Hi there, I have yet another log problem that's got me tearing my hair out. e^(2x)-e^(x+3)-e^(x+1)+e^4=0. I tried taking e^4 to the RHS and taking the ln of both sides and got 0=0...wtf.

Results 1 to 9 of 9

- February 16th 2011, 02:49 PM #1

- Joined
- Feb 2011
- Posts
- 66

- February 16th 2011, 02:56 PM #2

- February 16th 2011, 03:01 PM #3

- Joined
- Mar 2010
- Posts
- 280

- February 16th 2011, 03:07 PM #4

- February 16th 2011, 05:21 PM #5

- Joined
- Feb 2011
- Posts
- 66

Hi there, firstly thank you topsquark, zzzoak and plato for looking at this. I have 2 queries, Plato your quadratic has (e^3+e^2)e^x but wouldn't that give e^(x+3)-e^(x+2)? And how do you go from this quadratic equation to the solution. I just can't see it. I feel like throwing my computer out of the window!

- February 16th 2011, 05:36 PM #6

- Joined
- Feb 2011
- Posts
- 66

- February 17th 2011, 06:20 AM #7

- Joined
- Apr 2005
- Posts
- 16,242
- Thanks
- 1795

No, it wouldn't. How did that "+" in (e^3+ e^2)(e^x) become a "-"?

Wait, I think I get it, do you then factorise to get (e^x - e^3)(e^x - e^1)=0 ?

Also how are you getting the formula to display properly?

You can double click on any formula to see the code used. Also, there is a "Latex tutorial" area under the main menu.

- February 17th 2011, 06:44 AM #8

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 11,866
- Thanks
- 744

- February 17th 2011, 10:36 AM #9

- Joined
- Feb 2011
- Posts
- 66