Question

Помогите решить уравнение пожалуйста

Answer 1

1)
4^(x + 1) + 7•2^x - 2 = 0
4•(2^x)^2 + 7•2^x - 2 = 0
2^x = t > 0
4t^2 + 7t - 2 = 0
D = 49 + 32 = 81
t = (-7 (+/-) 9)/8 = {-2; 1/4}
2^x = 1/4
x = -2

2)
2^x + 3•2^(x -3) = 22
2^x + 3/8 • 2^x = 22
11/8 • 2^x = 22
2^x = 16
x = 4