Can I solve this with a Lambert Function?

Question

New to W-Functions and do not understand it properly. How do I solve this equation? I know about numerical solutions (or graph solution), but I'm interested in pure algebraic solution if it exists.

$x^2 + 2^x = 100$

Answer 1

Starting from: $$x^2 + 2^x = 100$$ we can rewrite: $$(x-10)(x+10)=-e^{x\ln 2}$$ $$x=-10-\frac{e^{x\ln 2}}{x-10}$$ Applying Lagrange inversion we obtain: $$x=-10-\sum_{n=1}\left.D^{n-1}\frac{e^{xn\ln 2}}{n!(x-10)^n}\right|_{x=-10}$$ Fortunately the term $D^{n-1}\frac{e^{xn\ln 2}}{n!(x-10)^n}$ can be simplified by means of Bessel polynomials. $$x=-10-\sum_{n=1}\frac{(n-1)!}{k!(n-1-k)!n!}\left.D^{n-1-k}e^{xn\ln 2}D^{k}(x-10)^{-n}\right|_{x=-10}$$ $$x=-10-\sum_{n=1}\frac{(n-1+k)!}{k!(n-1-k)!n!}\left.{(n\ln 2)}^{n-1-k}e^{xn\ln 2}(-1)^k(x-10)^{-n-k}\right|_{x=-10}$$ Remembering the Bessel polynomials: $$B_n(x)=\sum_{k=0}^n\frac{(n+k)!}{(n-k)!k!}\left(\frac{x}{2}\right)^k$$ See: http://en.wikipedia.org/wiki/Bessel_polynomials We can rewrite the Lagrange series: $$x=-10-\sum_{n=1}{(n\ln 2)}^{n-1}\frac{1}{n!}e^{-10n\ln 2}(-20)^{-n}B_{n-1}\left(-\frac{2}{n\ln 2(-20)}\right)$$ $$x=-10-\sum_{n=1}\frac{1}{nn!\ln 2 }\left((n\ln 2)e^{-10\ln 2}\over (-20)\right)^{n}B_{n-1}\left(-\frac{2}{n\ln 2(-20)}\right)$$

References Other similar Math.stackexchange thread: Lagrange Bürmann Inversion Series Example

I'm writing a paper on solution of transcendental equations of the form $$(x-a)(x-b)=\lambda e^x$$ and similar equations. A draft (some typo errors I have to fix and some partes to extend, but the work is essentially correct ) is here: http://arxiv.org/pdf/1501.00138v2 ("Generalization of Lambert W-function, Bessel polynomials and transcendental equations")