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If x is an integer, is xx<2^x ?
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18 Dec 2012, 07:31
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If x is an integer, is xx < \(2^x\) ? (1) x < 0 (2) x = 10 Edit: Formating
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18 Dec 2012, 07:35
Walkabout wrote: If x is an integer, is xx<2^x ?
(1) x < 0 (2) x= 10 If x is an integer, is xx<2^x ?Notice that the RHS (right hand side) of the expression is always positive (\(2^x>0\)), but the LHS is positive when \(x>0\) (\(x>0\) > \(x*x=x^2\)), negative when \(x<0\) (\(x<0\) > \(x*x=x^2\)) and equals to zero when \(x={0}\). (1) x < 0. According to the above \(x*x<0<2^x\). Sufficient. (2) x = 10. The same here \(x*x=100<0<\frac{1}{2^{10}}\). Sufficient. Answer: D.
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Re: If x is an integer, is xx<2^x ?
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Updated on: 27 Jul 2015, 05:44
reza52520 wrote: If x is an integer, is xx<2^x ?
(1) x < 0 (2) x = 10 Question : Is xx<2^x ?Statement 1: x < 0For x to be Negative LHS i.e. xx will always be NEGATIVE and 2^x will be positive for any value of x i.e. xx<2^x will always be true SUFFICIENTStatement 1: x = 10For x =10 LHS i.e. xx will always be NEGATIVE (100) and 2^x will be positive for given x (1/2^10) i.e. xx<2^x will always be true SUFFICIENTAnswer: option D
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Re: If x is an integer, is xx<2^x ?
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24 Apr 2016, 14:22
This question can be solved as follows.
stmt1) it says that x /x/ < 2^x. and also we are told that x< 0. So if x is zero and the abs of x is always positive then we know that x/x/ will be negative. In addition to that, we know that 2^negative number will be positive because it will be in the form of 1/2^x, it will be less than 1 but it will be greater than a negative number. So stmt1 is SUFF.
stmt2) this is a repetition of stmt1 because the left side is negative and the right side is positive. SUFF.
Answer is D!



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If x is an integer, is xx<2^x ?
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Updated on: 12 Nov 2019, 17:12
Walkabout wrote: If x is an integer, is xx<2^x ?
(1) x < 0 (2) x = 10 Target question: Is xx< 2^x ? Given: x is an integer Statement 1: x < 0 In other words, x is NEGATIVE So, xx = (NEGATIVE)(NEGATIVE) = (NEGATIVE)(POSITIVE) = NEGATIVE IMPORTANT: 2^x will be POSITIVE for all values of x. Since xx must be NEGATIVE, and since 2^x must be POSITIVE, we can be certain that xx< 2^x Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: x = 10 So, xx = (10)(10) = (10)(10) = 100 = a NEGATIVE On the other hand, 2^x = 2^(10) = 1/(2^10) = some POSITIVE number Since xx is NEGATIVE, and since 2^x must be POSITIVE, we can be certain that xx< 2^x Since we can answer the target question with certainty, statement 2 is SUFFICIENT Answer:
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Re: If x is an integer, is xx<2^x ?
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22 Oct 2017, 15:14
Hi,
For statement 1 I tested values i.e. x = 1 or x = 2.This is more so a question regarding reciprocals and inequalities. If x = 2, then 2 (2) = 2^2. Then, this is equal to 2 (2) = 1/2^2. In the second step where I converted 2^2 to 1/2^2  would I have to also flip the other side to become 1 / 2 (2) or is that wrong?
Thanks,
infinitemac



