1. ## Exponential function disguised as a quadratic

$\displaystyle 5^{2x} -130^x + 25 = 0$

This is what I tried:

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

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

$\displaystyle 1^x = y$

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

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?

2. ## 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.

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

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

If so then you're meant to factorise it as $\displaystyle (5^x - 25)(5^x - 1) = 0$. Now solve for x.

4. 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 $\displaystyle y=5^x$ and solve the quadratic $\displaystyle y^2 - 26y + 25 = 0$to see how he got his factoring, then you are basically done.