Помогите решить уравнение пожалуйста
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