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Re: If x is an integer, is xx<2^x ?
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23 Oct 2017, 23:31
infinitemac wrote: Hi,
For statement 1 I tested values i.e. x = 1 or x = 2.This is more so a question regarding reciprocals and inequalities. If x = 2, then 2 (2) = 2^2. Then, this is equal to 2 (2) = 1/2^2. In the second step where I converted 2^2 to 1/2^2  would I have to also flip the other side to become 1 / 2 (2) or is that wrong?
Thanks,
infinitemac No. The right hand side is \(2^{(2)}\), which is the same as \(\frac{1}{2^2}\) but the left hand side (2*2) stays the same. Negative powers:\(a^{n}=\frac{1}{a^n}\) Important: you cannot rise 0 to a negative power because you get division by 0, which is NOT allowed. For example, \(0^{1} = \frac{1}{0}=undefined\). 8. Exponents and Roots of Numbers Check below for more: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps.
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Re: If x is an integer, is xx<2^x ?
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28 Oct 2017, 12:31
Walkabout wrote: If x is an integer, is xx<2^x ?
(1) x < 0 (2) x = 10 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Since we have 1 variables and 0 equation, D could be the answer most likely. Condition 1) Since x < 0 and x≥0, xx≤0. 2^x > 0 Thus xx < 2^x. This is sufficient. Condition 2) Since x = 10, xx = (10)*10 = 100 < 0 And 2^(10) = 1/(2^10) > 0 Thus xx < 2^x This is also sufficient. Therefore, D is the answer. For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both con 1) and con 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. D is most likely to be the answer using con 1) and con 2) separately according to DS definition. Obviously, there may be cases where the answer is A, B, C or E.
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Re: If x is an integer, is xx<2^x ?
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05 Dec 2017, 07:16
Bunuel wrote: Walkabout wrote: If x is an integer, is xx<2^x ?
(1) x < 0 (2) x= 10 If x is an integer, is xx<2^x ?Notice that the RHS (right hand side) of the expression is always positive (\(2^x>0\)), but the LHS is positive when \(x>0\) (\(x>0\) > \(x*x=x^2\)), negative when \(x<0\) (\(x<0\) > \(x*x=x^2\)) and equals to zero when \(x={0}\). (1) x < 0. According to the above \(x*x<0<2^x\). Sufficient. (2) x = 10. The same here \(x*x=100<0<\frac{1}{2^{10}}\). Sufficient. Answer: D. what is wrong in my approach : x x < 2^x x *sqrt(x^2) < 2^x square on both sides, x^2 * x^2 < 2^2x x^4 < 2^2x given 1 stmt, x as ve, always x^4 > 2^2x, whereas I know i am making some mistake. are we not allowed to take square on both sides?



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Re: If x is an integer, is xx<2^x ?
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05 Dec 2017, 07:24
Avinash_R1 wrote: Bunuel wrote: Walkabout wrote: If x is an integer, is xx<2^x ?
(1) x < 0 (2) x= 10 If x is an integer, is xx<2^x ?Notice that the RHS (right hand side) of the expression is always positive (\(2^x>0\)), but the LHS is positive when \(x>0\) (\(x>0\) > \(x*x=x^2\)), negative when \(x<0\) (\(x<0\) > \(x*x=x^2\)) and equals to zero when \(x={0}\). (1) x < 0. According to the above \(x*x<0<2^x\). Sufficient. (2) x = 10. The same here \(x*x=100<0<\frac{1}{2^{10}}\). Sufficient. Answer: D. what is wrong in my approach : x x < 2^x x *sqrt(x^2) < 2^x square on both sides, x^2 * x^2 < 2^2x x^4 < 2^2x given 1 stmt, x as ve, always x^4 > 2^2x, whereas I know i am making some mistake. are we not allowed to take square on both sides? We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (the same for taking an even root of both sides of an inequality). Here xx is negative if x is negative, so we cannot square. RAISING INEQUALITIES TO EVEN/ODD POWER1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (the same for taking an even root of both sides of an inequality).For example: \(2<4\) > we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) > we can square both sides and write: \(x^2<y^2\); But if either of side is negative then raising to even power doesn't always work. For example: \(1>2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are nonnegative. 2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example: \(2<1\) > we can raise both sides to third power and write: \(2^3=8<1=1^3\) or \(5<1\) > \(5^3=125<1=1^3\); \(x<y\) > we can raise both sides to third power and write: \(x^3<y^3\). Adding, subtracting, squaring etc.: Manipulating Inequalities. 9. Inequalities For more check Ultimate GMAT Quantitative Megathread
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Re: If x is an integer, is xx<2^x ?
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13 Dec 2017, 20:48
Hi All, We're told that X is an integer. We're asked if XX < 2^X. This is a YES/NO question. We can answer it with a bit of Number Property knowledge. 1) X < 0 With Fact 1, we know that X is NEGATIVE. By definition, that means... XX = (Neg)Neg = Negative 2^(Negative) = Positive Thus, XX will ALWAYS be less than 2^X and the answer to the question is ALWAYS YES. Fact 1 is SUFFICIENT 2) X = 10 With the value of X, we can absolutely answer the question (we would just need to plug in that value: Is (10)10 < 2^(10)? The answer to the question IS yes, but we don't have to actually do that work. There would be just one answer to the question, so it doesn't really matter what that one answer is. Fact 2 is SUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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If x is an integer, is xx<2^x ?
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13 Sep 2019, 22:14
Walkabout wrote: If x is an integer, is xx < \(2^x\) ? (1) x < 0 (2) x = 10 Edit: Formating Given: x is an integer Asked: Is xx < \(2^x\) ? (1) x < 0 x = x x^2 < 2^x x^2 < 0 2^x >0 x^2 < 2^x \(xx < 2^x\) SUFFICIENT (2) x = 10 x = 10 xx = 10 * 10 = 100 \(2^x = 2^{10} = \frac{1}{1024}\) \(100 < \frac{1}{1024}\) \(xx < 2^x\) SUFFICIENT IMO D



