3*4^x + 6^x = 2*9^x
3*4^x/9^x + 6^x/9^x = 2
3*(4/9)^x + (6/9)^x = 2
3*(4/9)^x + (2/3)^x - 2 = 0
Пусть (2/3)^x = t
3t^2 + t - 2 = 0
D = 1 + 24 = 25
t1 = ( - 1 + 5)/6 = 4/6 = 2/3
t2 = ( - 1 - 5)/6 = - 6/6 = - 1
Обратная замена
(2/3)^x = 2/3
x = 1
(2/3)^x = - 1
Нет решений
Ответ
1