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\end{equation}  We could also use the set \begin{equation}  \sum_{k=1}^\infty e=\sum_{k=1}^\infty  \frac{1}{k!} \end{equation} This gives the problem \begin{equation}  a=\sum_{k=1}^\infty \frac{1}{(k!)^x}  \end{equation}  and the solution for $x$ is \begin{equation}  x=\log_{[e]}(a)  \end{equation}