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 An alternate power series representation for ln(x) JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/09/2011, 01:02 AM (05/08/2011, 08:28 PM)bo198214 Wrote: Oh thats just: $\frac{1}{e^{an}}(x-e^a)^n=\frac{(x-e^a)^n}{(e^a)^n} = \left(\frac{x-e^a}{e^a}\right)^n = \left(\frac{x}{e^a} - 1\right)^n$ Oh that's so simple! Thanks for the help. « Next Oldest | Next Newest »

 Messages In This Thread An alternate power series representation for ln(x) - by JmsNxn - 05/07/2011, 08:41 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/07/2011, 09:43 PM RE: An alternate power series representation for ln(x) - by bo198214 - 05/07/2011, 10:45 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/07/2011, 11:20 PM RE: An alternate power series representation for ln(x) - by bo198214 - 05/08/2011, 01:38 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/08/2011, 07:54 PM RE: An alternate power series representation for ln(x) - by bo198214 - 05/08/2011, 08:28 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/09/2011, 01:02 AM

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