if x and y are non zero integers is x^y < y^x?
1. x=y^2
2. y >2
the explanation in the OG is not very clear. Can someone please explain. thanks.
x^y < y^x?
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- Rich@VeritasPrep
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1. x = y^2
Substitute in the prompt:
x^y < y^x ?
(y^2)^y < y ^ (y^2) ?
y^(2y) < y^(y^2) ?
If y=2, then the answer is NO, because the two terms are equal. ( 2^(2*2) = 2^(2^2) )
If y=3, then the answer is YES, because 3^(2*3) < 3^(3^2) or 3^6 < 3^9.
INSUFFICIENT
2. y>2
If y is a positive odd number and x is negative, then x^y will be negative and y^x will be positive, and thus x^y < y^x, leading to a YES.
But if x=2 and y=4, then x^y = y^x, and the answer is NO.
INSUFFICIENT
Together, 1 and 2 tell us that y>2 and x=y^2.
Go back to the expression we derived from Statement 1: y^(2y) < y^(y^2) ?
If y=3, then we have 3^6 < 3^9, which is true.
If y=4, then we have 4^8 < 4^16, which is true.
If y=5, then we have 5^10 < 5^25, which is true.
Notice that as y gets bigger, the distance between the two terms being compared grows larger and larger and larger. Therefore, no matter what value of y we choose, the inequality will always be true.
SUFFICIENT
Answer: C
Substitute in the prompt:
x^y < y^x ?
(y^2)^y < y ^ (y^2) ?
y^(2y) < y^(y^2) ?
If y=2, then the answer is NO, because the two terms are equal. ( 2^(2*2) = 2^(2^2) )
If y=3, then the answer is YES, because 3^(2*3) < 3^(3^2) or 3^6 < 3^9.
INSUFFICIENT
2. y>2
If y is a positive odd number and x is negative, then x^y will be negative and y^x will be positive, and thus x^y < y^x, leading to a YES.
But if x=2 and y=4, then x^y = y^x, and the answer is NO.
INSUFFICIENT
Together, 1 and 2 tell us that y>2 and x=y^2.
Go back to the expression we derived from Statement 1: y^(2y) < y^(y^2) ?
If y=3, then we have 3^6 < 3^9, which is true.
If y=4, then we have 4^8 < 4^16, which is true.
If y=5, then we have 5^10 < 5^25, which is true.
Notice that as y gets bigger, the distance between the two terms being compared grows larger and larger and larger. Therefore, no matter what value of y we choose, the inequality will always be true.
SUFFICIENT
Answer: C
Rich Zwelling
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GMAT Instructor, Veritas Prep
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@Rich, by completing st(1) we may also get to the condition LHS{2y} ~ {y^2}RHS, deduce y>0 and stop here with assurance. You still suggest plugging in the numbers?
further st(2) Is not Sufficient, BUT combined st(1&2) we get y>0 and y>2 Sufficient.
further st(2) Is not Sufficient, BUT combined st(1&2) we get y>0 and y>2 Sufficient.
raz1024 wrote:1. x = y^2
Substitute in the prompt:
x^y < y^x ?
(y^2)^y < y ^ (y^2) ?
y^(2y) < y^(y^2) ?
If y=2, then the answer is NO, because the two terms are equal. ( 2^(2*2) = 2^(2^2) )
If y=3, then the answer is YES, because 3^(2*3) < 3^(3^2) or 3^6 < 3^9.
INSUFFICIENT
2. y>2
If y is a positive odd number and x is negative, then x^y will be negative and y^x will be positive, and thus x^y < y^x, leading to a YES.
But if x=2 and y=4, then x^y = y^x, and the answer is NO.
INSUFFICIENT
Together, 1 and 2 tell us that y>2 and x=y^2.
Go back to the expression we derived from Statement 1: y^(2y) < y^(y^2) ?
If y=3, then we have 3^6 < 3^9, which is true.
If y=4, then we have 4^8 < 4^16, which is true.
If y=5, then we have 5^10 < 5^25, which is true.
Notice that as y gets bigger, the distance between the two terms being compared grows larger and larger and larger. Therefore, no matter what value of y we choose, the inequality will always be true.
SUFFICIENT
Answer: C
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