My Math Forum logarithmic problem

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 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 $\frac{e^{3x}}{e^{5x}}=2$ tale Ln of both sides $3x-5x= ln[2]$ solve for x $x=-\frac{ln[2]}{2}$

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