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"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?

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

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

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  • $\begingroup$ I never said why $(-2)^0=0 I said why this not 1 $\endgroup$ – Time rub Feb 13 '18 at 14:19
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    $\begingroup$ What I meant was there's no difference between $(-n)^{even}$ and $n^{even}$, so their values should be the same. $\endgroup$ – Mehrdad Zandigohar Feb 13 '18 at 14:25
  • $\begingroup$ please see magdiragdag's comment in the bottom $\endgroup$ – Time rub Feb 13 '18 at 14:34
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    $\begingroup$ 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 $\endgroup$ – Mehrdad Zandigohar Feb 13 '18 at 14:38
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    $\begingroup$ 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 $\endgroup$ – Mehrdad Zandigohar Feb 13 '18 at 14:49
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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

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

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

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

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

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

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