clark's answer is surely a great and simple one. Thomas Andrews' hint is another great one. Here's a more complicated answer that shows more, that there are entire intervals of numbers that would not be generated.
Let $x$ be a generator of the cyclic group $\mathbb{R}^*$. If $|x| = 1$, then all powers of $x$ satisfy $|x^n| = 1$. So, $|x| < 1$ or $|x| > 1$.
If $|x| < 1$, then $|x^{-1}| > 1$ and $x^{-1}$ is also a generator. So, assume $|x| > 1$.
If $|x| > 1$, then $|x| = 1 + \epsilon$ for some $\epsilon > 0$. Any positive power of $x$ will satisfy $|x|^n = (1 + \epsilon)^n > 1 + \epsilon$. Any negative power of $x$ will satisfy $|x|^{-n} = (1 + \epsilon)^{-n} < (1+ \epsilon)^{-1}$.
So, there are entire intervals that are never achieved.