# Does $\exp(\log x + \log y) = \exp(\log xy)$ hold always for complex $x$ and $y$?

Let $\log z$ be the usual principal branch of the complex logarithm. I know that some of the "laws" of logarithms of real numbers do not hold. For example, $\log x + \log y$ could not equal to $\log x y$. Is this not true if the logarithms are in the position of exponents? What I say is $\exp(\log x + \log y) = \exp(\log xy)$ and $\exp(x \log y)=\exp(\log y^x)$ seems to hold for complex $x$ and $y$. Is it right? Could someone give a counter example or a rigorous proof?

• Yes, both sides are equal to $xy$ (except in cases where $xy=0$ and both sides are undefined). Do you allow both sides undefined when you say "always holds"? – GEdgar Jul 13 '13 at 16:09
• @GEdgar, Yes of course. – asofas Jul 13 '13 at 16:15

$\log x$ has lots of values, but they all differ by multiples of $2\pi i$. Then $$e^{\log x+\log y + 2\pi i} = xy e^{2\pi i} = xy$$ so I think that is true.
However, the various values of $x\log y$ differ by multiples of $2\pi x i$. That means, after exponentiation, the various values differ by ratios of $\exp(2\pi x i)$. So $\exp(x\log y)$ doesn't have a unique value.
On the other hand, $\exp(x\log y)$ could well be the definition of the various values of $y^x$.
If $x$, $y$ and $x+y$ are all in the domain of logarithm, then from the definition we have: $$e^{\log(xy)} = xy$$ so if we had $e^{a+b} = e^a e^b$, then we would have: $$e^{\log x + \log y} = e^{\log x} e^{\log y} = x y$$ so what you should think about is if that equation holds in general or not? and why.