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If x and y are nonzero integers, is x^y < y^x? (1) x =
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27 Oct 2005, 16:28
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If x and y are nonzero integers, is x^y < y^x?
(1) x = y^2
(2) y > 2
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Re: Exponents/inequalities problem from QR 2nd DS 121
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04 Aug 2010, 06:52
jpr200012 wrote: If x and y are nonzero integers, is \(x^y < y^x\)?
(1) \(x = y^2\) (2) \(y > 2\) If x and y are nonzero integers, is \(x^y < y^x\)?(1) \(x = y^2\) > if \(x=y=1\), then \(x^y=1=y^x\), so the answer would be NO BUT if \(y=3\) and \(x=9\), then \(x^y=9^3<y^x=3^9\), so the answer would be YES. Not sufficient. (2) \(y>2\). No info about \(x\), not sufficient. (1)+(2) From (1) \(x = y^2\), thus the question becomes: is \((y^2)^y<y^{(y^2)}\)? > is \(y^{2y}<y^{(y^2)}\)? Now, since from (2) \(y=integer>2\), then \(2y\) will always be less than \(y^2\), therefore \(y^{2y}\) will be less than \(y^{(y^2)}\). Sufficient. Answer: C. jpr200012 wrote: This is a tricky question. I think it relies on you misapplying the rule: \((x^a)^b = x^{ab}\). Is this only valid if a and b are constants?
Example: (1) \(x = y^2\);
\(x^y < y^x\) therefore, \((y^2)^y < y^{y^2}\). How do you simplify this? The guide shows to \(y^{2y} < y^{y^2}\). The left hand side makes sense to me.
Why would \(y^{y^2}\) not simplify to \(y^{2y}\) also? Plugging in numbers, it makes sense. I just want to understand the concept. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\). So: \((a^m)^n=a^{mn}\); \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\). Hope it's clear.
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I would pick A
x = y^2
so y^2^y = y^y^2 which i think is
y^2y = y^2y
so they are equal and you can answer the question



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Joined: 05 Oct 2005
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i think its C
y^2y < y^y2
2y < y2
1 is Insuff b/c if y<=2, the answer is no, whereas when y>2, the answer is yes.
2) alone is insuff b/c it doesn't say anything about x (we need to know whether it is greater or less than 2.
Together, they are SUFF.
C



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A
coz 1 ^2 < 1 ^2 .. No it is not.. False and it stands false for anyother value too.



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Quote: coz 1 ^2 < 1 ^2 .. No it is not.. False and it stands false for anyother value too.
I dont think its A, here is my reasoning.
Q: is x^y < y^x?
(1) x = y^2
if x is 4 and y is 2, statement 1 is correct but look at what happens.
4^2 < 2^4
statement is now TRUE
if x is 4 and y is 2
4^2 < 2^4
statement is not true  they are equal.



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Valley,
I agreed with C, knew it couldnt be A because of 2 and 4. Would be nice to see the OE though just to see whether we all thought the right way.



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xennie  here is the OE...
It is helpful to note that (x^r)^s = x^(rs)
(1) From this, x = y^2, so by substitution then x^y = (y^2)^y or y^(2y), and y^x = y^(y^2). Comparing x^y to y^x can then be done by comparing y^(2y) to y^(y^2), or simply comparing the exponents 2y to y^2. If, for example, y = 2, then 2y = 4 and y^2 = 4, and then x^y would equal y^x. If, however, y = 3, then 2y = 6 and y^2 = 9, and so x^y would be less than y^x; NOT SUFFICIENT.
(2) It is known that y > 2, but no information about x is given; NOT SUFFICIENT.
If both (1) and (2) are taken together, then 2y is compared to y^2 (1) and from (2) it is known that y > 2, so 2y will always be less than y^2. Therefore, x^y < y^x.
<b>The correct answer is C.</b>



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Re: Exponents/inequalities problem from QR 2nd DS 121
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04 Aug 2010, 06:21
This is a tricky question. I think it relies on you misapplying the rule: \((x^a)^b = x^{ab}\). Is this only valid if a and b are constants?
Example: (1) \(x = y^2\);
\(x^y < y^x\) therefore, \((y^2)^y < y^{y^2}\). How do you simplify this? The guide shows to \(y^{2y} < y^{y^2}\). The left hand side makes sense to me.
Why would \(y^{y^2}\) not simplify to \(y^{2y}\) also? Plugging in numbers, it makes sense. I just want to understand the concept.



