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Updated on: 04 Jul 2019, 01:08
Originally posted by gettinit on 04 Oct 2010, 19:14.
Last edited by Sajjad1994 on 04 Jul 2019, 01:08, edited 1 time in total.
Last edited by Sajjad1994 on 04 Jul 2019, 01:08, edited 1 time in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
50% (02:00) correct
50%
(02:04)
wrong
based on 681
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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
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
Difficulty Level: 700
05 Oct 2010, 03:04
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gettinit
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.
General Discussion
04 Oct 2010, 23:39
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Not sure if this question is copied correctly, almost looks like it should be "which is not always true"
