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?

It is true for every number, except for $0$ itself where there is no choice more natural than another. — Arnaud Mortier, Feb 13 '18 at 14:02
Ask your friend why he thinks so... And he will finally conclude $(-2)^0=1$ — Mehrdad Zandigohar, 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. — Magdiragdag, Feb 13 '18 at 14:23

Mehrdad Zandigohar · Accepted Answer · 2018-02-13T14:47:01.790

2

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

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

edited Feb 13 '18 at 14:47

answered Feb 13 '18 at 14:10

Mehrdad Zandigohar

1,638

I never said why $(-2)^0=0 I said why this not 1 – Time rub Feb 13 '18 at 14:19
1

What I meant was there's no difference between $(-n)^{even}$ and $n^{even}$, so their values should be the same. – Mehrdad Zandigohar Feb 13 '18 at 14:25
please see magdiragdag's comment in the bottom – Time rub Feb 13 '18 at 14:34
1

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 – Mehrdad Zandigohar Feb 13 '18 at 14:38
how can be 0<-2 – Time rub Feb 13 '18 at 14:43
We can't tell 0 odd or even as it can be divided by anything and not by itself – Time rub Feb 13 '18 at 14:46
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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: https://en.wikipedia.org/wiki/Parity_of_zero @Timerub – Mehrdad Zandigohar Feb 13 '18 at 14:49
$(-2)^0 > 0$ so it is not $0<-2$ but $0<|-2|$ @Timerub – Mehrdad Zandigohar Feb 13 '18 at 14:50
many many thanks for the answer to you – Time rub Feb 13 '18 at 15:12
@Timerub glad I could help. :) – Mehrdad Zandigohar Feb 13 '18 at 15:38

score 1 · Answer 2 · answered Feb 13 '18 at 14:03

1

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

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

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

4 Answers4