logarithms

• December 18th 2007, 09:01 PM
CenturionMonkey
logarithms
I think this is supposed to go here...anyways..

I need help with this log.

Solve for x
5^(2x) - 5^(x) - 20 = 0

please and thank you...I've been stuck on it for a half hour now.
• December 18th 2007, 09:16 PM
earboth
Quote:

Originally Posted by CenturionMonkey
I think this is supposed to go here...anyways..

I need help with this log.

Solve for x
5^(2x) - 5^(x) - 20 = 0

please and thank you...I've been stuck on it for a half hour now.

Hello,

use the substitution $y = 5^x$ and your equation becomes:
$5^{2x} - 5^x - 20 = 0~\longrightarrow~y^2-y-20=0$

This is a quadratic equation in y. Solve for y and don't forget to re-substitute to calculate x. I've got x = 1
• December 18th 2007, 09:18 PM
CenturionMonkey
ahh I see. Thanks a lot, yep the answer sheet says it's 1.