Given $a,b,c$ positive numbers, is there a function $h$ (no trivial) such that $$h(ax+b) = c\, h(x)$$ for all $x>0$ ?.
I know that taking $h(x)= c^{\frac{x}{b}}$ makes $h(x+b) = c \, h(x)$, and that taking $h(x)= c^{\frac{\ln(x)}{\ln(a)}}$ makes $h(ax)=c\,h(x)$ but i don't see how to combine both conditions.
Any help will be appreciated
On
This is a specific case of Shr\"oder's equation,
$$f(x)=cf(\alpha(x))=c^nf(\alpha^n(x))$$
We can solve this problem by first solving for $\alpha^n(x)$, where $\alpha(x)=ax+b$. Some playing around shows that $$\alpha^n(x)=b+ab+a^2b+\cdots+a^{n-1}b+a^nx=b\frac{1-a^n}{1-a}+a^nx=\frac{b}{1-a}+\left(x-\frac{b}{1-a}\right)a^n$$ Now let's seperate the domain of $f$ into sets $E_{\gamma}$ where $x\in E_\gamma$ if and only if there is an integer $m$ such that $x=\alpha^m(\gamma)$. This forms an equivalence relation, but more importantly it seperates our domain into independent parts of $f$. For each of these, let $$f_\gamma(x)=\frac{f(x)}{f(\gamma)}$$ Then $$f_\gamma(\alpha^n(\gamma))=c^n$$ Solving for $a^n$ in $\alpha^n(\gamma)$ $$a^n=\left(\alpha^n(\gamma)-\frac{b}{1-a}\right)/\left(\gamma-\frac{b}{1-a}\right)$$ we define a new function $$g_\gamma\left(\left(x-\frac{b}{1-a}\right)/\left(\gamma-\frac{b}{1-a}\right)\right)=f_\gamma(x)$$ So $$g_\gamma(a^n)=c^n$$ At this point, note that we mearly seek any function which satisfies the above for all $n$, since these are the only values we care about. Thus, $$g_\gamma(x)=x^{\log_a(c)}$$ is one simple solution. But this works for all $\gamma$, so working backwords, we see that $$f(x)=f(\gamma)f_\gamma(x)=f(\gamma)\left(\left(x-\frac{b}{1-a}\right)/\left(\gamma-\frac{b}{1-a}\right)\right)^{\log_a(c)}$$ on each $E_\gamma$ domain.
If you would like to learn more about such problems, I recommend "Functional Equations and How to Solve Them" by Christopher G. Small.
There are many many such functions (except in the case $a=1,b=0,c\neq 1$). For convenience, let me first assume $a\neq 1$ and consider functions $h:\mathbb{R}\to\mathbb{R}$ which satisfy $$h(ax+b)=ch(x)$$ for all $x\in\mathbb{R}$, not just $x>0$. There is then a unique $d\in\mathbb{R}$ such that $ad+b=d$, namely $d=\frac{b}{1-a}$. Writing $f(y)=h(y+d)$ and substituting $x=y+d$ in the functional equation, we then see that the required functional equation on $h$ is equivalent to $f$ satisfying $$f(ay)=cf(y).$$ This equation has tons of solutions besides just an exponential solution $f(y)=c^{\log_a|y|}$ like you mentioned. Indeed, supposing $a>1$, then for any function $f_0:(-a,-1]\cup[1,a)\to\mathbb{R}$, we can extend to a function $f$ satisfying the functional equation on all of $\mathbb{R}$ by defining $f(y)=c^nf_0(y/a^n)$ for the unique $n\in\mathbb{Z}$ such that $y/a^n\in (-a,-1]\cup[1,a)$ when $y\neq 0$, and $f(0)=0$. (If $c=1$, then the value of $f(0)$ can also be assigned arbitrarily.) When $a>1$ the story is similar, just with $[-1,-a)\cup (a,1]$ instead of $(-a,-1]\cup[1,a)$.
The case $a=1$ similarly has many solutions unless $b=0$: given any function $h_0:[0,b)\to\mathbb{R}$, you can extend to $h$ satisfying the functional equation $h(x+b)=ch(x)$ by defining $h(x)=c^nh_0(x-nb)$ for the unique $n\in\mathbb{Z}$ such that $x-nb\in [0,b)$. If $a=1$ and $b=0$, then the problem is trivial: if $c\neq 1$ then obviously only the zero function works and if $c=1$ then any function works.