Does this transformation and reparametrization preserve non-analyticity of smooth functions

Question

A $C^{\infty}$ function $f : \mathbb{R} \to \mathbb{R}_+$ is non-analytic for some $x_0 \in \mathbb{R}$. Can we conclude that $x\mapsto\log(f(e^x))$ is non-analytic at $x = \log(x_0)$?

Answer 1

Assume that $f(y)>0$ in some neighborhood of $y_0\in{\mathbb R}$. If the function $g:=\log \circ f\circ \exp$ is analytic at $x_0:=\log y_0$ then $f=\exp\circ g\circ\log$ is analytic at $y_0$.