Задание 4 Решите уравнение. a) \(4^x = 64\); б) \(\left(\frac{2}{3}\right)^x \cdot \left(\frac{9}{8}\right)^x = \frac{27}{64}\); в) \(2^{2x-4}=64\); г) \(5^{x^2-3x}=5^{3x-8}\); д) \(\log_5 x=2\).

Решение:

1. \(4^x = 64\)
   \(4^x = 4^3\)
   \(x = 3\)
2. \(\left(\frac{2}{3}\right)^x \cdot \left(\frac{9}{8}\right)^x = \frac{27}{64}\)
   \(\left(\frac{2}{3} \cdot \frac{9}{8}\right)^x = \frac{27}{64}\)
   \(\left(\frac{18}{24}\right)^x = \frac{27}{64}\)
   \(\left(\frac{3}{4}\right)^x = \left(\frac{3}{4}\right)^3\)
   \(x = 3\)
3. \(2^{2x-4}=64\)
   \(2^{2x-4}=2^6\)
   \(2x-4 = 6\)
   \(2x = 10\)
   \(x = 5\)
4. \(5^{x^2-3x}=5^{3x-8}\)
   \(x^2-3x = 3x-8\)
   \(x^2 - 6x + 8 = 0\)
   \(D = (-6)^2 - 4 \cdot 1 \cdot 8 = 36 - 32 = 4\)
   \(x_1 = \frac{6+\sqrt{4}}{2} = \frac{6+2}{2} = 4\)
   \(x_2 = \frac{6-\sqrt{4}}{2} = \frac{6-2}{2} = 2\)
5. \(\log_5 x=2\)
   \(x = 5^2\)
   \(x = 25\)

Ответ: а) 3; б) 3; в) 5; г) 2; 4; д) 25.