# Exponential function disguised as a quadratic

• Apr 20th 2009, 10:16 PM
Phire
Exponential function disguised as a quadratic
$5^{2x} -130^x + 25 = 0$

This is what I tried:

$5(1^{2x}-26^x + 5) = 0$

$5(1^{2x}-26 * 1^x + 5) = 0$

$1^x = y$

$5(y^2-26y+5) = 0$

Solving the above quadratic doesn't give me the right answers, unless I solved it wrong, but I don't think I did. What did I do wrong?
• Apr 20th 2009, 11:34 PM
a little help
Fatal mistake, the way you are factoring the exponential pieces. You can not factor the 5 out like that, or the 1 like that.
• Apr 21st 2009, 05:42 AM
mr fantastic
Quote:

Originally Posted by Phire
$5^{2x} -130^x + 25 = 0$

[snip]

Are you sure this is the correct equation? I'll bet it's meant to be $5^{2x} - 26 \cdot 5^x + 25 = 0$ but you made an incorrect 'simplification' to this before posting.

If so then you're meant to factorise it as $(5^x - 25)(5^x - 1) = 0$. Now solve for x.
• Apr 21st 2009, 01:01 PM
Gamma
I came to the same conclusion today while at school, this problem has been bugging me for 24 hours. Me and 4 other grad students worked on it over lunch to no avail, no way the problem posted as is should be assigned in a pre calculus class, maybe a numerical methods course at best.

If it is as Mr. Fantastic suggested (and I hope it is) simply substitute $y=5^x$ and solve the quadratic $y^2 - 26y + 25 = 0$to see how he got his factoring, then you are basically done.