log ²5 (2x)-20log5 (2x)=21​

Используем метод замены.

log²5(2x) - 20log5(2x) - 21 = 0

Замена: log5(2x) = t

t²- 20t - 21 = 0

D = 400 + 84 = 484 = 22²

t1 = (20+22)/2 = 21

t2 = (20-22)/2 = -1

1) log5(2x) = 21

2x = 5^21

x = (5^21)/2

2) log5(2x) = -1

2x = 5^-1 = 1/5 = 0.2

x = 0.1