Can anyone tell me how to find number of solutions $(x+a)^x=b$? For example $(x+1)^x=-1$ has four complex solutions, $(x+3)^x=10$ has two solutions,one positive one negative, and $(x-4)^x=-10$ hasn't any solutions.
PS.Sorry for my bad English,I hope you understand my question.
How to solve $(x+a)^x=b$?
Well, to begin, as you already gather, we must allow for $x \in \mathbb{C}$. Then the question begs the question how do we define $z^x$ for complex $x$. Use the complex exponential function: $$ z^x = \exp (x \log (z)) $$ clearly $z=0$ is a problem. Moreover, this is a set of values due to the fact that $\log (z)$ is multiply valued. Consider then, $(x+a)^x=b$ becomes: $$ \exp (x \log (x+a)) =b $$ If $b \neq 0$ then it follows that, $$ x \log (x+a) = \log b $$ Assume $x \neq 0$ and divide by $x$, $$ \log (x+a) = \frac{1}{x}\log b. $$ Exponentiate, note $\exp (\log (z))=z$ $$ x+a = \exp \left( \frac{1}{x}\log b \right). $$ This is clearly transcendental, perhaps that was clear from the outset. But, at best we can hope to characterize solutions with a special function. Perhaps the Lambert W-function. According to (1) under "Generalizations" in the Wikipedia article, $$ e^{-cx} = a_o(x-r) $$ has solution $$ x = r + \frac{1}{c} \text{W} \left( \frac{ce^{-cr}}{a_o} \right)$$ This is lovely, but $e^{-cx}$ should look more like $e^{c/x}$ to match the problem we face. Perhaps a champion of the Lambert-$W$ function will appear and show us the light.
Sorry, I don't understand. So, $a_0=1$,$c=-\ln b$,$r = -a$; But there is 1/x in $\exp \left( \frac{1}{x}\log b \right)$and x in the left side of $e^{-cx} = a_o(x-r)$.
