Solution for $x$ of $\,y=(a+bx)^x$

Question

Solution for $x$ of $\,y=(a+bx)^x$

Is there an algebraic solution for $x$ to this equation? I've tried using the Lambert $W$ function and approximating it, but I've had no luck with that. Even an approximation of the solution would be appreciated.

Answer 1

With Lagrange reversion where $a,b,c,z$ can be complex:

$$(a+b z)^z=c\iff y\ln(a+b y)=\ln(c)+2\pi i m\implies z_m=\sum_{n=1}^\infty\frac{(\ln(c)+2\pi i m)^n}{n!}\left.\frac{d^{n-1}}{dx^{n-1}}\ln^{-n}(a+bx)\right|_{x=0},m\in\Bbb Z$$

Its summation identity is tedious to adjust for the Lagrange reversion solution to work. However, there is a Stirling S1 $S_n^{(m)}$ pattern.

$$\left.\frac{d^r}{dx^r}\ln^{-n}(a+bx)\right|_{x=0}=-\frac{S_r^{(1)}nb^r\ln^{-n-1}(a)}{a^r}+\frac{S_r^{(2)}(n+1)nb^r\ln^{-n-2}(a)}{a^r}-\frac{S_r^{(3)}(n+2)(n+1)nb^r\ln^{-n-3}(a)}{a^r}+\frac{S_r^{(4)}(n+3)(n+2)(n+1)n\ln^{-n-4}(a)b^r}{a^r}+\dots$$

Therefore a solution uses the pochhammer symbol $(u)_v$:

$$\boxed{(a+b z)^z=c\implies z_m=\sum_{n=1}^\infty\sum_{k=0}^{n-1}\frac{(-1)^k S_{n-1}^{(k)}(n)_k(\ln(c)+2\pi i m)^n}{n!\ln^{n+k}(a)}\left(\frac ba\right)^{n-1},m\in\Bbb Z}$$

shown here. If both sums are $\infty$, then they are interchangeable and if $a,b\in\Bbb R$, then $\bar z$ is a solution with $2\pi i m,\pm m\in\Bbb N$ corresponding to solutions $z$ with $2\pi i m,\mp m\in\Bbb N$. Finally, for smaller $|b|,|c|$ and larger $|a|$, $|m|$ can be larger with the sum converging to the correct root; usually only $m=0,\pm 1$ works for $|a|\approx |b|\approx |c|$.

It is easier to construct a series solution for the real root if wanted.

Answer 2

My answer is for real $a,b,x,y$. Ask if you want complex values.

$$(a+bx)^x=y$$ $$a+bx=y^\frac{1}{x}$$ $$a+bx=e^\frac{\ln(y)}{x}$$

This equation is an algebraic equation over the complex numbers in dependence of more than one algebraically independent monomials ($x,e^\frac{\ln(y)}{x}$). We don't know therefore how we can rearrange the equation for $x$ by only elementary functions/operations.

We want to look for applicability of Lambert W by presenting the equation in dependence of a variable $t$ and $e^t$.

$$\ln(a+bx)=\frac{\ln(y)}{x}$$ $x\to\frac{e^t-a}{b}$: $$t=\frac{b\ln(y)}{e^t-a}$$ $$e^t=\frac{at+b\ln(y)}{t}$$ $$\frac{t}{at+b\ln(y)}e^t=1$$ $$\frac{t}{t+\frac{b}{a}\ln(y)}e^t=a$$

We see, this equation cannot be solved in terms of Lambert W, but in terms of Generalized Lambert W:

$$\frac{t}{t-(-\frac{b}{a}\ln(y))}e^t=a$$ $$t=W\left(^{\ \ \ \ \ \ 0}_{-\frac{b}{a}\ln(y)};a\right)$$ $$x=\frac{1}{b}e^{W\left(^{\ \ \ \ \ \ 0}_{-\frac{b}{a}\ln(y)};a\right)}-\frac{a}{b}$$

So we have a closed form for $x$, and the series representations of Generalized Lambert W give some hints for calculating $x$.

[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