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

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)

.......

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 TI-89 with me here at work.

It should be more like x=3.025

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If it is homework, that means your most likely going to be learning the glorious unit circle next or soon after. :)

I'm right. Try checking again.

javascript:alert( Math.pow(0.5*Math.exp(27/5), 1/3) )

x = 4.801608

Thanks JKD. Seeing as I'm only 15, simple algebra to me is a little different than it is to you :p 2 years of math makes quite a difference, especially when I wasn't paying attention the day we learned about natural logs.

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

Yeah your right... silly google calculator. The windows one doesn't have the e constant.