When solving for solutions to $f(x)f(y) = f(x + y)$ (namely my proof attempt here which I am still unsure if it's correct to be honest), I eventually arrive at the result $f(x) = f(1)^x$.
So for example if $f(1) = 5$ then $f(x) = 5^x$ and we see that $5^x 5^y = 5^{x+y}$ as expected.
However I sometimes see people say that the solution is $f(x) = e^{kx}$ or something to this effect?
Where are these coming from? Isn't it the case that all $f(x)f(y) = f(x+y)$ implies is $f(x) = f(1)^x$, no more, no less, where $f(x)$ is some real-valued differentiable function, i.e. $f(1)$ is just some arbitrary constant real? Where is $k$ coming from when people state the result like this?
I could maybe see some argument like: Since $f(1) > 0$, let $k = \ln(f(1))$ which is also $e^k = f(1)$, so $f(x) = (e^k)^x = e^{kx}$ but then why couldn't we have just said $f(1) = e$ to begin with such that $f(x) = e^x$ without the $k$?
$f$ is a real function. Setting $x=y=1/2$ you get $f(1)=f(1/2)^2\ge 0$. If $f(1)=0$ you get the trivial solution $f=0$. Otherwise, you have $f(1)=e^k$ for some $k$. Any exponential function $g(x)=A^x$ with $A>0$ can be written as $g(x)=e^{kx}$ with $k=\ln A$.