Thread: One more exponential equation

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

Please help solve this equation.

Thank you very much

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.

Ticbol,

thank you. You explained it so well. It's great that we have people like you here ready to explain and help.