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 $x,y \in \mathbb{Z}$ 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.
