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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  
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.... Quote:
 
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.... 
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 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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