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Is y^x > x^y?

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Math Expert
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Is y^x > x^y?  [#permalink]

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New post 10 Sep 2017, 21:42
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

60% (01:15) correct 40% (01:26) wrong based on 88 sessions

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Is y^x > x^y?  [#permalink]

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New post 10 Sep 2017, 23:14
Is y^x > x^y?

(1) y > x
if y=2 x=1 => 2^1 > 1^ 2 => 2>1 true
y=2 x= -2 => 2^(-2) > (-2)^2 => 1/4 > 2 False
Not sufficient
(2) x is a positive number
so x>0 . Nothing about y.
if y=2 x=1 => 2^1 > 1^ 2 => 2>1 true
if y =-2 x=1 => (-2)^1 > (1)^(-2) => -2 > 1 false
So Not sufficient

1+2
y> x and x>0
( Its not given that numbers are integers so always check for fractions. If given x and y are integers equation might be different)
if y=2 x=1 => 2^1 > 1^ 2 => 2>1 true
y= 1/2 and x=1/4 => (1/2)^(1/4) > (1/4)^(1/2) => (1/2)^(1/4) > 1/2 True ( Well this step not required)
y =5 and x=3 => 5^ 3 > 3^5 => 125 > 243 False
Not Sufficient


Answer: E
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Is y^x > x^y?  [#permalink]

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New post 11 Sep 2017, 02:21
Bunuel wrote:
Is y^x > x^y?

(1) y > x
(2) x is a positive number


(1) y > x

Let y=2 & x =1.........2^1 > 1^2...........Answer is yes

Let y=4 & x =3.........4^3 > 3^4...........Answer is NO

Insufficient

(2) x is a positive number

Same examples above

Insufficient

Combine 1 & 2

Same examples above

Insufficient

Answer: E
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Re: Is y^x > x^y?  [#permalink]

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New post 25 Jul 2018, 07:53
Bunuel wrote:
Is y^x > x^y?

(1) y > x
(2) x is a positive number


Hi Bunuel. Is there a conceptual approach to solving this problem?
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Re: Is y^x > x^y?  [#permalink]

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New post 25 Jul 2018, 10:34
agarwalaayush2007 wrote:
Bunuel wrote:
Is y^x > x^y?

(1) y > x
(2) x is a positive number


Hi Bunuel. Is there a conceptual approach to solving this problem?


Hello

I personally dont think there is any conceptual approach, we have to try various cases and see for ourselves. But yes, if we do try and test various possible cases for x and y (and check in which cases x^y is greater, & in which cases y^x is greater), we can definitely use that knowledge in other questions. That knowledge could be useful.

But since you asked, there is a small related concept which comes to mind, and I will share it here, it might be useful some day who knows:

If both x and y are positive integers greater than 3, and if x < y, then:
x^y is always greater than y^x
(smaller number raised to a bigger power will be greater than the bigger number raised to smaller power)

If we try to apply this concept in this particular question, we can see that even after combining the two statements, if x=4 & y=5; then x^y > y^x
But we can check that if x=2 & y=3; then x^y < y^x (thats why the concept I gave is applicable for numbers greater than 3).
So in this question, the data is not sufficient.
Re: Is y^x > x^y? &nbs [#permalink] 25 Jul 2018, 10:34
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