February 10th, 2010, 05:03 PM  #1 
Newbie Joined: Feb 2010 Posts: 1 Thanks: 0  logarithmic problem
okay guys I need help on this one....solve for x using logs. 2e^3x=4e^5x 
February 10th, 2010, 07:03 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond  Re: logarithmic problem
With logs the following rules hold for a, b > 0: ln(a*b) = ln(a) + ln(b) ln(a/b) = ln(a)  ln(b) x*ln(a) = ln(a^x) Also: ln(e) = 1. So 2e^(3x) = 4e^(5x) ln(2) + ln(e^(3x)) = ln(4) + ln(e^(5x)) ln(2) + 3x = ln(4) + 5x 2x = ln(2)  ln(4) 2x = ln(1/2) x = ln(1/2)/2 x = ln(1/sqrt(2)) 
February 10th, 2010, 09:25 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,288 Thanks: 1968 
Or ln(?2).

April 10th, 2010, 03:24 PM  #4 
Senior Member Joined: Jan 2009 Posts: 344 Thanks: 3  Re: logarithmic problem tale Ln of both sides solve for x 

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