2^2x -2 x 2^x -8 = 0

I'm sure that is

2^(2x) -2(2^x) -8 = 0

In another form,

(2^x)^2 -2(2^x) -8 = 0 -----------(i)

(i) is a quadratic erquation in 2^x.

So if y = 2^x,

then (i) becomes

y^2 -2y -8 = 0

So,

(y -4)(y+2) = 0

y = 4 or -2

When y = 4,

2^x = 4

Take the natural logs of both sides,

x*ln(2) = ln(4)

x = ln(4) / ln(2)

x = 2ln(2) / ln(2)

x = 2 ----------------answer.

When y = -2,

2^x = -2

x*ln(2) = ln(-2)

There are no real values for logs of negative numbers, so reject this.