What common developments are there to solve $1+x=x^a$? I am facing the trinominal equation $1+x=x^a$ where $a$ is not necessarely an integer, but some positive real. Are there some common series developments of functions $x=V(a)$ that one can solve the equatios via $1+V(a)=V(a)^a$.
 A: If you plot $1+x$ and $x^a$ it will be clear that , for $0 \leqslant x$, 
there is one, and only one, real solution iff $1<a$.
That can be proven rigorously by analytical considerations.
Then, if you consider the equality witten as $$f(x)=1+x-x^a=0$$
you can apply the Newton-Raphson method and find a sequence
which will approximate the root.
 
----  Addendum    ----
Following your comment, consider that you can write a series expansion of the $f(x)$ above around $x=1$, as: $$\begin{gathered}
  f(x) = \left( {2 + \left( {x - 1} \right)} \right) - \left( {1 + \left( {x - 1} \right)} \right)^{\;a}  =  \hfill \\
   = \left( {2 + \left( {x - 1} \right)} \right) - \sum\limits_{0\, \leqslant \,k} {\left( \begin{gathered}
  a \\ 
  k \\ 
\end{gathered}  \right)\left( {x - 1} \right)^{\;k} }  =  \hfill \\
   = 1 - \left( {a - 1} \right)\left( {x - 1} \right) - \sum\limits_{2\, \leqslant \,k} {\left( \begin{gathered}
  a \\ 
  k \\ 
\end{gathered}  \right)\left( {x - 1} \right)^{\;k} }  \hfill \\ 
\end{gathered} 
$$
A: Hint: Under certain conditions (see Fixed-point iteration)
You can rewrite $x=x^a-1$, in which $\phi(x)=x^a-1$. A recursive solution is given by 
$$x_{n+1}=\phi(x_n).$$
