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

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 Explanation

+1 from my side too

IMO C too..............

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.