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Re: Exponents/inequalities problem from QR 2nd DS 121
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05 Aug 2010, 02:25
Posting this msg here even though i sent a private msg to youfor the benefit of others here.
Hi Bunuel, apprecite ur wonderful explanation. I am having trouble in DS question where x & y are termed as nonzero integers.
What is the best way to analyze instances where x & y are are NEGATIVE integers. I see that u have not analyzed this possibility. is there a trick to be sure that this is not needed as u have solved in this case?
Please enlighten. Thanks.



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Re: Exponents/inequalities problem from QR 2nd DS 121
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05 Aug 2010, 03:12
ramanankris wrote: Posting this msg here even though i sent a private msg to youfor the benefit of others here.
Hi Bunuel, apprecite ur wonderful explanation. I am having trouble in DS question where x & y are termed as nonzero integers.
What is the best way to analyze instances where x & y are are NEGATIVE integers. I see that u have not analyzed this possibility. is there a trick to be sure that this is not needed as u have solved in this case?
Please enlighten. Thanks. On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another. For statement (1) I got YES answer and then NO answer with positive numbers, so my goal to prove that this statement was not sufficient was reached, hence there was no need to try negative numbers. Hope it's clear.
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Re: Exponents/inequalities problem from QR 2nd DS 121
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10 Aug 2010, 02:34
Thanks Bunuel, you are an inspiration.



