Question

Решите логарифмеческое уравнение под в) и г)

Answer 1

(㏒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