Решите логарифмеческое уравнение под в) и г)

(㏒₂4 + log₂x)² + (log₂2 + log₂x)² = 1 (2 + log₂x)² + (1 + log₂x)² = 1 4 + 4 log₂x +  log²₂x + 1 + 2 log₂x +  log²₂x - 1 = 0 2 log²₂x + 6 log₂x + 4 = 0  log₂x = t t² + 3t + 2 = 0 t = -2  или  t = -1  log₂x = -2    log₂x = -1 x = 2⁻²          x = 2⁻¹ x = 1/4          x = 1/2 г) log₆(5 + 6⁻ˣ) = x+1 6ˣ⁺¹ = 5 + 6⁻ˣ     | * 6ˣ 6²ˣ⁺¹ = 5·6ˣ + 1 6ˣ = t 6·t² - 5t - 1 = 0 D = 25 + 24 = 49 t = 1   или   t = -1/6 6ˣ = 1         6ˣ = 1/6 x = 0           x = -1