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April 25th, 2014, 10:38 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, 10:52 AM  #2 
Newbie Joined: Apr 2014 From: London Posts: 13 Thanks: 1 
btw, correct answer is : (e^(2x))/((e^2x) + 3)

April 25th, 2014, 10:58 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.

April 25th, 2014, 10:58 AM  #4 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 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} 
April 25th, 2014, 10:58 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^2x3)*(1/e^2x3))= (1/e^2x)/(1/e^2x  3)= (1/e^2x 3 +3)/(1/e^2x 3)= 3/(1/e^2x  3) 
April 25th, 2014, 10:59 AM  #6 
Newbie Joined: Apr 2014 From: Albania Posts: 8 Thanks: 2 
finaIy : 3/(e^(2x)3)

April 25th, 2014, 11:02 AM  #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, 04:13 PM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
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, 06:36 PM  #9  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, The! Quote:
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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