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March 1st, 2007, 09:13 AM   #1
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proving log(a)

an identity of log(a) = lim as x->0 of [(a^x-1)/x].
I want to prove that, and I can only think about de-log both sides, then I end up with a = lim [esp(a^x/x-1/x)] ... otherwise, I can't think of anything. How would you prove that?

This identity is interesting, because it kind of entails how much faster does an exponential change faster than log... just sort of by a linear factor! LOL Do you think this interpretation is valid??
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March 1st, 2007, 09:19 AM   #2
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One thing also I thought about is to devide both sides by log(a), then end up needing to prove lim (a^x-1)/log(a^x) = 1. Somehow though... using l'hopital's rule, the left lim becomes..-1 rather than 1... strange...
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March 1st, 2007, 09:45 AM   #3
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Re: proving log(a)

Quote:
Originally Posted by ^e^
an identity of log(a) = lim as x->0 of [(a^x-1)/x].
I want to prove that, and I can only think about de-log both sides, then I end up with a = lim [esp(a^x/x-1/x)] ... otherwise, I can't think of anything. How would you prove that?
By definition, this is the derivative of a^x evaluated at x = 0.

Quote:
Originally Posted by ^e^
This identity is interesting, because it kind of entails how much faster does an exponential change faster than log... just sort of by a linear factor! LOL Do you think this interpretation is valid??
No.
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March 1st, 2007, 11:15 AM   #4
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yeah, it's the definition at 0, bu tthen... taking the limit, we are trying to prove that its close to the falue when x is close to 0 right?
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March 1st, 2007, 02:29 PM   #5
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So convert it to a power series, etc.
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March 1st, 2007, 04:03 PM   #6
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Quote:
Originally Posted by ^e^
yeah, it's the definition at 0, bu tthen... taking the limit, we are trying to prove that its close to the falue when x is close to 0 right?
Perhaps I was too terse. Here are the details.

If f(x) = a^x then f'(x) = a^x * ln(a), so f'(0) = ln(a).
But f'(0) = lim (x->0) (f(x)-f(0))/(x-0) = lim (x->0) (a^x - 1)/x.
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March 1st, 2007, 09:16 PM   #7
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oh I see! thank you!!
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