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# If n is a positive integer and x does not equal zero, is x^n

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Intern
Joined: 19 Jun 2009
Posts: 28

Kudos [?]: 122 [0], given: 1

If n is a positive integer and x does not equal zero, is x^n [#permalink]

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28 Jul 2009, 10:15
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0% (00:00) correct 0% (00:00) wrong based on 3 sessions

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If n is a positive integer and x does not equal zero, is x^n > x^(n+1)?

1) x < 1

2) n is even.

Kudos [?]: 122 [0], given: 1

Manager
Joined: 16 Apr 2009
Posts: 231

Kudos [?]: 101 [0], given: 10

Schools: Ross

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28 Jul 2009, 10:28
Quote:
If n is a positive integer and x does not equal zero, is x^n > x^(n+1)?

1) x < 1

2) n is even.

is x^n > x^(n+1)?

1) x < 1

x can be between 0 and 1 or x can be negative
If x is b/w 0 and 1, then x^n > x^(n+1)
if x is negative, x^n > x^(n+1) - here it depends on n
not suff

2) n is even

not suff because no information on x

Together

If x is b/w 0 and 1 and n is even then x^n > x^(n+1)
ex 0.1 -->x^2=0.01
0.1 -->x^2+1=x^3=0.001
0.01>0.001

if x is negative and n is even then x^n > x^(n+1)
ex x=-2-->x^2=4
-2 -->x^2+1=x^3=-8

Hence C
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Kudos [?]: 101 [0], given: 10

SVP
Joined: 05 Jul 2006
Posts: 1750

Kudos [?]: 432 [0], given: 49

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28 Jul 2009, 11:41
amitgovin wrote:
If n is a positive integer and x does not equal zero, is x^n > x^(n+1)?

1) x < 1

2) n is even.

Is??
x^(n)-x^(n+1)>0
ie: is??
x^n(1 - x)>0 ie is both x^n and (1-x) have the same sign?? only possibloe if we have 2 conditions

1>x , n is even....................

obviously each alone is not suff

both are

C

Kudos [?]: 432 [0], given: 49

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