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April 25th, 2014, 10:38 AM   #1
The
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Simplify

simplify: (e^(-4x)+3e^(-2x))/(e^(-4x)-9)
i tried long division, but I'm not getting the correct answer
little hand here?
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April 25th, 2014, 10:52 AM   #2
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btw, correct answer is : (e^(-2x))/((e^-2x) + 3)
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April 25th, 2014, 10:58 AM   #3
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You could start by substituting y = e^(-2x) which would turn this into an ordinary rational function.
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April 25th, 2014, 10:58 AM   #4
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{e^(-4x) + 3e^(-2x)}/{e^(-4x) - 9}
{(e^(-2x))^2+ 3e^(-2x)}/[{e^(-2x)}^2 - 3^2]
[e^(-2x){e^(-2x) + 3}]/[{e^(-2x) - 3}{e^(-2x)+3}]
e^(-2x)/{e^(-2x) - 3}
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April 25th, 2014, 10:58 AM   #5
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(e^(-4x)+3e^(-2x))/(e^(-4x)-9) =

(1/e^4x +3/e^2x)/(1/e^4x - 9) =

((1/e^2x)*(1/e^2x +3))/((1/e^2x)^2 - (3^2)) =

((1/e^2x)*(1/e^2x+3))/((1/e^2x-3)*(1/e^2x-3))=

(1/e^2x)/(1/e^2x - 3)=

(1/e^2x -3 +3)/(1/e^2x -3)=

3/(1/e^2x - 3)
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April 25th, 2014, 10:59 AM   #6
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finaIy : 3/(e^(-2x)-3)
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April 25th, 2014, 11:02 AM   #7
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Thank you sir,
i really like what you did there. I did not see common factor!
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April 25th, 2014, 04:13 PM   #8
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Dunno...but Klea's doesn't seem right to me...

I get 1 / (1 - 3e^(2x)) ; just another way to express the given solution...

Btw Mr.The, you can check by substituting amounts for e and x: try 1 and 1
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April 25th, 2014, 06:36 PM   #9
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Hello, The!

Quote:
$\displaystyle \text{Simplify: }\:\frac{e^{-4x}+3e^{-2x}}{e^{-4x}-9}$

Factor: $\displaystyle \:\frac{e^{-2x}(e^{-2x} + 3)}{(e^{-2x} - 3)(e^{-2x} + 3)}$

Cancel: $\displaystyle \:\frac{e^{-2x}(\cancel{e^{-2x} + 3})}{(e^{-2x} - 3){(\cancel{e^{-2x} + 3})}} \;=\; \frac{e^{-2x}}{e^{-2x}-3}$

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