Ответ:** \(x = 1\) --- **5. \(4^{x} - 6 \cdot 2^{x} + 8 = 0\) (3 балла)** **Решение:** \[ 4^x = (2^2)^x = 2^{2x} \] Пусть \(y = 2^x\), тогда уравнение становится: \[ y^2 - 6y + 8 = 0 \] Решим квадратное уравнение: \[ y = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2} \Rightarrow y = 4 \text{ или } y = 2 \] Возвращаемся к \(2^x\): \[ 2^x = 4 \Rightarrow 2^x = 2^2 \Rightarrow x = 2 \] \[ 2^x = 2 \Rightarrow 2^x = 2^1 \Rightarrow x = 1 \] **Ответ:** \(x = 1\) или \(x = 2\) --- **6. \(9^{x} + 3 \cdot 3^{x} - 18 = 0\) (3 балла)** **Решение:** \[ 9^x = (3^2)^x = 3^{2x} \] Пусть \(y = 3^x\), тогда уравнение становится: \[ y^2 + 3y - 18 = 0 \] Решим квадратное уравнение: \[ y = \frac{-3 \pm \sqrt{9 + 72}}{2} = \frac{-3 \pm \sqrt{81}}{2} = \frac{-3 \pm 9}{2} \] Получаем два корня: \[ y = \frac{6}{2} = 3 \quad \text{и} \quad y = \frac{-12}{2} = -6 \] Поскольку \(y = 3^x > 0\), принимаем только \(y = 3\): \[ 3^x = 3 \Rightarrow 3^x = 3^1 \Rightarrow x = 1 \] **Ответ:** \(x = 1\) --- Если у вас есть дополнительные вопросы или требуется подробное объяснение какого-либо шага, пожалуйста, сообщите!

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Bonus Knowledge

To solve these equations, it's often handy to express them with the same base since most of them involve powers. For instance, \( 32 \) can be rewritten as \( 2^5 \) in the first equation. This creates a basis to equate the exponents directly, making it a breeze to find \( x \). Similarly, \( 81 \) is \( 3^4 \) and \( \frac{1}{3} \) is \( 3^{-1} \), so manipulating these ensures clarity in isolating \( x \). As you tackle these problems, common mistakes can often occur through miscalculations when converting bases or exponents. Always double-check that any logarithmic manipulations maintain the integrity of equations. Don't forget to look out for extraneous solutions, especially in cases where squaring or re-expressing introduces possible false positives. And remember, patience is key—math is as much an art as it is a skill!

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