How do I solve the following equation? I understand it is not possible with regular arithmetic steps:
$$xb^x = a$$
Provided $b \in \mathbb{R}^+$ and $a \in \Bbb{R}$. What would be the formal explanation telling this cannot be solved by regular arithmetics, and what would be a good way to solve it?
Write the equation as $$ x\,e^{(\log b)x}=a. $$ Multiply the equation by $\log b$: $$ (\log b)x\,e^{(\log b)x}=a\log b. $$ The Lambert function $W(z)$ is defined as the inverse of $z\,e^z$: $$ (\log b)x=W(a\log b). $$ Finally $$ x=\frac{W(a\log b)}{\log b}. $$