If $f(nx)=f(x)^n$ then $f(x+y)=f(x)f(y)$ If $f:\mathbb{R}\to\mathbb{R}_{+}$ is injective monotone function and if, $f(nx)=f(x)^n$ for all $n\in\mathbb{Z}$ and for all $x\in\mathbb{R}$, then,
$$f(x+y)=f(x)f(y),\text{ for all }x,y\in\mathbb{R}$$
It is easy to check the equality if $y$ is a multiple of $x$.
I thought about proving the inequalities, $f(x+y)\le f(x)f(y)$ and $f(x+y)\ge f(x)f(y)$, but I couldn't get to them.
Assuming that $f$ is increasing and $x<y$, then $y=kx+r$, for some $k\in\mathbb{Z}$ and $0< r\le \lvert x \rvert$, so
$$f(x+y)=f(x+kx+r)=f(x(k+1)+r)\ge f(x(k+1))=f(x)^{k+1}=f(x)f(x)^k=f(x)f(kx)$$
That is, $f(x+y)\ge f(x)f(kx)$, but the problem is that $f(kx)<f(y)$ so I can't continue the demonstration.

If $i$ is an irrational number is a limit of a sequence $r_n$ of rational numbers
$$ i= \lim_{n \rightarrow +\infty} r_n .$$
Thus
$$\forall N \in \mathbb N \quad \exists n_0 \in \mathbb N  \quad \forall n \in \mathbb N \quad  n \ge n_0  \Rightarrow  |i-r_n| \le \frac{1}{N}
 \Rightarrow  \quad |ix-r_nx| \le \frac{|x|}{N} \quad -\frac{|x|}{N} + r_n\cdot x \le ix \le \frac{|x|}{N} + r_n \cdot x.$$
If $x \ge 0$ then
$$f(x)^{r_n-\frac{1}{N}} \le f(ix) \le f(x)^{r_n + \frac{1}{N}}.$$
This implies that
$$f(i\cdot x) = f(x)^{i}.$$
The case $x \le 0$ is similar.
