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February 10th, 2010, 04:03 PM   #1
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logarithmic problem

okay guys I need help on this one....solve for x using logs.

2e^3x=4e^5x
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February 10th, 2010, 06:03 PM   #2
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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))
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February 10th, 2010, 08:25 PM   #3
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Or -ln(?2).
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April 10th, 2010, 02:24 PM   #4
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Re: logarithmic problem

tale Ln of both sides

solve for x

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