Solve 2\(^{2x}\) – 5(2\(^x\)) + 4 = 0
-
x = 0 and 2
- x = 1 and 2
- x = 0 and 1
- x = 2 and 4
The correct answer is: A
Explanation
2\(^{2x}\) - 5(2\(^x\)) + 4 = 0
2\(^{(x)2}\) - 5(2\(^x\)) + 4 = 0
let p = 2\(^x\)
p\(^2\) - 5(p) + 4 = 0
p\(^2\) -4p - p + 4 = 0
p(p - 4) - 1(p - 4) = 0
(p - 1)(p -4) = 0
p = 1 or 4
But p = 2\(^x\) = 1 or 4
2\(^x\) = 2\(^2\)
x = 2
2\(^x\) = 2\(^0\) = 1
x = 0
Therefore, the values of x = 0 and 2