We know that $2^0$=1 but why $(-2)^0$=not 1

Question

"Any number to the power $0$ is $1$"- this is what I am taught. But my friend says that it is not true for negative numbers. Why? Well my friend said if you think $y=(-2)^0$ then $ln(y) = 0*ln(-2)$ then my friend said $ln(-2)$ is not valid. So we do not get a value. In case of $y= 2^0$ ; $lny= 0*ln2$, so $lny = 0$ then $y=1$. And one more question came into my mind that if $y= - 2$ then $lny= ln(-2)$ how is this possible?

Answer 1

Zero is an even number, and $(anything)^{even}>0$

$\ln({(-2)}^0) =\ln({|-2|}^0) = 0 \ln(|-2|)=0$

Answer 2

Note that for $\forall a\neq0$ and $n\in\mathbb{N}$ by definition

$$a^0=a^{n-n}=\frac{a^n}{a^n}=1$$

see also the related MSE OP

Answer 3

$(B)^0 = (B)^{a-a}$ $= B^a / B^a= 1$

$(-B)^{a-a} = -B^a / -B^a = 1 $

I don't know why your friend said that !!

Answer 4

$$(-2)^3=(-2)\cdot (-2)\cdot (-2)=-8$$ $$(-2)^2=(-2)\cdot (-2)=4$$ $$(-2)^1=(-2)=-2$$ $$(-2)^0=?$$

In each step you divide by $-2$ to get to the next step. What do you think $(-2)^0$ should be?