2^2x - 5 x 2^x +6 = 0. Solve for x?

Answer 1

let 2^x=t
theequation t^2-5t+6=0
(t-3)(t-2)=0
t=3 or 2
2^x=3
x=ln3/ln2
2^x=2 x=1
so soln set 1 or ln3/ln2

Answer 2

2^2x - 5 x 2^x +6 = 0
let y = 2^x

So y^2 - 5y + 6 = 0
(y - 3)( y - 2) = 0

y = 2 or 3

ie 2^x = 2 OR 2^x = 3
Thus x = 0 or log3/log2 (2^x = 3 Take logs. So log(2^x) = xlog2 = log3)

Answer 3

if by this you mean

2^(2x - 5) * 2^(x + 6) = 0
2^((2x - 5) + (x + 6)) = 0
2^(2x - 5 + x + 6) = 0
2^(3x + 1) = 0

Since any number raised to anything will never give you 0, this problem has no real solution.