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For all real numbers a and b, where a ⋅ b =/ 0, let a◊b = ab

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For all real numbers a and b, where a ⋅ b =/ 0, let a◊b = ab  [#permalink]

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New post Updated on: 04 Jul 2019, 01:08
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Difficulty:

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Question Stats:

52% (01:58) correct 48% (02:02) wrong based on 330 sessions

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For all real numbers \(a\) and \(b\), where \(ab\neq{0}\), let \(a@b=a^b\). Then which of the following MUST be true?

I. \(a@b=b@a\)

II. \((-a)@(-a) =\frac{(-1)^{-a}}{a^a}\)

III. \((a@b)@c=a@(b@c)\)

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only

Source: Nova GMAT
Difficulty Level: 700

Originally posted by gettinit on 04 Oct 2010, 19:14.
Last edited by SajjadAhmad on 04 Jul 2019, 01:08, edited 1 time in total.
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Re: For all real numbers a and b, where a ⋅ b =/ 0, let a◊b = ab  [#permalink]

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New post 05 Oct 2010, 03:04
3
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gettinit wrote:
For all real numbers a and b, where a ⋅ b =/ 0, let a◊b = ab . Then which of the
following must be true?
I. a◊b = b◊a
II. (−a)◊(−a)= (−1)^−a / a^a
III. ( a◊b)◊c = a◊(b◊c)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only


Please explain the math in II in detail. Thank you.


The question is as follows:

For all real numbers \(a\) and \(b\), where \(ab\neq{0}\), let \(a@b=a^b\). Then which of the following MUST be true?

I. \(a@b=b@a\)

II. \((-a)@(-a) =\frac{(-1)^{-a}}{a^a}\)

III. \((a@b)@c=a@(b@c)\)

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only


I. \(a@b=b@a\) --> \(a@b=a^b\) and \(b@a=b^a\), these 2 expressions are not always equal: \(2^3\neq{3^2}\);

II. \((-a)@(-a)=\frac{(-1)^{-a}}{a^a}\) --> \((-a)@(-a)=(-a)^{-a}=(-1*a)^{-a}=-1^{-a}*a^{-a}=\frac{-1^{-a}}{a^a}\) --> \(\frac{-1^{-a}}{a^a}=\frac{-1^{-a}}{a^a}\), so this statement is always true;

III. \((a@b)@c=a@(b@c)\) --> \((a@b)@c=(a^b)^c=a^{bc}\) and \(a@(b@c)=a^{(b^c)}=a^{b^c}\) these 2 expressions are not always equal: \(2^{2*3}=2^6=64\neq2^{2^3}=2^8=256\).

Answer: B (II only).

Notes for III:

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

So:
\((a^m)^n=a^{mn}\);

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\).


Hope it helps.
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Re: For all real numbers a and b, where a ⋅ b =/ 0, let a◊b = ab  [#permalink]

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New post 04 Oct 2010, 23:39
Not sure if this question is copied correctly, almost looks like it should be "which is not always true"
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Re: For all real numbers a and b, where a ⋅ b =/ 0, let a◊b = ab  [#permalink]

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New post 06 Oct 2010, 19:02
Thanks Bunuel very informative and helpful. One question as I am new here, how did you get the symbols to show up in the question? I obviously did not know how to do this.

Kudos for you my friend!
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Re: For all real numbers a and b, where a ⋅ b =/ 0, let a◊b = ab  [#permalink]

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New post 07 Oct 2010, 00:48
gettinit wrote:
Thanks Bunuel very informative and helpful. One question as I am new here, how did you get the symbols to show up in the question? I obviously did not know how to do this.

Kudos for you my friend!


http://gmatclub.com/forum/writing-mathematical-symbols-in-posts-72468.html#p536379
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Re: For all real numbers a and b, where a ⋅ b =/ 0, let a◊b = ab  [#permalink]

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New post 10 Feb 2014, 23:43
The question is written wrong. Can this be written correctly please !
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Re: For all real numbers a and b, where a ⋅ b =/ 0, let a◊b = ab  [#permalink]

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New post 10 Feb 2014, 23:48
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Re: For all real numbers a and b, where a ⋅ b =/ 0, let a◊b = ab  [#permalink]

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New post 15 Sep 2019, 05:13
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: For all real numbers a and b, where a ⋅ b =/ 0, let a◊b = ab   [#permalink] 15 Sep 2019, 05:13
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