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.

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- Feb 16th 2011, 01:49 PM #1

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- Feb 16th 2011, 01:56 PM #2

- Feb 16th 2011, 02:01 PM #3

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- Feb 16th 2011, 02:07 PM #4

- Feb 16th 2011, 04:21 PM #5

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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!

- Feb 16th 2011, 04:36 PM #6

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- Feb 17th 2011, 05:20 AM #7

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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.

- Feb 17th 2011, 05:44 AM #8

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Hello, flashylightsmeow!

$\displaystyle \text{Solve for }x\!:\;\;e^{2x}-e^{x+3}-e^{x+1}+e^4\:=\:0$

We have: .$\displaystyle e^{2x} - e^3\!\cdot\!e^x - e\!\cdot\!e^x + e^4 \;=\;0 $

. . . . . . . . $\displaystyle e^{2x} - (e^3+e)e^x + e^4 \;=\;0 $

Factor:. . . . . $\displaystyle (e^x - e)(e^x - e^3) \;=\;0 $

Got it?

- Feb 17th 2011, 09:36 AM #9

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