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

"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?

• It is true for every number, except for $0$ itself where there is no choice more natural than another. Feb 13 '18 at 14:02
• @Magdiragdag I think you meant $(-2)^0=1$, instead of $0$.
– user228113
Feb 13 '18 at 14:05
• Yes I really meant that @G. Sassatelli and Magdiragdag Feb 13 '18 at 14:07
• Ask your friend why he thinks so... And he will finally conclude $(-2)^0=1$ Feb 13 '18 at 14:07
• $\ln(a^b) = b \ln(a)$ only holds for $a > 0$, so the whole reasoning breaks down right at the start. Feb 13 '18 at 14:23

Zero is an even number, and $(anything)^{even}>0$
$\ln({(-2)}^0) =\ln({|-2|}^0) = 0 \ln(|-2|)=0$
• I never said why $(-2)^0=0 I said why this not 1 Feb 13 '18 at 14:19 • What I meant was there's no difference between$(-n)^{even}$and$n^{even}$, so their values should be the same. Feb 13 '18 at 14:25 • please see magdiragdag's comment in the bottom Feb 13 '18 at 14:34 • Yes, I see. I am afraid a minor error he has made and that is: the negative number has an even power, so$a>0$naturally and it satisfies the conditions for taking an$ln$from it. Please refer to the first line of my answer. @Timerub Feb 13 '18 at 14:38 • It is even because one way to prove that is you can rewrite it like$2k$and not$2k+1$. You can refer to this article for full details: en.wikipedia.org/wiki/Parity_of_zero @Timerub Feb 13 '18 at 14:49 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 $$(-2)^3=(-2)\cdot (-2)\cdot (-2)=-8$$ $$(-2)^2=(-2)\cdot (-2)=4$$ $$(-2)^1=(-2)=-2$$ $$(-2)^0=?$$ In each step you divide by$-2$to get to the next step. What do you think$(-2)^0$should be?$(B)^0 = (B)^{a-a}= B^a / B^a= 1(-B)^{a-a} = -B^a / -B^a = 1 \$