Finding the solution value of $x$ for $2^xa - bx - c = 0$

Question

I have to find a positive value of $x\geq 0$ that satisfies the given equation $$2^xa - bx - c = 0$$ where $a,b,c > 0$ are given constants. I know that this will most likely use the Lambert W function, however I have never learned that in my classes and everytime I look it up I get confused. If someone could help me walk through this I would really appreciate it!

Answer 1

Let $y=-\left(x+\frac{c}{b}\right)\ln 2$. Then $x=-\frac{y}{\ln 2}-\frac{c}{b}$, so $2^xa-bx-c=0$ iff $$2^{-\frac{y}{\ln 2}-\frac{c}b}a=2^xa=bx+c=-\frac{by}{\ln 2}.$$ That is $$ye^y=y2^{\frac{y}{\ln 2}}=-2^{\frac{c}{b}}\frac{a}b\ln 2.$$ Hence $y=W\left(-2^{\frac{c}{b}}\frac{a}b\ln 2\right)$, where $W$ is a branch of the Lambert $W$ function. Therefore $$x=-\frac{W\left(-2^{\frac{c}{b}}\frac{a}b\ln 2\right)}{\ln 2}-\frac{c}b.$$

Answer 2

Rearrange to \begin{eqnarray*} e^{x \ln 2}=bx+c. \end{eqnarray*} Raise this to the power of $b / \ln 2$ and multiply by $e^c$ \begin{eqnarray*} e^{bx +c}=(bx+c)^{b / \ln 2} e^c . \end{eqnarray*} Now let $X=bx+c$, $B=b / \ln 2$ and $C=e^c$ \begin{eqnarray*} e^{X}=(X)^{B} C . \end{eqnarray*} Take the $B^{th}$ root and rearrange \begin{eqnarray*} -\frac{X}{B} e^{-X/B}=-\frac{1}{BC^{1/B}}. \end{eqnarray*} Now recall the Lambert $W$ function is defined by $we^w=z$ gives $w=W(z)$. So we have \begin{eqnarray*} -\frac{X}{B} = W(-\frac{1}{BC^{1/B}}). \\ \end{eqnarray*} Just need to sub back ...