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.
 A: If $x,y \in \mathbb{Z}$ then
$$
f(x+y) = f((x+y)1) = f(1)^{x+y} = f(1)^x f(1)^y = f(x1) f(y1) = f(x)f(y).
$$
If $x,y \in \mathbb{Q}$ then there is $n\in\mathbb{N}$ such that $nx,ny\in\mathbb{Z}.$ Then
$$
f(x+y)^n = f(n(x+y)) = f(nx+ny) = f(nx) f(ny) = f(x)^n f(y)^n = (f(x)f(y))^n,
$$
so $f(x+y)=f(x)f(y).$
If $x,y\in\mathbb{R}$ then take sequences $x_k^+, x_k^-, y_k^+, y_k^- \in \mathbb{Q}$ such that $x_k^+\to x+,\ x_k^-\to x-,\ y_k^+\to y+,\ y_k^-\to y-$. The result then follows from density of $\mathbb{Q}$ in $\mathbb{R}$ and from monotonicity. Try to complete this step yourself.
A: 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.
A: Let $$f(\frac{p}{q}\cdot x) =f(p\cdot\frac{x}{q})=f(\frac{x}{q})^p.$$
But
$$f(x)=f(q\cdot\frac{x}{q})=f(\frac{x}{q})^q.$$
Hence
$$f(x)^{\frac{1}{q}}= f(\frac{x}{q}).$$
Thus
$$f(\frac{p}{q}\cdot x)= f(x)^{\frac{p}{q}}.$$
A: I would like to approach this a little bit differently - and possibly inferiorly as I will be looking only at the rational numbers so it could still be the case that the property does not hold for $x=\pi$ or some other irrational number - I will first proof that the only class of functions that satisfies the property
$$  f(nx) = f(x)^n \, \forall \,x \in \mathbb{Q}, n \in \mathbb{N} $$
is the class
$$f(x) = f(1)^x \,\forall \,x \in \mathbb{Q}$$
I will do this by first observing that
$$f\left(\frac{1}{n}\right) = \sqrt[n]{f(1)} \, \forall \, n \in \mathbb{N}$$
and combine that property with the original property to show that
$$ x = \frac{p}{q} \implies f(x) = \sqrt[q]{f(1)}^{p} = f(1)^x \, \forall \, p \in \mathbb{Z}, q \in \mathbb{N}$$
which would be equivalent to saying
$$ f(x) = f(1)^x \, \forall \, x \in \mathbb{Q}$$
since any rational number can be written as the quotient of an integer and a natural number
The first property will be proven by contradiction: let us assume that
$$\exists \, n \in \mathbb{N} : f\left(\frac{1}{n}\right) \neq \sqrt[n]{f(1)}$$
from that it would follow that
$$f(1) = f\left(\frac{n}{n}\right) = f\left(\frac{1}{n}\right)^n \neq \sqrt[n]{f(1)}^n = f(1)$$
This is a contraciction, therefore it must be so that
$$f\left(\frac{1}{n}\right) = \sqrt[n]{f(1)} \, \forall \, n \in \mathbb{N}$$
Now for our second step, let us pick two values $p \in \mathbb{Z}, q \in \mathbb{N}$ with $x = \frac{p}{q}$ making $x$ any rational number
$$ f(x) =  f\left(\frac{p}{q}\right) = f\left(\frac{1}{q}\right)^p = \sqrt[q]{f(1)}^{p} = f(1)^{p/q} = f(1)^x$$
Now, we have proven that $f(xn) = f(x)^n \implies f(x) = f(1)^x$ for any rational number $x$, because $f(1)$ is a constant we will give it a new name, $r$ to get that $f(x)$ must be of the form $r^x$. To this we can apply power rules $$f(x+y) = r^{x+y} = r^x r^y = f(x)f(y)$$
Therefore, we have shown that $$f(nx) = f(x)^n \, \forall x \in \mathbb{Q}, n \in \mathbb{Z} \implies f(x+y) = f(x)f(y) \, \forall x, y \in \mathbb{Q}$$
A: Following the ideas presented by Mohamed Cheddadi and Martin van IJcken, it is possible to ask the question again in the following way,

Let $f:\mathbb{R}\to\mathbb{R}_{+}$ be a monotone injective function. The statements are equivalent:

*

*$f(nx)=f(x)^n$, for all $n\in\mathbb{Z}$ and for all $x\in\mathbb{R}$

*$f(x)=a^x$, for all $x\in\mathbb{R}$, where $a=f(1)$

*$f(x+y)=f(x)\cdot f(y)$, for all $x,y\in\mathbb{R}$

The implication of the original question is $1. \Rightarrow 3.$, but for that we must proof $1. \Rightarrow 2.$ and then $2. \Rightarrow 3.$
$\boxed{1. \Rightarrow 2.}$ 
Assuming that $1.$ is valid, note that if $r =\frac{m}{n}, m\in\mathbb{Z},n\in\mathbb{N}$, we have $f(rx)=f(x)^r$ for all $x\in\mathbb{R}$. Indeed, with $nr=m$,
$$f(rx)^n=f(nrx)=f(mx)=f(x)^m$$
So
$$f(rx)^n=f(x)^m\Rightarrow (f(rx)^n)^{\frac{1}{n}}=(f(x)^m)^{\frac{1}{n}}\Rightarrow f(rx)=f(x)^{\frac{m}{n}}$$
So, if $a=f(1)$, we have to
$$f(r)=f(r\cdot 1)=f(1)^r=a^r \text{ for all } r\in\mathbb{Q}$$
Assuming without loss of generality that $f$ is an increasing function, therefore
$$1=f(x)^0 = f(0\cdot x)=f(0)<f(1)=a\Longrightarrow 1<a$$
Let's suppose by absurdity that there is $x_0 \in\mathbb{R}$, such that $f(x_0)\neq a^{x_0}$. Say $f(x_0)<a^{x_0}$. Then we can choose an $r\in\mathbb{Q}$ such that
$$f(x_0)<a^r<a^{x_0}\Rightarrow f(x_0)<f(r)<a^{x_0}$$
As $f$ is increasing, $x_0<r$, but on the other hand, like $a>1$, then if $a^r < a^{x_0}$ we have $r<x_0$, absurd!.
$\boxed{2. \Rightarrow 3.}$ 
Assuming that $2.$ is valid,
$$f(x+y)=f(1\cdot (x+y))=f(1)^{x+y}=f(1)^x \cdot f(1)^y = f(x)\cdot f(y)$$
$\boxed{3. \Rightarrow 1.}$ 
Is trivial
