# prove $f(x)f(y)=f(xy), f(1)=1 \iff f(x)=x^k (k\ real)$ [duplicate]

prove $$f(x)f(y)=f(xy), f(1)=1 \iff f(x)=x^k (k\ real)$$ for $$f:\mathbb{R}^+\to \mathbb{R}^+$$
I find $$f(a^r)=f(a)^r$$ for rational r, but I cannot move to the next step.
## marked as duplicate by José Carlos Santos calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 25 at 14:02
• Do we additionally assume that $f$ is continuous? Otherwise the statement seems false to me. If you have continuity, then you can use a density argument. – Bib-lost Sep 25 at 13:57
• If $g(x)$ is a discontinuous solution to Cauchy's Functional Equation then $f(x)=e^{g(\ln x)}$ satisfies your functional equation. – lulu Sep 25 at 14:01