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Inequality

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Inequality

by GHong14 » Thu Jan 27, 2011 9:33 pm
x^y < y^x is x < y?

x=y^2

Sufficient or not?
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by Anurag@Gurome » Fri Jan 28, 2011 12:03 am
GHong14 wrote:x^y < y^x is x < y?

x = y²

Sufficient or not?
Take, y = 3 => x = 9 ----> x > y
Take, y = 0.5 => x = 0.25 ----> x < y

NOT sufficient.
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by cyrwr1 » Fri Jan 28, 2011 10:22 am
Can you clarify the question ? thanks
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by roshnipat1610 » Mon Feb 07, 2011 1:07 am
Dont' we have to take into account this condition that x^y < y^x?

Continuing with Anurag's numbers:

If y = 3 , x = 9

(9) ^ 3 < (3) ^ 9

True

If y = 1/2 x = 1/4

(1/4)^ (1/2) < (1/2) ^(1/4)

1/2 < (1/2)^(1/4)

True

Hence this condition is insuff.
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by Night reader » Mon Feb 07, 2011 5:49 am
GHong14 wrote:x^y < y^x is x < y?

x=y^2

Sufficient or not?
statement: x=y^2 --> Sqrt(x)=|y|. Condition for y>0 i) when y is even, x >y; ii) when y is odd, x<y. We are let to decide with more than one statement, hence Not Sufficient.
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by shashank03 » Thu Feb 10, 2011 2:35 am
y^2y<y^x
1<y^(x-2y)
so
x-2y>0
and hence x>2y
which means the condition is sufficient
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by earnest10 » Fri Feb 11, 2011 7:18 am
GHong14 wrote:x^y < y^x is x < y?

x=y^2

Sufficient or not?
since x = y^2

x ^y < y^x will be :

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

so , 2y = y^2 ; 2 < y

since y > 2 ... it confirms that y is +ve and greater than 2.

so it is sufficient.
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by Night reader » Fri Feb 11, 2011 8:49 am
statement: x=y^2 --> Sqrt(x)=|y|.
x=9, y={-3;3} --> x^y < y^x= 9^3 < 3^9 BUT 9^-3 > -3^9, hence not Sufficient
earnest10 wrote:
GHong14 wrote:x^y < y^x is x < y?

x=y^2

Sufficient or not?
since x = y^2

x ^y < y^x will be :

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

so , 2y = [<] y^2; //2y<y^2 --> sqrt(2y)< |y|, it's not correct to say that 2<y, because |y| = {-ve;+ve}, for example y=|3| while 2*3<3*3, 2*(-3)>(-3)*(-3); so it is not always possible for 2 < y// 2 < y

since y > 2 ... it confirms that y is +ve and greater than 2.

so it is sufficient.
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