As the question states, show that the property exhibited can only be satisfied by a logarithmic function i.e no other family of functions can satisfy the above property.
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3What about the zero function? – Ángel Mario Gallegos Jan 21 '16 at 15:13
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@5xum, corrected thanks! – TheChetan Jan 21 '16 at 15:14
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3You can think of the zero function as $\log_\infty$. :P (Don't actually do this.) – Noah Schweber Jan 21 '16 at 15:19
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2See https://shreevatsa.wordpress.com/2013/04/08/the-functional-equation-fxy-fxfy/ . I think you must assume continuity. – Ethan Bolker Jan 21 '16 at 15:19
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Continuity is necessary.
If $F(x+y)=F(x)+F(y)$, for all $x,y$ and $F$ discontinuous (such $F$ exist due to the Axiom of Choice, and in particular, the fact that $\mathbb R$ over $\mathbb Q$ possesses a Hamel basis) and $f(x)=F(\log x)$, then $f(xy)=f(x)+f(y)$, and $f$ is not logarithmic!
Yiorgos S. Smyrlis
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With the assumption that $f$ is continuous, compute the partial derivatives; firstly wrt to $x$ then $y$ \begin{align} yf'(xy) &=f'(x) \\ xf'(xy) &=f'(y) \\ \end{align} Equating terms in $f(xy)$ we have $$\frac{1}{y}f'(x)=\frac{1}{x}f'(y)$$ Or $$xf'(x)=yf'(y)$$ Now, if you know anything about these equations you will notice something interesting about both sides of the equation above, allowing you to make the crucial step in obtaining a very simple ODE to solve; of which the solution should be the one you are looking for.