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Re: Is, x^y<y^x?
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25 May 2011, 11:54
The answer is C Statement 1: Insufficient x=y^2 tells us that x is positive, but it tells us nothing about y. For example if y=1 then x=1. Therefore 1^1=1^1 and x^y=y^x If y=2 then x=4. Therefore 4^2<2^4 Since we cannot get a definite yes or no from this statement, it is INSUFFICIENT Statement 2: Insufficient y>2 This tells us nothing about x. If x=1 and y=4, then 1^4>4^1 If x=5 and y=3, then 5^3<3^5 Since we cannot get a definite yes or no from this statement, it is INSUFFICIENT Putting both statements together we know that y>2 and x=y^2 If y=4, then x=16, then 16^4<4^16 (16^4 = 4^8). No matter which integers you choose x^y < y^x, so Statements 1 and 2 together are SUFFICIENT. The answer is C.
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Re: Is, x^y<y^x?
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26 May 2011, 01:03
a check for y = +1 and x = 1. giving different values for the equation. Hence not sufficient. b check for y = 3 and x = 1  9 giving different values for the equation. Not sufficient. a+b y=3, x= 9 y=4 x = 16 give same value meaning LHS = RHS in fact. hence C.
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Re: Is, x^y<y^x?
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26 May 2011, 01:04
if x and y are nonzero integers is, x^y < y^x? (1) x = y^2 (2) y > 2 x^y < y^x Stmt1: x=y^2 \(x^y > y^2^y= y^(2y)\) \(y^x > y^(y^2)\) Is y^(2y) < y^(y^2) ? Take log both side 2y logy < y^2logy ? Canceling log y Is 2y < y^2 ? Is 2<y ? i.e Is y>2? We don't know the value of y. Hence not sufficient. Stmt2: y>2 Not sufficient. Combining, from Stmt2: we know that y>2 . Hence Stmt1: Is y>2 can be answered taking Stmt1 and Stmt2 together. OA C.
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Re: Is, x^y<y^x?
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26 May 2011, 04:10
jamifahad wrote: if x and y are nonzero integers is, x^y < y^x? (1) x = y^2 (2) y > 2
x^y < y^x Stmt1: x=y^2 \(x^y > y^2^y= y^(2y)\) \(y^x > y^(y^2)\) Is y^(2y) < y^(y^2) ? Take log both side 2y logy < y^2logy ? Canceling log y Is 2y < y^2 ? Is 2<y ? i.e Is y>2? We don't know the value of y. Hence not sufficient.
Stmt2: y>2 Not sufficient.
Combining, from Stmt2: we know that y>2 . Hence Stmt1: Is y>2 can be answered taking Stmt1 and Stmt2 together.
OA C. Nice solve. However, GMAT Math does not require the knowledge of logarithms. Definitely can help on the test, but for people who haven't touched a logarithm since high school it's not necessary to relearn them to answer this question.
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Re: Is, x^y<y^x?
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26 May 2011, 04:43
jamifahad wrote: Is 2y < y^2 ? Is 2<y ? i.e Is y>2?
The rephrase is not complete. \(2y<y^2\) \(y^22y>0\) \(y(y2)>0\) ************************ If y>0; y2>0; Is y>2 OR y<0 *********************** However, "y<0" actually doesn't hold true for x^y<y^x (for y=1) ********************* The only point I am trying to make here is that solving through logarithm may give us undesired results. What if x^y or y^x is negative. Then, taking the logarithms would be wrong!!! *******************************************
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Re: Is, x^y<y^x?
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27 May 2011, 02:21
On another look, From stmt1: \(x=y^2\), so\(y=\sqrt{x}\). Since \(\sqrt{x}\)will always be a positive number, y will always be positive. Substituting, in \(x^y < y^x\), \(x^sqrt(x)\) < \(\sqrt{x} ^ x\) Now we can safely take log as both sides are positive. \(sqrt(x)logx < xlog sqrt(x)\) \(sqrt(x) logx < x/2 log x\) Is \(sqrt(x) < x/2\) ? Is \(2 < sqrt(x)\) ? Cannot be determined. Not sufficient. Stmt2: y>2. Not sufficient. Combining, from stmt1 \(y=\sqrt{x}\) From stmt2: y>2. i.e \(\sqrt{x}\) >2. Hence \(2 < sqrt(x)\) ? . Yes. OA C.
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Re: Is, x^y<y^x?
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27 May 2011, 04:21
jamifahad wrote: On another look, From stmt1: \(x=y^2\), so\(y=\sqrt{x}\). Since \(\sqrt{x}\)will always be a positive number, y will always be positive. Substituting, in \(x^y < y^x\), \(x^sqrt(x)\) < \(\sqrt{x} ^ x\) Now we can safely take log as both sides are positive. \(sqrt(x)logx < xlog sqrt(x)\) \(sqrt(x) logx < x/2 log x\) Is \(sqrt(x) < x/2\) ? Is \(2 < sqrt(x)\) ? Cannot be determined. Not sufficient.
Stmt2: y>2. Not sufficient.
Combining, from stmt1 \(y=\sqrt{x}\) From stmt2: y>2. i.e \(\sqrt{x}\) >2. Hence \(2 < sqrt(x)\) ? . Yes.
OA C. x=y^2, Just because x is positive doesn't mean y has to be positive. Even powers always yield a positive number, even if the base is negative. For example, if x=4 then y can be 2 or 2. The only thing we can determine from statement 1 is that x is positive. y can be either positive or negative. If y is negative then x^y would be 1/x^y. 1/x^y may or may not be greater than y^x. y=3, x=9 vs y=3, x=9. 1/9^3>3^9. 9^3<3^9.
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Re: If x and y are nonzero integers, is x^y < y^x? (1) x =
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19 Dec 2015, 00:09
Bunuel Would this approach be correct? statement 1x = y^2 y^2 is positive. So x is positive. But we aren't sure about the magnitude of y. So Statement 1 is insufficient. Statement 2: Insufficient since no relation between x and y Combining the two, statement two assures that y is positive. Hence, C
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Re: If x and y are nonzero integers, is x^y < y^x? (1) x = &nbs
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