User:Cryptanalysis

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a(aa) = b
(aa)loga = logb
(aa) = logb / loga
aloga = log(logb / loga)
aloga = log(logb) − log(loga)
a = (log(logb) − log(loga)) / loga

a(aa) = b
(aa)loga = logb
(aa) = logb / loga
a = (logb / loga)(1 / a)

(logb / loga)(1 / a) = (log(logb) − log(loga)) / loga
--Cryptanalysis 06:49, 16 February 2006 (UTC) \sum_{n=0}^p\frac{m^2\,n} {3^m\left(m\,3^n+n\,3^m\right)}

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\sum_{i=1}^{\infty}{1 \over i^2} This user is an advanced mathematician.
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