Change of base formula for Tetration
#37
I suppose it goes without mentioning that you simply reverse the process to take logarithms of very large numbers.

\( \text{Let }z=\log_c(\log_c(x))\\
\begin{eqnarray}
\text{Then }\log_c(\log_c(a^x)) & = & c^{z}+\log_c(\log_c{a}))
\end{eqnarray}
\)

which entails:

\( \text{Let }w=\log_c(\log_c(b^x))\\
\begin{eqnarray}
\text{Then }\log_c(\log_c(\log_b(b^x))) & = & \log_c(w-\log_c(\log_c{b}))
\end{eqnarray}
\)

Here we see the tools to iteratively exponentiate in base a, then iteratively take logarithms in base b.
~ Jay Daniel Fox
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Messages In This Thread
Change of base formula for Tetration - by jaydfox - 08/12/2007, 06:39 AM
RE: Change of base formula for Tetration - by jaydfox - 08/31/2007, 04:11 AM
RE: Parabolic Iteration - by jaydfox - 08/15/2007, 09:19 PM
RE: Parabolic Iteration - by bo198214 - 08/15/2007, 09:30 PM
RE: Parabolic Iteration - by jaydfox - 08/15/2007, 11:41 PM
RE: Parabolic Iteration - by bo198214 - 08/16/2007, 08:17 AM
RE: Parabolic Iteration - by jaydfox - 08/16/2007, 05:51 PM
RE: Parabolic Iteration - by bo198214 - 08/16/2007, 06:40 PM
RE: Parabolic Iteration - by jaydfox - 08/16/2007, 09:47 PM
RE: Parabolic Iteration - by bo198214 - 08/16/2007, 10:07 PM

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