Generalisation of the functional equation f(xy) = f(x) + f(y)

Question

If I understand correctly (I am a high school student), when proving the existence of $\ln x$ in real analysis, we consider the functional equation $$ f(xy) = f(x) + f(y)$$ I understand that there are many questions on Stackexchange on this functional equation, but I was wondering whether anything interesting occurs when we consider $$f(xyz) = f(x) + f(y) + f(z)$$ or even $$f(xyz\dots) = f(x) + f(y) +f(z) + \dots$$ $\ln x$ seems to work for this too, but do any other functions?

Thanks in advance.

Answer 1

There are two obvious solutions:

$f(x) = 0$

and

$f(x) = \log_b (x)$

I am not exactly sure, but this could be a possibility too:

Let $x = 2^{(p/q)} \cdot b$ with $b$ being not of the form $2^{(p/q)}$, then

$f(x) = p/q$

If $x$ is irrational, then $f(x)$ is not defined.

This function would be, of course, discontinues everywhere.

-

I might have found another possibility, but it is hard to turn it into a formula.

The required condition implies that $f(x^n) = n f(x)$. Let's now define the function for natural numbers:

If $x$ is prime, then $f(x) = 1$. Of course $f(1) = 0$. Now, by applying the rule above, we can get do any power of a prime. Any other composite number can be split into it's prime factors and therefore be deducted.

Now we can get to rational numbers as well. For example:

"What is $f(\frac{2}{3})$?" Answer: $f(9 \cdot \frac{2}{3}) = f(6)$ which leads to $f(9) + f(\frac{2}{3}) = f(6)$ and therefore $f(\frac{2}{3}) = 0$

Finally, $f(p/q) = f(p) - f(q)$, so this can be expanded to rational numbers as well.

Answer 2

There is no way you can have

$$f(xy) = f(x)+f(y)$$

without having

$$f(xyz) = f(x)+f(y)+f(z)$$ since, treating $xy$ as one factor, we have: $$f((xy)z) = f(xy)+f(z) = f(x)+f(y)+f(z)$$

In a similar way you can by induction show that for $n$ factors you can reduce down to $n-1$ factors and that the $2$ factor case is true.