$\DeclareMathOperator\elim{lim}\DeclareMathOperator\Lim{Lim}\DeclareMathOperator\lmb{lmb}\DeclareMathOperator\Lmb{Lmb}\DeclareMathOperator\mts{mts}$There are two similar functions; they determine the dependence of the values ​​of similar equations

\begin{gather*} x = L e^{s x} \iff x = \elim_s(L) \\ x = e^{L x^s} \iff x = \lmb_s(L). \end{gather*}

For positive parameters, in the region of convergence of their power series, both have two positive roots.

Through power series we can only obtain zero roots of these functions \begin{gather*} \Lim_{s, 0}(L) = \mts_{0, 0, s}(L) = L + 2s\frac{L^2}{2!} + (3s)^2\frac{L^3}{3!} + (4s)^3\frac{L^4}{4!} + \dotsb \\ \Lmb_{s, 0}(L) = \mts_{0, 1, s}(L) = 1 + L + (1 + 2s)\frac{L^2}{2!} + (1 + 3s)^2\frac{L^3}{3!} + (1 + 4s)^3\frac{L^4}{4!} + \dotsb. \end{gather*}

We need to find an analytical method, a formula for obtaining the first root of zero. It is known that the dependence of these roots is the same for both functions: $$ \frac{\Lim_{s, 1}(L)}{\Lim_{s, 0}(L)} = \left(\frac{\Lmb_{s, 1}(L)}{\Lmb_{s, 0}(L)}\right)^s. $$

I tried this: let's take special cases $s=1$, and $L=.01$ and $L=.1$.

\begin{gather*} a=\Lim_{1, 0}(.1) = 0.1118325591589629648 \\ b=\Lmb_{1, 0}(.1) = 3.5771520639572972184 \\ b/a=31.9866780377671590 \\ A=\Lim_{1, 0}(.01) = 0.01010152719853875327 \\ B=\Lmb_{1, 0}(.01) = 6.47277512439400469474 \\ B/A=640.7719344982143479. \end{gather*} It is very similar that in the formula that shows the dependence of the first root on the zero root, the natural logarithm of $1/L$ divided by $L$ is used \begin{gather*} 10\ln10=23.02585092994 \\ 100\ln100=460.5170185988091368 \end{gather*} There are still some factors and/or terms missing.