Talk:2008 IMO Problems/Problem 4

There are many interesting properties of $f$ one can prove. The most interesting could be

\[f(xy) = f(x)f(y)\ \forall_{x,y \in \mathbb{R}^+}\]

So f must be an automorphism of the group $\mathbb{R}^+, \times$.

If you assume (or can prove) that f must be continuous, then $f$ must be of the form $f(x) = x^a$ for $a\ne 0$. (the only continuous automorphisms of the group)

Plugging this into the original functional equation gives $a=\pm 1$.

I would like to see an alternative solution along these lines.

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