How to solve a power/polynomial equation

Question

Is there a method to solve such equations : $AB^{x} + C x = D$ ? with $A$, $B$, $C$ and $D$ real parameters and $x\in \mathbb{R}$ and $B>0$?

Answer 1

First, three degenerate cases;

1. If $C=0$ and $B=1$ then the equation reduces to $A=D$, and there is nothing to solve.
2. If $C=0$ and $B\neq1$ then $AB^x=D$ and hence $x=\frac{\ln{\frac{D}{A}}}{\ln{B}}$.
3. If $C\neq0$ and $B=1$ then $A+Cx=D$ and hence $x=\frac{D-A}{C}$.

If $C\neq0$ and $B\neq1$ then setting $a:=\frac{A}{C}$ and $d:=\frac{D}{C}$ yields $$aB^x=d-x.$$ Setting $y:=d-x$ and $v:=B^{-1}$ and $u:=aB^d$ then yields $$y=aB^{d-y}=aB^dB^{-y}=uv^y=ue^{y\ln{v}},$$ where $e$ denotes Euler's constant. Finally setting $z:=-y\ln{v}$ and rearranging yields the equation $$ze^z=-u\ln{v},$$ which has no 'elementary' closed form solution in general. The solution to this is precisely what the Lambert W function denotes. That is, by definition we have $z=W(-u\ln{v})$. Substituting everything back yields $$-\left(\frac{D}{C}-x\right)\ln{B^{-1}}=W\left(-\frac{A}{C}B^{\frac{D}{C}}\ln{B^{-1}}\right),$$ and isolating $x$ shows that $$x=\frac{D}{C}-\frac{1}{\ln{B}}W\left(\frac{A}{C}B^{\frac{D}{C}}\ln{B}\right).$$ Finding exact values of the Lambert W function in terms of elementary functions is in general not possible, though of course there are practical algorithms for numerical approximations.