When does $(-1)^x$ give reals solutions?

Question

For any real x, how can I know if $(-1)^x\in\mathbb R$?
We easely guess that if x is an integer, $(-1)^x$ has a real solution.
If $x = \frac{1}{2n+1}$ with $n\in\mathbb Z$, it also gives a real solution.
So then I wonder if... is that all? Did I list all the x that gives a real solution or is it possible that a specific let's say irrational number gives a real number?

Answer 1

First assume that $-\pi < \arg(z) \le \pi$. Since $z=-1$ is pure negative real number then ${\rm{Arg(-1)}}=\pi$. Let us write \begin{align} \left( { - 1} \right)^x = e^{x\log \left( { - 1} \right)} = e^{x\left[ {\ln \left| { - 1} \right| + i\arg \left( { - 1} \right)} \right]} = e^{x\left[ {0 + i(2n+1)\pi } \right]} \end{align} which means $\cos \left( {(2n+1)\pi x} \right) + i\sin \left( {(2n+1)\pi x} \right) \in \mathbb{R}$ iff $(2n+1)\pi x=k\pi, k,n\in \mathbb{Z}$. Thus $x=\frac{k}{2n+1},k,n\in\mathbb{Z}$. So that $\cos \left( {k\pi} \right) + i\sin \left( {k\pi} \right)=(-1)^k$.

Answer 2

$$\begin{eqnarray*} (-1)^x = e^{i\pi(2n+1)x}; n\in \mathbb Z \end{eqnarray*}$$ can only be real if $(2n+1)x \in \mathbb Z$

So you would be correct to assert that the only real solutions occur for rational numbers having odd denominator.