Can we find a function $f : R \to R$ such that $$f(mx) = m^x.f(x)$$ provided $f(x)$ is continuous and differentiable in $[1,\infty) \subset R$ and $m \in N$ ?
My understanding :
We know that a homogeneous function would be defined in the similar manner with any integer $m \neq 0$, $h(mx) = m^k.h(x)$ where k is a non-zero integer. But, since power of $m$ is variable in $f$ , It should be a non-homogeneous function.
Thank you
Edit : I have added some relatable tags, I am not sure where this question would correctly fit in and the question seems to get lost in the never-ending wave of questions.
As per the comment above such an $f$ must be zero if the relation is assumed for all $m \ge 1$ and $f$ continuous.
We have $f(m)=mf(1)$ but for $x=m=2$ we get $4f(1)=f(4)=4f(2)=8f(1)$ so $f(1)=0$ hence $f(m)=0$ for all $m$ integer at least $1$ but then if $n/m>1$ is a fraction we get $f(n/m)=(n/m)^mf(n)=0$ and by continuity $f=0$
If we want the relation only for a fixed $m$ then for $m \ge 2$ the function $m^{\frac{x}{m-1}}$ works