Is $2^{\log_2(-5)}$ defined?

Question

As far as I know for $\log_2 x$ to be defined $x$ must be higher than 0. However when I enter $2^{\log_2(-5)}$ into wolframalpha it gives result $-5$. Is it mistake?

Answer 1

Yes...and, no.

There are ways to define $\log_2(-5)$. They require knowing something about complex numbers. And if you use one of those ways to define $\log_2(-5)$, then $2^{\log_2(-5)}=-5$.

Now, if WA says $2^{\log_2(0)}=0$, then I'll be worried.

Answer 2

It is not a mistake, though you are right to be worried. What they are using is the complex logarithm. It is possible to define logarithms $\log z$ of any nonzero complex number $z$, just by saying that $\log z$ is a complex number $w$ with the property that $e^{w} = z$. Unfortunately there are infinitely many such solutions, so there are infinitely many logarithms $\log z$ of $z$. By definition, though, they all have the property that $e^{\log z} = z$, which is what wolframalpha gave you.