Прошу помощи! Решите уравнение:

(log2 (4x)) ^2+log2 (x^2/8) = 8

ОДЗ: x>0

(log2 (4) + log2 (x)) (log2 (4) + log2 (x)) + log2 (x^2) - log2 (8) = 8

(2+log2 (x)) * (2+log2 (x)) + 2log2 (x) - 3=8

4+2log2 (x) + 2log2 (x) + (log2 (x)) ^2+2log2 (x) = 11

(log2 (x)) ^2+6log2 (x) - 7=0

Замена: log2 (x) = t

t^2+6t-7=0

D=6^2-4*1 * (-7) = 64

t1 = (-6-8) / 2=-7

t2 = (-6+8) / 2=1

Обратная замена:

1) log2 (x) = - 7

x=2^-7

x=1/128

2) log2 (x) = 1

x=2

Ответ: 1/128; 2