TITLE: Prove that $x^2$ and $x^3$ are topological conjugated on $[0, \infty)$.
QUESTION [8 upvotes]: As in the topic. I have to prove that the functions $x ^ 2$ and $x ^ 3$ are topologically conjugated. I tried to write it out by definition: $f (x ^ 2) = f (x) ^ 3$ and choose $f (x) = x ^ a$, but unfortunately it doesn't work. It's my beginnings in this field, so I do not have much experience yet.
Do you have any hints?

REPLY [3 votes]: We can consider the function $f:[0,\infty)\to[0,\infty)$ defined by
$$f(x)=
\begin{cases}
e^{\log(x)^{\log(3)/\log(2)}}&x>1\\
e^{-(-\log(x))^{\log(3)/\log(2)}}&0<x\leq 1\\
0&x=0
\end{cases}$$
To find this function, can be useful to note that $x^2$ is topologically conjugated to $2x$ through $\log$ function and similarly $x^3$ is conjugated to $3x$.
Moreover $x\to x^{\log(3)/\log(2)}$ is a conjugation between $2x$ and $3x$.
More precisely, we have the commutative diagram below where $h(x)=\operatorname{sign}(x)|x|^{\log(3)/\log(2)}$