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# If n is an integer and x^n x^(-n) = 0, what is the value of

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If n is an integer and x^n x^(-n) = 0, what is the value of [#permalink]  13 Dec 2008, 22:54
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If n is an integer and x^n – x^(-n) = 0, what is the value of x ?
(1) x is an integer.
(2) n ≠ 0
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Joined: 29 Aug 2007
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Re: DS - Integer [#permalink]  13 Dec 2008, 23:06
vivektripathi wrote:
If n is an integer and x^n – x^(-n) = 0, what is the value of x ?

(1) x is an integer.
(2) n ≠ 0

tough...........

x^n – x^(-n) = 0
x^n – 1/x^n = 0
x^2n – 1= 0
x^2n = 1

1: x could have any value depending on n, whose value could also be anything.
2: if n is not equal to 0, n could be 1 or something else.
When n = 1, x could be 1 or -1. still insuff.

togather also repets st 2. so not suff.
E.
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Re: DS - Integer [#permalink]  13 Dec 2008, 23:18
E.

1)if n =0 , x can be any number (not suff)
2) pick no. (n,x) = (-1,1) and (-1,-1) both value are true (not suff)
(1) and (2) together are not suff as well
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Re: DS - Integer [#permalink]  14 Dec 2008, 05:40
x^n - x^-n = 0
=> x^2n = 1

put log on both sides
2n logx = log1 => 2n logx = 0

Now by 2nd statment n # 0, that means logX shd be zero.
so X = 1.

Hence 2nd statement alone is suffiecent .

Re: DS - Integer   [#permalink] 14 Dec 2008, 05:40
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