How can we find functions that satisfy $f(x\cdot y)=f(x)f(y)$? 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:)
 A: 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.
A: Even assuming continuity and a domain of $\Bbb R$, there are at least two solutions: $$\tag1f(x)=\sqrt[4]{|x|} $$
and $$\tag2f(x)=\operatorname{sgn}(x)\cdot\sqrt[4]{|x|}. $$
What can we be sure of?
From $f(0\cdot 81)=f(0)\cdot f(81)$ we conclude $f(0)=0$. From $f(1\cdot 81)=f(1)\cdot f(81)$ we conclude $f(1)=1$. 
On the other hand, if $x\ne 0$, then $f(x)\ne 0$ because $f(x)\cdot f(\frac{81}x)=f(81)\ne 0$. 
From $f((-1)\cdot(-1))=f(-1)\cdot f(-1)$  we conclude $f(-1)=\pm1$ (and as the example solutions show, both are indeed possible). As $f(-x)=f(-1)f(x)$, we need only consider positive $x$, once we have picked a value for $f(-1)$. (note tha tin particular $|f(-x)|=|f(x)|$).
Now define $g(x)=\ln |f(e^x)|$. Then 
$$g(x+y)= \ln|f(e^{x+y})|=\ln|f(e^xe^y)|=\ln|f(e^x)f(e^y)|=\ln|f(x)|+\ln|f(y)|=g(x)+g(y).$$
This is a well-known equation, and under very mild additional conditions (such as: $g$ is continuous at at least one point, or: it is possible to write down an explicit expression for $g$), it follows that $g(x)=x\cdot g(1)$. Thus, for positive $x$, we find (under just as mild conditions)
$$|f(x)|= e^{g(\ln x)}=e^{g(1)\ln x}=x^{g(1)}\qquad\text{for }x>0$$
and as $f(81)=3$, we must have $g(1)=\frac14$. The sign of $f$ might still flicker wildly, but invoking our "mild conditions" once again, we find that $(1)$ and $(2)$ are the only solutions.
A: If $f$ has domain over the integers, then let $g$ be any mapping of the primes into the reals. Let $x \geqslant 2$ be an integer. Then $f(x)=g(p_1)^{a_1}g(p_2)^{a_2}\cdots g(p_n)^{a_n},$ where $x={p_1}^{a_1}{p_2}^{a_2}\cdots {p_n}^{a_n}$ is the prime factorization of $x.$
