Solve $(1-(-a*\ln(bx))^n)x=\delta$ for $x$

Question

Should be a simple question, but don't know how to solve this: Given $a,b,x,\delta>0$ with $\ln(bx)<0$, I want to solve $$(1-(-a*\ln(bx))^n)x=\delta$$ for $x$, where $n\in\mathbb N$. How do I approach this?

As suggested in the comments, we may set $A:=-(-a)^d$, $t:=xb$ and $B:=\delta b$ so that the equation reads $$t(1+A\ln^n(t))=B,$$ which we could try to solve for $t$.

I'm especially interested in $n=2$, if that's easier to solve.

Answer 1

$$t(1+A\log^n(t))=B$$ Let $t=e^y$ and assumue $y>0$ $$e^y \left(A y^n+1\right)=B\quad \implies \quad e^{-y}=\frac{A y^n+1}{B}$$ that is to say $$e^{-y}=\frac A B \prod_{k=1}^n (y-r_i)$$ the $r_i$ being the solutions of $A y^n+1=0$.

In principle, the solution could be given in terms of the generalized Lambert function (have a look at equation $(4)$).

Answer 2

$$t(A\ln^N(t)+1)=B\iff w=e^\frac{2\pi i m}N\sqrt[N]{\frac BAe^{-w}-\frac 1A},m=0,\dots,N$$

Using Lagrange reversion:

$$t_k=\exp\left(\sum_{n=1}^\infty\frac{e^\frac{2\pi i m n}N}{A^\frac nN n!}\left.\frac{d^{n-1}}{dw^{n-1}}\left(B e^{-w}-1\right)^\frac nN\right|_0\right)$$

Binomial series for $|t|<|B|$ and evaluating derivatives gives:

$$\bbox[2px,border:2px dashed gray ]{t_m=\exp\left(\sum_{n=1}^\infty\sum_{k=0}^\infty\binom{\frac nN}k\frac{(-1)^kB^{\frac nN-k}}{A^\frac nNn!} \left(k-\frac nN\right)^{n-1} e^\frac{2\pi i mn}N\right)}$$

shown here

Answer 3

$$(1-(-a\ln(bx))^n)x=\delta$$ $$-(-a)^n\ln(bx)^nx+x=\delta$$

We see, this is a polynomial equation of more than one algebraically independent monomials ($\ln(bx),x$). The main theorem in [Ritt 1925] cannot help in these cases: We cannot read an elementary inverse function over a non-discrete domain directly from the equation because we don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (operations) we can read from the equation.

We want to try to solve the equation in terms of Lambert W.

$$(1-(-a\ln(bx))^n)x=\delta$$ $$x-(-a\ln(bx))^nx=\delta$$ $$x-(-a)^n\ln(bx)^nx=\delta$$ $x\to\frac{e^t}{b}$: $$-\frac{(-a)^nt^n-1}{b}e^t=\delta$$

We see, we cannot solve this equation in terms of Lambert W but in terms of Generalized Lambert W.

$$((-a)^nt^n-1)e^t=-b\delta$$ $$\left(t^n-\frac{1}{(-a)^n}\right)e^t=-\frac{b}{(-a)^n}\delta$$ $$(t-t_1)(t-t_2)...(t-t_n)e^t=-\frac{b}{(-a)^n}\delta$$

$t_1,t_2,...,t_n$ are the solutions of the equation $t^n-\frac{1}{(-a)^n}=0$. $$t=W\left(^{t_1,t_2,...,t_n}_{};-\frac{b}{(-a)^n}\delta\right)$$ $$x=\frac{1}{b}e^{W\left(^{t_1,t_2,...,t_n}_{};-\frac{b}{(-a)^n}\delta\right)}$$

So we have a closed form for $x$, and the series representations of Generalized Lambert W give some hints for calculating $x$.
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[Mező 2017] Mező, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mező/Baricz 2017] Mező, I.; Baricz, Á.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018