Prove that the logarithm of the roots of $\varphi(x)=e^{P(\log(x))}-1$ are the roots of any polynomial, $P(x),$ for $\Re(x).$
For example, the logarithm of the roots of $\varphi(x)=e^{\log^3(x)-5\log^2(x)+1}-1$ are the roots of $P(x)=x^3-5x^2+1.$
My attempt:
If one thinks of the roots of $\varphi(x)$ as images of the roots of $P(x)$ under some transformation $f$, then this implies that $\varphi(x)$ is some image of $P(x)$ under $f.$ So it suffices to construct a diffeomorphism $f:\Bbb R^2\to \Bbb R^2$ that maps $P(x)$ to $\varphi(x)$ showing that $P(x)$ and $\varphi(x)$ have the same properties. $f:\Bbb R^2\to\Bbb R^2$ with $f(x,y)=(e^x,e^y)$ clearly puts the roots of $P(x)$ in bijection with the roots of $\varphi(x)$ after translating $-1$ units down. So $\varphi(x)$ is a polynomial up to isomorphism.
This has a bit simpler solution. Rearranging,
$$e^{P(\log x)} = 1 \implies P(\log x) = 0$$
which gives the desired result immediately.