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Better approach for such questions

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Better approach for such questions [#permalink] New post 18 Aug 2011, 14:25
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] New post 18 Aug 2011, 22:08
Expert's post
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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Looking for modern lighting........ [#permalink] New post 19 Aug 2011, 08:34
Our modern furniture line consists of chairs, sofas, loveseats, loungers, tables and accessories for individuals, design professionals and corporations of all sizes. We are commited on bringing you tremendous value into the marketplace.

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Looking for modern lighting........   [#permalink] 19 Aug 2011, 08:34
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