Algebra Pre-Algebra and Basic Algebra Math Forum

 April 25th, 2014, 11:38 AM #1 Newbie   Joined: Apr 2014 From: London Posts: 13 Thanks: 1 Simplify simplify: (e^(-4x)+3e^(-2x))/(e^(-4x)-9) i tried long division, but I'm not getting the correct answer little hand here? thanks April 25th, 2014, 11:52 AM #2 Newbie   Joined: Apr 2014 From: London Posts: 13 Thanks: 1 btw, correct answer is : (e^(-2x))/((e^-2x) + 3) April 25th, 2014, 11:58 AM #3 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms You could start by substituting y = e^(-2x) which would turn this into an ordinary rational function. Thanks from topsquark April 25th, 2014, 11:58 AM #4 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 911 Thanks: 64 Math Focus: सामान्य गणित {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} Thanks from topsquark and The April 25th, 2014, 11:58 AM #5 Newbie   Joined: Apr 2014 From: Albania Posts: 8 Thanks: 2 (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) Thanks from The April 25th, 2014, 11:59 AM #6 Newbie   Joined: Apr 2014 From: Albania Posts: 8 Thanks: 2 finaIy : 3/(e^(-2x)-3) April 25th, 2014, 12:02 PM #7 Newbie   Joined: Apr 2014 From: London Posts: 13 Thanks: 1 Thank you sir, i really like what you did there. I did not see common factor! April 25th, 2014, 05:13 PM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 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 April 25th, 2014, 07:36 PM   #9
Math Team

Joined: Dec 2006
From: Lexington, MA

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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: $\displaystyle \:\frac{e^{-2x}(\cancel{e^{-2x} + 3})}{(e^{-2x} - 3){(\cancel{e^{-2x} + 3})}} \;=\; \frac{e^{-2x}}{e^{-2x}-3}$ Tags simplify Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Shamieh Algebra 2 September 12th, 2013 07:05 AM JediZiro Algebra 2 August 5th, 2013 05:49 AM perfect_world Algebra 3 July 29th, 2013 04:58 AM Valar30 Algebra 4 June 11th, 2011 10:56 PM xdeathcorex Calculus 2 August 31st, 2010 02:37 PM

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