Question

2*4^(2x)-17*4^(x)+8=0 Решите пожалуйста

Answer 1

Пусть 4^x = t, t > 0
2t^2 - 17t + 8 = 0
t = (17 +- √(289 - 64)) / 4 = (17 +- √225) / 4 = (17 +- 15) / 4
t1 = 32 / 4 = 8
4^x = 8
x = log4(8) = 3/2
t2 = 1/2
4^x = 1/2
x = log4(1/2) = -1/2
Ответ: 3/2; -1/2.