3. Решить уравнение: 1) log₅ (2x-1) = 2 2) log₂(x-2)+ log₂ x = 3 3) log₀.³ x + log₉ x = 14

Решение:

1. \( \log_5 (2x-1) = 2 \)
   \( 2x-1 = 5^2 \)
   \( 2x-1 = 25 \)
   \( 2x = 26 \)
   \( x = 13 \)
   Проверка: \( 2(13) - 1 = 26 - 1 = 25 > 0 \).
2. \( \log_2 (x-2) + \log_2 x = 3 \)
   \( \log_2 (x(x-2)) = 3 \)
   \( x(x-2) = 2^3 \)
   \( x^2 - 2x = 8 \)
   \( x^2 - 2x - 8 = 0 \)
   \( D = (-2)^2 - 4(1)(-8) = 4 + 32 = 36 \)
   \( x_1 = \frac{2 + \sqrt{36}}{2} = \frac{2+6}{2} = 4 \)
   \( x_2 = \frac{2 - \sqrt{36}}{2} = \frac{2-6}{2} = -2 \)
   Проверка: \( x > 0 \) и \( x-2 > 0 \). Значит, \( x > 2 \). Подходит только \( x = 4 \).
3. \( \log_{\frac{1}{3}} x + \log_9 x = 14 \)
   \( \log_{3^{-1}} x + \log_{3^2} x = 14 \)
   \( -\log_3 x + \frac{1}{2} \log_3 x = 14 \)
   \( (\frac{1}{2} - 1) \log_3 x = 14 \)
   \( -\frac{1}{2} \log_3 x = 14 \)
   \( \log_3 x = -28 \)
   \( x = 3^{-28} = \frac{1}{3^{28}} \)
   Проверка: \( x > 0 \).

Ответ: 1) 13; 2) 4; 3) \(\frac{1}{3^{28}}\).