March 1st, 2007, 09:13 AM  #1 
Member Joined: Nov 2006 From: Vancouver, Canada Posts: 46 Thanks: 0  proving log(a)
an identity of log(a) = lim as x>0 of [(a^x1)/x]. I want to prove that, and I can only think about delog both sides, then I end up with a = lim [esp(a^x/x1/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?? 
March 1st, 2007, 09:19 AM  #2 
Member Joined: Nov 2006 From: Vancouver, Canada Posts: 46 Thanks: 0 
One thing also I thought about is to devide both sides by log(a), then end up needing to prove lim (a^x1)/log(a^x) = 1. Somehow though... using l'hopital's rule, the left lim becomes..1 rather than 1... strange...

March 1st, 2007, 09:45 AM  #3  
Member Joined: Dec 2006 From: St. Paul MN USA Posts: 37 Thanks: 0  Re: proving log(a) Quote:
Quote:
 
March 1st, 2007, 11:15 AM  #4 
Member Joined: Nov 2006 From: Vancouver, Canada Posts: 46 Thanks: 0 
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?

March 1st, 2007, 02:29 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,977 Thanks: 1851 
So convert it to a power series, etc.

March 1st, 2007, 04:03 PM  #6  
Member Joined: Dec 2006 From: St. Paul MN USA Posts: 37 Thanks: 0  Quote:
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))/(x0) = lim (x>0) (a^x  1)/x.  
March 1st, 2007, 09:16 PM  #7 
Member Joined: Nov 2006 From: Vancouver, Canada Posts: 46 Thanks: 0 
oh I see! thank you!!


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