Given $a\ln\ln x = \ln\ln(bx)$. Seeking any general form for the isolation of $x$. Perhaps in terms of Lambert $W$. I'm not really interested in any specific value solution or Newton's method. I tried solving for $x$ in terms of $W(x)$ but was unable.
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1You can raise $e$ to both sides to obtain $[\ln(x)]^a=\ln(x)+\ln(b)$. Making the substitution $y:=\ln(x)$, we obtain $$y^a-y-\ln(b)=0$$ Which is not going to have a solution using the Lambert function. Indeed, I don't think there would be a closed form solution using standard functions. – nathan.j.mcdougall Nov 07 '19 at 23:34
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Thank you Nathan. – Anaxagoras Nov 08 '19 at 00:28
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2@nathan.j.mcdougall Lagrange's inversion formula will give a root of $x^a - x + b$ as a series for $a \in \mathbb N$ and for values of $b$ where the series converges. Due to continuity, the series still gives a root for non-integral $a$. The series is a Fox-Wright function $$b , {1 \hspace {-2.5px} \Psi{\hspace {-1px} 1}} {\left( b^{a - 1} \middle | {(1, a) \atop (2, a - 1)} \right)}.$$ – Maxim Nov 15 '19 at 00:16
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@Maxim. Exactly what I needed. It's beautiful, thank you – Anaxagoras Jan 26 '23 at 19:00