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\end{equation}  giving \begin{equation}  x=\mathrm{log}_b(a)  \end{equation}  We can expand this concept by considering the base $b$ as a sequence of size $1$. We then generalised to any sized set as \begin{equation}  a=b_1^x + b_2^x + \cdots b_n^x = \sum_{i=1}^n b_i^x  \end{equation}  to have a solution \begin{equation}  \mathrm{log}_{[b_i]}(a)=x  \end{equation}