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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 $k$?

user681336
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  • By taking $f(1)=e$ you are getting just one function which solves the equation. You are asked to find all functions which solve the given equation. – Kavi Rama Murthy Jun 12 '19 at 05:44
  • @KaviRamaMurthy So could I not then say "Let $k =\log_2(f(1))$ then $2^k = f(1)$ and therefore $f(x) = f(1)^x = (2^k)^x = 2^{kx}$? Or any real base $b$, really? – user681336 Jun 12 '19 at 06:00
  • I guess what I mean is, what's the motivation to even do that in the first place? Could we not just leave it as $f(x) = f(1)^x$ and say it satisfies any real $f(1)$? – user681336 Jun 12 '19 at 06:02

2 Answers2

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$f$ 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 $k=\ln A$.

GReyes
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  • Why couldn't we have just done $A = e$? – user681336 Jun 12 '19 at 05:20
  • Your equation (plus some continuity assumption) determines the solution up to the value at some point, say $f(1)$, but this value can be any non-negative number in principle. In other words, your equation has infinitely many solutions, depending on one free parameter. You can take the parameter to be $f(1)$ or the parameter $k$ above. – GReyes Jun 12 '19 at 05:23
  • So both are "equivalent" ways to express things? I can either leave it as $f(x) = f(1)^x$ for some real $f(1)$ or express as $f(x) = e^{kx}$ for some real $k$? Or $f(x) = 3^{kx}$ for some real $k$, etc, any base? – user681336 Jun 12 '19 at 06:05
  • That is exactly right. The answer could be: either $f$ is identically zero or $f(x)=A^x$ where $A=f(1)$ is some positive real number (equivalently $f(x)=e^{kx}$ where $k$ is some real number). Yes, you can choose any base. – GReyes Jun 12 '19 at 06:12
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We have that $e^{kx}$ is actually just $(e^k)^x$, which yields all possible $a^x$ (in your example of $5^x$, $k= \ln 5$ yields $e^{kx} = 5^x$). We're just changing how we describe the solutions, it's not any different.

auscrypt
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  • It's all the same as $f(x) = f(1)^x$ then correct? Just that here $f(1)$ is some unknown real whereas in $f(x) = e^{kx}$ we're just "moving it" to some unknown real $k$? – user681336 Jun 12 '19 at 06:08