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# Better approach for such questions

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Intern
Joined: 30 Jun 2011
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Better approach for such questions [#permalink]

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18 Aug 2011, 15:25
Each of the following equations has at least one solution EXCEPT

a) –2^n = (–2)^-n
b) 2^-n = (–2)^n
c) 2^n = (–2)^-n
d) (–2)^n = –2^n
e) (–2)^-n = –2^-n
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Re: Better approach for such questions [#permalink]

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18 Aug 2011, 23:08
sm021984 wrote:
Each of the following equations has at least one solution EXCEPT

a) –2^n = (–2)^-n
b) 2^-n = (–2)^n
c) 2^n = (–2)^-n
d) (–2)^n = –2^n
e) (–2)^-n = –2^-n

The first thing that comes to mind here is that n = 0 and n = 1 will satisfy many of these equations since a^0 = 1 and (-a)^0 = 1. Also, (-a)^1 = -a and -a^1 = -a

n = 0 satisfies options (b) and (c)
n = 1 satisfies options (d) and (e)
Neither satisfies option (a) and hence, by elimination, answer is (a).
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19 Aug 2011, 09:34
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Looking for modern lighting........   [#permalink] 19 Aug 2011, 09:34
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