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If $f(xy)=f(x)+f(y)$, then \[f(x+y)-f(xy)=f\left(\frac{x+y}{xy}\right)\]or \[f(a)-f(b)=f\left(\frac{a}{b}\right)\]for some $a$ and $b$. It is well known that only the logarithmic function $f(x)=\log_c{d}$ for some $c$ and $d$ satisfies this. We also notice that it satisfies the condition $f(xy)=f(x)+f(y)$, so we are done. $\blacksquare$

S.Das93 wrote: If $f(xy)=f(x)+f(y)$, then \[f(x+y)-f(xy)=f\left(\frac{x+y}{xy}\right)\]or \[f(a)-f(b)=f\left(\frac{a}{b}\right)\]for some $a$ and $b$. It is well known that only the logarithmic function $f(x)=\log_c{d}$ for some $c$ and $d$ satisfies this. We also notice that it satisfies the condition $f(xy)=f(x)+f(y)$, so we are done. $\blacksquare$ I think that you have to show the other side, too

CDE method.

Can you explain in more detail?

Insert a new variable $z$ and try to double count!

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The proof naturally splits into two parts. Part 1: Assume that $f(xy)=f(x)+f(y)$. That's the easy part. Note that $f(x+y)-f(x)-f(y)=f(x+y)-f(xy)=f(\dfrac{x+y}{xy})=f(\dfrac{1}{x}+\dfrac{1}{y}),$ as desired. Part 2: Assume that $f(x+y)-f(x)-f(y)=f(\dfrac{1}{x}+\dfrac{1}{y})$. Note that by taking $y=1/x$, we obtain that $f(1/x)=-f(x)$. The trick is to introduce a third variable, $z$. Note that $f(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})=f(\dfrac{1}{x}+\dfrac{1}{\dfrac{yz}{y+z}})=$ $=f(x+\dfrac{yz}{y+z})-f(x)-f(\dfrac{yz}{y+z})=f(\dfrac{xy+yz+zx}{y+z})-f(x)+f(\dfrac{y+z}{yz})=$ $=f(\dfrac{xy+yz+zx}{y+z})-f(x)+f(y+z)-f(y)-f(z)$ Therefore, by computing $f(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})$ symmetrically, we obtain $f(\dfrac{xy+yz+zx}{y+z})+f(y+z)=f(\dfrac{xy+yz+zx}{z+x})+f(z+x),$ for all $x,y,z>0$. Now, let $y+z=k$ and $z+x=\ell$. Then, $xy+yz+zx=\ldots=k \ell -z^2,$ and so $xy+yz+zx$ takes all values $<k\ell$. Thus, $f(\dfrac{A}{k})+f(k)=f(\dfrac{A}{\ell})+f(\ell),$ for all $A<k\ell$. Taking $\ell=1$ here, we obtain that $f(\dfrac{A}{k})+f(k)=f(A),$ for all $A<k$. Therefore, $f(xy)=f(x)+f(y)$ for all $x,y$ such that at least one of them is $<1$. Note that if at least one of them is equal to $1$, we obviously have $f(xy)=f(x)+f(y)$ (in light of $f(1)=0$, which is obtained if we put $x=y=1$ in the initial relation). Now, if $x,y>1,$ we have $f(xy)=-f(\dfrac{1}{xy})=-(f(\dfrac{1}{x})+f(\dfrac{1}{y}))=f(x)+f(y),$ as desired. Thus, in all cases we obtain that $f(xy)=f(x)+f(y),$ and so we may finish.

Cool solution, what was your motivation on introducing the third variable?

Beautiful Problem!

We get the equation $f\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) =f\left(\frac{xy+yz+zx}{y+z}\right)+f(y+z)-f(x)-f(y)-f(z)$ and therefore by interchanging variables we get $f\left(\frac{xy+yz+zx}{y+z}\right)+f(y+z)=f\left(\frac{xy+yz+zx}{z+x}\right)+f(x+z)$ now take $y+z=u$ and $z+x=v$ so $\sum_{\text{cyc}} xy = uv - z^2 $ and therefore this achieves all values less than $uv$. from this It is easy to get that $f(xy)=f(x)+f(y)$ where at least one is less than $1$, now by using $f(x)+f\left(\frac{1}{x}\right)=0$ this gives $f(xy)=f(x)+f(y)$.