# What does $(-2)^x$ really mean?

I think we can all agree that $(-2)^{-1}=-1/2,(-2)^0=1,(-2)^1=-2,(-2)^2=4$
But what does the function $f=(-2)^x$ really mean? It is defined on the integers based on how most people understand exponents, but on the real numbers it's not so easy...

• It means whatever you define it to mean. There are a number of approaches to defining it, or you can leave it undefined. If $x,y$ are complex, then $x^y$ can take infinitely many values, in general, and you often have to pick one. – Thomas Andrews Mar 31 '16 at 19:19
• if $x$ is a real number (not integer) such a term doesn't exist, example $$(-2)^{\sqrt{2}}$$ – Dr. Sonnhard Graubner Mar 31 '16 at 20:35