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Thread: Natural Log Question

03032004, 08:39 PM #1
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Natural Log Question
I realize that this may not be the appropriate forum for this kind of question, but it was the closest that I could find. Seeing how many of the people around here are very talented mathematicians, I was wondering if someone could solve this for me, and relate to me the steps in which to do so.
5 ln (2x^3) = 27
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03032004, 10:32 PM
#2
Umm... simple algebra? Is this a homework question?
5 ln (2x^3) = 27
ln (2x^3) = 27/5
2x^3 = e^(27/5)
x^3 = 0.5 e^(27/5)
x = (0.5 e^(27/5))^(1/3)
.......
03032004, 10:55 PM
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Yeah but that doesn't come out right.
If you do that: x = (0.5 e^(27/5))^(1/3)
Then x=1.73730502
So then 5*ln(2(1.73730502)^3)=18.6822
Unless I'm entering the thing in wrong into the windows calculator. I don't have my trusty TI89 with me here at work.
It should be more like x=3.025

If it is homework, that means your most likely going to be learning the glorious unit circle next or soon after.
OracleGuy
03032004, 11:18 PM
#4
I'm right. Try checking again.
javascript:alert( Math.pow(0.5*Math.exp(27/5), 1/3) )
x = 4.801608
03042004, 02:00 AM
#5
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Thanks JKD. Seeing as I'm only 15, simple algebra to me is a little different than it is to you 2 years of math makes quite a difference, especially when I wasn't paying attention the day we learned about natural logs.
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03042004, 03:49 AM
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Yeah your right... silly google calculator. The windows one doesn't have the e constant.Originally posted by jkd
I'm right. Try checking again.
javascript:alert( Math.pow(0.5*Math.exp(27/5), 1/3) )
x = 4.801608
OracleGuy