Benedict Irwin edited section_Fermat_Take_a_notable__.tex  over 8 years ago

Commit id: e528b654592b2687addb691ca081851e8da00b1c

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n=\log_{[a,b]}(c^n)  \end{equation}  wlog, if $a$ is $1$, then we can say \begin{equation}  n=\ln(b)\cdot\ln(c^n -1) n=\frac{\ln(c^n -1)}{\ln b}  \end{equation}