# How would you solve this log?

• Dec 2nd 2007, 07:28 PM
Darkhrse99
How would you solve this log?
$2^{2x} +2^{x+2} -12=0$ They want you to solve the eqation.

since it's a subtraction you divide. I multiplied 2x(x+2) and got $2x^2frac/-12$

$2x^2/12=0$

I don't think I'm even close, but am not sure how to go from here.

Thanks

Jason
• Dec 3rd 2007, 06:10 AM
topsquark
Quote:

Originally Posted by Darkhrse99
$2^{2x} +2^{x+2} -12=0$ They want you to solve the eqation.

Let's rewrite this as
$2^{2x} + 2^2 \cdot 2^x - 12 = 0$

$2^{2x} + 4 \cdot 2^x - 12 = 0$

This is a quadratic in $2^x$ as Jhevon said.

For further clarity, let's define $y = 2^x$. Then your equation becomes:
$y^2 + 4y - 12 = 0$

-Dan