I previously mentioned a proposed "multi-stage logarithm" function, and managed to come up with a generalisation of the function.
Originally, the multi-logarithm was defined as:
$a_0^x+a_1^x+...+a_n^x=b$
$x=\text{Log}\left(b\mid a_0, a_1, ..., a_n\right)$
But now suppose we add a multipler in front of each exponent:
$m_0\epsilon_0^x+m_1\epsilon_1^x+...+m_n\epsilon_n^x=\alpha$
$x=\text{Log}\left(\alpha\,\middle|\begin{array}{ccc} m_0, m_1, ..., m_n\\ \epsilon_0, \epsilon_1, ..., \epsilon_n \end{array}\right)$ for all $m_i\ge 0.$ If all $m_i=0$, this function is undefined, as division by zero is undefined.
For example, the solution to
$2^x+5*e^x+3^x+\pi*7^x+\sqrt{2}*10^x=1000$
would be
$x=\text{Log}\left(1000\,\middle|\begin{array}{ccc} 1,\, 5,\, 1,\, \pi,\, \sqrt{2} \\ 2,\, e,\, 3,\, 7,\, 10 \end{array}\right)$,
and
$2^x+e^x+3^x+\pi^x+5*6^x=4560$
yields
$x=\text{Log}\left(4560\,\middle|\begin{array}{ccc} 1,\, 1,\, 1,\, 1,\, 5 \\ 2,\, e,\, 3,\, \pi,\, 6 \end{array}\right)$
Special cases include: (correct me if I'm wrong)
$\displaystyle\text{Log}\left(\alpha\,\middle|\begin{array}{ccc} m, m, ..., m \\ \epsilon_0, \epsilon_1, ..., \epsilon_n \end{array}\right)=\text{Log}\left(\frac{\alpha}{m}\,\middle|\,\epsilon_0, \epsilon_1, ..., \epsilon_n\right)$ for $m>0$ (when all $m_i$ are equal),
$\displaystyle\text{Log}\left(\alpha\,\middle|\begin{array}{ccc} m_0, m_1, ..., m_n \\ \epsilon, \epsilon, ..., \epsilon \end{array}\right)=\log_{\,\epsilon}\left(\frac{\alpha}{\sum_{i=0}^{n}m_i}\right)$ for $\epsilon>0$ (when all $\epsilon_i$ are equal)
$\displaystyle\text{Log}\left(\alpha\,\middle|\,\begin{array}{ccc} \epsilon_0^\beta, \epsilon_1^\beta, ..., \epsilon_n^\beta \\ \epsilon_0, \epsilon_1, ..., \epsilon_n \end{array}\right)=\text{Log}\left(\alpha\,\middle|\,\epsilon_0, \epsilon_1, ..., \epsilon_n\right)-\beta$ for all real numbers $\beta,$
and if all $m_i$ are equal and all $\epsilon_i$ are equal, then the expression reduces to $\log_{\,\epsilon}\left(\frac{\alpha}{nm}\right)$.
How would one go about evaluating this function? A series expansion would only have a finite convergence radius, but what would that series be? Or a definite integral? How would this function be evaluated outside the radius of convergence, and if any of the $m_i$ are negative? What problems might arise if any of the $m_i$ are negative?
"Series expansion" means a series that determines $x$ in terms of $\alpha,$ all $m_i$, and all $\epsilon_i$.