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if x and y are non zero integers, is x^y<y^x?

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Source: — Data Sufficiency |
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by rahulg83 » Sun Jun 21, 2009 10:11 pm
Statement 1 : x=y^2

therefore we need to find out whether x^y<y^x

or y^(2y)<y^(y^2)

now compare 2y and y^2, we don't know whether y is positive, or negative, or between 0 and 1, any of the above (L.H.S. or R.H.S.) could be greater than other. Insufficient

Statement 2: y>2, but what is x? insufficient

Combining two statements: if y>2, y^2 will surely be greater than 2y. hence our answer is YES, y^(2y)<y^(y^2)

C

HTH!!
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by nitya34 » Mon Jun 22, 2009 4:08 am
Good Explanation

+1 from my side too
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by abhinav85 » Mon Jun 22, 2009 6:21 am
IMO C too..............
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by atulkumar79 » Thu Jun 25, 2009 11:39 am
rahulg83 wrote:Statement 1 : x=y^2

therefore we need to find out whether x^y<y^x

or y^(2y)<y^(y^2)

now compare 2y and y^2, we don't know whether y is positive, or negative, or between 0 and 1, any of the above (L.H.S. or R.H.S.) could be greater than other. Insufficient

Statement 2: y>2, but what is x? insufficient

Combining two statements: if y>2, y^2 will surely be greater than 2y. hence our answer is YES, y^(2y)<y^(y^2)

C

HTH!!
good explaination. Just a minor correction to the above explaination, we are given x and y are integers.
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