(log5x)^2 - 2 - 3log125x = 0
(log5x)^2 - 2 - 3log5^3(x) = 0
(log5x)^2 - 2 - 3/3*log5(x) = 0
(log5x)^2 - log5(x) - 2= 0
Пусть log5(x) = t, тогда
t^2 - t - 2 = 0
D = 1 + 8 = 9
t1 = ( 1 + 3)/2 = 2
t2 = (1 - 3)/2 = - 1
Обратная замена
log5(x) = 2
x = 5^(2)
x = 25
log5(x) = - 1
x = 5^(-1)
x = 0,2
Ответ
0,2; 25