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REPORT ABUSIVE REPLY
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Message Subject
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Serious Math Question. Need a serious math nerd to help explain this to me. (Involves Calc)
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Poster Handle
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Anonymous Coward |
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Post Content
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Suppose you try calculating a few logs and see what happens:
(here l is the natural log function)
l(1/1)
0
l(1/10)
-2.30258509299404568401
l(1/100)
-4.60517018598809136803
l(1/1000)
-6.90775527898213705205
l(1/10000)
-9.21034037197618273607
l(1/1000000)
-13.81551055796427410410
l(1/1000000000)
-20.72326583694641115616
l(1/1000000000000)
-27.63102111592854820821
you can see that the result of the natural log function gets smaller as n gets bigger. It sure looks like it might be approaching negative infinity, doesn't it?
Quoting: Anonymous Coward 37573475 Ahh, Thank you! My calculus teacher doesn't let us use a calculator on anything. So, I'm so used to going without. I think I've been doing homework for too long today. It might be beer thirty. :) hey i didn't use a calculator! i used a computer and ran the gnu "bc" program: [ link to www.gnu.org] if you think about what logarithms and exponents mean, i think you will figure out that the log of a number between zero and one has to get smaller as the number approaches zero. think about the meaning of the number 10^-2 in terms of logarithms. then think about 10^-3, 10^-4, and so on. then relate that to 1/100, 1/1000, 1/10000, and so on. then relate the fractions 1/100 and so on to the original calculus question, and it will hopefully be pretty obvious what the correct answer is.
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