algebra precalculus - How can we find functions that satisfy $f(xcdot y)=f(x)f(y)$?

Today I've encountered a question like The following;

If function $f$ satisfies $f(xy)=f(x)f(y)$ and $f(81)=3$ then find The value of $f(2)$?
What baffles me about this question is that I have to find The equation of the function in order to find $f(2)$ because $2$ is not a divisor of $81$ , using The property I found out that $f(3)=\sqrt[4]{3}$ and wondered if the function could be $f(x)=\sqrt[4]{x}$ (it satisfies the equation up there) but I do not know whether f gives $2$ a value like this or not. And there are many other functions that can be found.

So The question is how can İ get myself out of this ugly situation, and how can I find other $f$ functions that satisfy the constraints? What I am asking is not to prove that these functions are in type of $x^n$ I am trying to get what $f(2)$ is and see also whether this question is deficient ör cannot ve solved with The given details.

Thank you:)

Answer

Yes, the function is $f(x)=x^{\frac 14}$. You have shown that it works, but you need to show that it is the only function that works. I believe you should have been told that the function is continuous. As you say, you know that $f(9)=\sqrt 3, f(3)=\sqrt[4]3$. You can extend this to show that for any $x$ that is $3$ raised to a dyadic rational power (one of the form $\frac a{2^b}$) you have $f(x)=x^{\frac 14}$. Then argue that those numbers are dense in the positive reals so by continuity you have $f(x)=x^{\frac 14}$ for all positive reals.