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Re: If x is an integer, is xx<2^x ?
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26 Jan 2020, 11:32
(1) x < 0 > x is ve so xx>ve and 2^x >+ve (2) x = 10> will give correct value D



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Re: If x is an integer, is xx<2^x ?
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22 Feb 2020, 03:30
Bunuel wrote: Walkabout wrote: If x is an integer, is xx<2^x ?
(1) x < 0 (2) x= 10 If x is an integer, is xx<2^x ?Notice that the RHS (right hand side) of the expression is always positive (\(2^x>0\)), but the LHS is positive when \(x>0\) (\(x>0\) > \(x*x=x^2\)), negative when \(x<0\) (\(x<0\) > \(x*x=x^2\)) and equals to zero when \(x={0}\). (1) x < 0. According to the above \(x*x<0<2^x\). Sufficient. (2) x = 10. The same here \(x*x=100<0<\frac{1}{2^{10}}\). Sufficient. Answer: D. Hi Bunuel, Please correct me if I am wrong 1) xx<2^x x=1 so, 11<2^1 1<1/2 which will always be true for any negative value of x and also with positive values it will remain true and sufficient 2) condition 2 is already giving fix value so it will be sufficient
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Re: If x is an integer, is xx<2^x ?
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22 Feb 2020, 13:25
Farina wrote: Bunuel wrote: Walkabout wrote: If x is an integer, is xx<2^x ?
(1) x < 0 (2) x= 10 If x is an integer, is xx<2^x ?Notice that the RHS (right hand side) of the expression is always positive (\(2^x>0\)), but the LHS is positive when \(x>0\) (\(x>0\) > \(x*x=x^2\)), negative when \(x<0\) (\(x<0\) > \(x*x=x^2\)) and equals to zero when \(x={0}\). (1) x < 0. According to the above \(x*x<0<2^x\). Sufficient. (2) x = 10. The same here \(x*x=100<0<\frac{1}{2^{10}}\). Sufficient. Answer: D. Hi Bunuel, Please correct me if I am wrong 1) xx<2^x x=1 so, 11<2^1 1<1/2 which will always be true for any negative value of x and also with positive values it will remain true and sufficient 2) condition 2 is already giving fix value so it will be sufficient Hi Farina, Most of what you wrote is correct. This DS question is based on a couple of Number Properties: when X is NEGATIVE.... XX will ALWAYS be NEGATIVE 2^(X) will ALWAYS be POSTIIVE This means that the answer to the question is ALWAYS YES and Fact 1 is SUFFICIENT. However, what you noted about POSITIVE values is not true. For example.... when X = 1... 11 = 1 2^1 = 2 so the answer to the question "is XX < 2^X?".... is YES. when X = 2... 22 = 4 2^2 = 4 so the answer to the question "is XX < 2^X?".... is NO. This means that when X is POSITIVE, the answer to the question changes and this situation would be INSUFFICIENT. GMAT assassins aren't born, they're made, Rich
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Re: If x is an integer, is xx<2^x ?
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22 Feb 2020, 17:47
Walkabout wrote: If x is an integer, is xx < \(2^x\) ? (1) x < 0 (2) x = 10 Edit: Formating #1. Clearly LHS is always ve and RHS is always +ve, so statement is xx < \(2^x\) always true. Sufficient. #2. Similar to #1. LHS is ve while RHS yields a +ve number. Sufficient OA D



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Re: If x is an integer, is xx<2^x ?
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22 Feb 2020, 22:49
EMPOWERgmatRichC Thank you for pointing out my mistake. I am really struggling in inequalities, hopefully ill come out of this phase soon with the help of you experts of Gmat club! Posted from my mobile device
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Re: If x is an integer, is xx<2^x ?
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26 Feb 2020, 19:17
This doesn't hold true when x = 2. The question should be modified as <= right?
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Re: If x is an integer, is xx<2^x ?
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26 Feb 2020, 22:21
Shirisha995 wrote: This doesn't hold true when x = 2. The question should be modified as <= right?
Posted from my mobile device How can x be 2? (1) says x < 0 and (2) says x= 10...
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