Let $h$ be a differentiable function where $h(xy) = h(x)h(y), \forall x>0$. Let $h(1)=1$ be an initial condition. Prove $\exists c$ such that $h(x)=x^c$. I've tried differentiating both sides of the original relation, but I seem to be doing something wrong, as I cannot pull anything relevant out of it. Could anyone steer me in the right direction?
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Hint: Let $g(x) = \ln(h(e^x))$. What functional equation does $g(x)$ satisfy?
universalset
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I hate to sully the simplicity of my other answer, so new answer it is! Here is a hint for a different way to do it:
Take derivatives with respect to $y$ to obtain $xh^\prime(xy) = h^\prime(y)h(x)$. Now let $y=1$ to obtain $xh^\prime(x) = h^\prime(1)h(x)$. You now have a separable differential equation for $h$.
universalset
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Unfortunately I don't think I'm allowed to use Diff. Eq. methods for this problem. But I will note that $h'(x)=\frac{h'(1)h(x)}{x}$. Maybe with this bit of info, I could do something else? I was thinking that I could create $f(x)=\frac{h(x)}{x^c}$, differentiate $f(x)$, and use this bit of info? It seems that it would work, except only if $h'(1)=c$. Any way around this? – Quentin Tarantino Dec 06 '13 at 17:30
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Well, if you let $c=h^\prime(1)$ and then differentiate your $f(x)$, you should be able to show that the result is equal to $0$ by using that equation for $h^\prime(x)$. Then you get that $f(x)$ is a constant, and it's not hard to finish from there. – universalset Dec 06 '13 at 20:12