I know that x' = kx is exponential growth. I've tried to come up with some solutions.
My first solution is x' = -x^3, which has an asymptotically stable point at x=0. It is approached more slowly, but does this still count as exponential?
Another possible solution is x' = -x/(1+x) which is asymptotically stable at x=0. While it is also approached more slowly, is this slow enough to not be exponential?
Not sure how to proceed with this problem.
Since you want to have something like $x(t)=t^{-p}$ for $t\to\infty$, instead of exponential dependence, consider that $x(t)=t^{-p}$ satisfies $x'(t) = -pt^{-p-1} = -px^{1+1/p}$. Thus, $x'=-x^\alpha$ with $\alpha>1$ does the job. The choice $\alpha=3$ (hence $p=1/2$) is convenient because one does not have to worry about how to raise negative $x$ to power $\alpha$ then.