My Math Forum Another exponential equation....

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 October 24th, 2013, 08:07 AM #1 Newbie   Joined: Oct 2013 Posts: 6 Thanks: 0 Another exponential equation.... Ok, so this time it is: e^2x - 3e^-x = 2 So far I get: e^2x - 3(1/3e^x) -3 = 2 ------to take away -x e^x(e^2x) - 3 = 2 e^3x = 5 lne^3x = ln5 3xlne = ln5 3x = ln5 x = ln5/3 .....is that right????
October 24th, 2013, 08:16 AM   #2
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Re: Another exponential equation....

Quote:
 e^2x - 3e^-x = 2
Are you sure it's not e^x - 3e^-x = 2?

 October 24th, 2013, 08:22 AM #3 Newbie   Joined: Oct 2013 Posts: 6 Thanks: 0 Re: Another exponential equation.... I wrote the problem wrong....should be... e^2x - 3e^x - 10 = 0 ...more than wrong, really wrong...
 October 24th, 2013, 08:28 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Another exponential equation.... $e^{2x}\,-\,3e^x\,-\,10\,=\,0 \\ $$e^x\,-\,5$$$$e^x\,+\,2$$\,=\,0 \\ e^x\,=\,5\,\Rightarrow\,x\,=\,\ln5 \\ e^x\,=\,-2\,\Rightarrow\,\text{ No solution.}$
 October 24th, 2013, 08:38 AM #5 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 Re: Another exponential equation.... gregs method is better , another approach is ... (1) Let e^x = t $e^{2x} \= \ e^x \cdot e^x \ = \ t^2$ $e^{2x} \ - \ 3e^x \ - \ 10 \= \ 0$ $t^2 \ - \ 3t \ - \ 10 \= \ 0$ This is factorable (if it wasn't you could use a quadratic formula) Solve for t , then use (1) to solve for x , then check the solutions by substituting for x in your original problem.
 October 24th, 2013, 08:47 AM #6 Newbie   Joined: Oct 2013 Posts: 6 Thanks: 0 Re: Another exponential equation.... Thanks guys!!!...I got it right once I wrote the the question correct

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