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If x and y are nonzero integers, is x^y < y^x ? 1) x =

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Manager
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If x and y are nonzero integers, is x^y < y^x ? 1) x = [#permalink] New post 31 Dec 2006, 13:33
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If x and y are nonzero integers, is x^y < y^x ?

1) x = y^2
2) y > 2

I dont agree with the official answer. Could someone please explain?
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 [#permalink] New post 31 Dec 2006, 13:55
(C) for me. :)

From 1

o If y=-3, then x = 9 and 9^-3 = 1/729 > (-3)^9 = -(3)^9
o If y=3, then x=9 and 9^3 = 279 < 3^9 = 9*9*9*9*3

INSUFF.

From 2
y > 2
We know nothing about x.

To be sure:
o x=1 ; y=3 >>> 1^3 = 1 < 3^1 = 4
o x=-1 ; y=4 >>> (-1)^4 = 1 > 4^-(1) = 1/4

INSUFF.

Both (1) & (2):
We are in similar cases of the y=3 and x=9.

SUFF.

Other way (Out of scope)
x^y < y^x
<=> y*ln(x) < x*ln(y) as x > 0 and y > 0 as well as ln() is a incresing function when x or y inscreases.
<=> y/x < ln(y)/ln(x)
<=> y/y^2 < ln(y)/ln(y^2) as x=y^2
<=> 1/y < 1/2*ln(y)/ln(y)
<=> 1/y < 1/2
<=> y > 2... the condition of 2 ;) :)
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 [#permalink] New post 31 Dec 2006, 19:36
Brilliant explanation! You're a champ!

Thanks very much Fig - :-D
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 [#permalink] New post 01 Jan 2007, 01:35
DreamMBA wrote:
Brilliant explanation! You're a champ!

Thanks very much Fig - :-D


Perhaps not a champ :p

U are welcome and happy New Year :)
  [#permalink] 01 Jan 2007, 01:35
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