1. ## Calculus homework problem

I'm having a little trouble with this calculus problem I got for homework. I've been working on it for over an hour and haven't gotten anywhere with it.

Solve for x: ln(x) + ln(x-4) = 5

The best I could come up with was raising both sides to e to get rid of the natural logs, but then ended up with x^2 - 4x = e^5. Any advice/answers?

2. $\displaystyle x^2 - 4x - e^5 = 0$

Does the quadratic formula ring any bells ?

3. Yeah, I got to that, but then how do you solve for x? The QF only works for ax^2 + bx + c, doesn't it? That would make it a bit harder since the e is raised to the 5th. I could do it if it was raised to the 4th, but since it's to the 5th I've got nothing.

Or did you mean use the e^5 as the c in the QF?

4. $\displaystyle e^5$ is just the constant term, so yes you can use the quadratic formula and you can use it as a "c".

5. Originally Posted by Chop Suey
$\displaystyle e^5$ is just the constant term, so yes you can use the quadratic formula and you can use it as a "c".
Ok, thanks. That makes more sense now.

6. e^a where a is any real number is a constant, for future reference