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# If x and y are nonzero integers, is x^y<y^x?

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If x and y are nonzero integers, is x^y<y^x? [#permalink]

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18 Jun 2011, 09:20
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If x and y are nonzero integers, is x^y<y^x?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-nonzero-integers-is-x-y-y-x-168279.html
[Reveal] Spoiler: OA

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Re: If x and y are nonzero integers, is x^y<y^x? [#permalink]

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18 Jun 2011, 09:31
1
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Baten80 wrote:
If x and y are nonzero integers, is x^y<y^x?
(1) x = y^2
(2) y>2

St1. Let Y = 1 X = 1
1^1 is not less than 1^1................. insufficient ( even 2..2^4 is not less than 4^2)

St2: nothing abt X ... insuffcient

1&2 we know Y >2 and integer
Y = 3 X = 9
9^3< 3^9... true

sufficient hence C
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Re: If x and y are nonzero integers, is x^y<y^x? [#permalink]

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18 Jun 2011, 10:17
Baten80 wrote:
If x and y are nonzero integers, is x^y<y^x?
(1) x = y^2
(2) y>2

(1) x=y^2 ..... basically translates the original equation to....

(y.y)^y < y^(y.y)

if y = 1, is x^y<y^x? No!
if y = 3, is x^y<y^x? Yes!

Hence insufficient.

(2) y>2 ...... lets say y = 3....translates the original equation to....

x^3 < 3^x

if x = 1, is x^3 < 3^x ? No!
if x = 4, is x^3 < 3^x ? Yes!

Hence insufficient.

Together:

we know the original translates into (y.y)^y < y^(y.y)

and based on (2), the minimum value for y = 3....so lets plug it in

(3.3)^3 < 3^(3.3) ? <<<<< - will be the same regardless of what value you plugin for y as long as its > 2

9^3 < 3^9 .... without doing this long calculated simply simplify the exponents

3^2(3) < 3^3

eliminate the base 6 < 9 ??? YES!!!!

Hence (C)
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Last edited by 386390 on 18 Jun 2011, 10:40, edited 2 times in total.
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Re: If x and y are nonzero integers, is x^y<y^x? [#permalink]

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18 Jun 2011, 10:25
1
KUDOS
x^y<y^x
1)
x=y^2
so put in this x^y<y^x
obtained
y^2y<y^(y^2)
or simplified
2y<y^2

not you can check
y=1, 2<1 - incorrect
y=2, 4<4 - incorrect
y=3, 6<9 - correct

so insufficient

2)
y>2

insufficient because no information about X

1)+2)

y=3, 6<9 - correct

Sufficient
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Re: If x and y are nonzero integers, is x^y<y^x? [#permalink]

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18 Jun 2011, 10:34
sudhir18n wrote:
Baten80 wrote:
If x and y are nonzero integers, is x^y<y^x?
(1) x = y^2
(2) y>2

St1. Let Y = 1 X = 1
1^1 is not less than 1^1................. insufficient ( even 2..2^4 is not less than 4^2)

St2: nothing abt X ... insuffcient

1&2 we know Y >2 and integer
Y = 3 X = 9
9^3< 3^9... true

sufficient hence C

See you missed y is equal to could be 3 and thus x = 9 for option 1.
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Re: If x and y are nonzero integers, is x^y<y^x? [#permalink]

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20 Apr 2016, 13:27
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Re: If x and y are nonzero integers, is x^y<y^x? [#permalink]

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21 Apr 2016, 00:00
Baten80 wrote:
If x and y are nonzero integers, is x^y<y^x?

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

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

(1) x = y^2 --> if $$x=y=1$$, then $$x^y=1=y^x$$, so the answer would be NO BUT if $$y=3$$ and $$x=9$$, then $$x^y=9^3<y^x=3^9$$, so the answer would be YES. Not sufficient.

(2) y > 2. No info about $$x$$, not sufficient.

(1)+(2) From (1) $$x = y^2$$, thus the question becomes: is $$(y^2)^y<y^{(y^2)}$$? --> is $$y^{2y}<y^{(y^2)}$$? Now, since from (2) $$y=integer>2$$, then $$2y$$ will always be less than $$y^2$$, therefore $$y^{2y}$$ will be less than $$y^{(y^2)}$$. Sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-nonzero-integers-is-x-y-y-x-168279.html
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Re: If x and y are nonzero integers, is x^y<y^x?   [#permalink] 21 Apr 2016, 00:00
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