Is it possible to find an explicit expression for the root(s) (except $x=0$) for the following function
$$f(x)= x-2 + 2b^x$$
where $0\leq b \leq 1$. Numerically this is no problem at all. But what about an explicit expression? I suppose this is an transcendental equation?
EDIT: Thanks to Gerry Myerson, I learned that the answer is given by the Lambert W function, i.e. roots of $f$, denoted $\gamma$ are given by
$$\gamma = 2- \frac{W_k(2b^2 \log{b})}{\log{b}}$$
where $k\in \mathbb{Z}$ and $W_k$ is the $k$'th branch of the Lambert W function.
Related question:
I numerically observed that the $k=0$ branch has maximal real part for a range of $k$'s. Can this be right and be proven by simple properties?
In case you are still interested, I can still solve this:
$$0=x-2+2b^x$$
Use substitution $x-2=A$.
$$0=A+2b^{A+2}=A+2b^2b^A$$
$$A=-2b^2b^A=-2b^2e^{\ln(b)A}$$
$$1=-2b^2A^{-1}e^{\ln(b)A}$$
$$1=-\frac12b^{-2}Ae^{-\ln(b)A}$$
Substitute $-\frac12b^{-2}=X$
$$1=XAe^{-\ln(b)A}$$
$$X^{-1}=Ae^{-\ln(b)A}$$
$$-X^{-1}\ln(b)=-\ln(b)Ae^{-\ln(b)A}$$
$$-\ln(b)A=W(-X^{-1}\ln(b))$$
$$A=\frac{W(-X^{-1}\ln(b))}{-\ln(b)}=X^{-1}e^{W(-X^{-1}\ln(b))}$$
$$A=-2b^2e^{W(-X^{-1}\ln(b))}$$
$$x-2=-2b^2e^{W(-X^{-1}\ln(b))}$$
$$x=-2b^2e^{W(-X^{-1}\ln(b))}+2$$