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• ### $450 Tuition Credit & Official CAT Packs FREE December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### FREE Quant Workshop by e-GMAT! December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. # x^n = x^(n+2) for any integer n. Is it true that x > 0?  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Manager Joined: 14 Oct 2014 Posts: 66 Location: United States GMAT 1: 500 Q36 V23 x^n = x^(n+2) for any integer n. Is it true that x > 0? [#permalink] ### Show Tags 21 Dec 2014, 19:41 1 11 00:00 Difficulty: 95% (hard) Question Stats: 46% (01:49) correct 54% (02:09) wrong based on 411 sessions ### HideShow timer Statistics x^n = x^(n+2) for any integer n. Is it true that x > 0? (1) x = x^2 - 2 (2) 2x < x^5 ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 51218 x^n = x^(n+2) for any integer n. Is it true that x > 0? [#permalink] ### Show Tags 21 Jan 2015, 01:59 3 16 x^n = x^(n+2) for any integer n. Is it true that x > 0? For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or -1. (1) x = x^2 - 2 --> x^2 - x - 2 = 0 --> (x - 2)(x + 1) = 0 --> x = 2 or x = -1. Since we know that x can only be 0, 1, or -1, then x = -1 < 0. Sufficient. (2) 2x < x^5 --> from the possible values of x (0, 1, -1), only -1 satisfies 2x < x^5, thus x = -1 < 0. Sufficient. Answer: D. _________________ ##### General Discussion Senior Manager Joined: 13 Jun 2013 Posts: 275 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0? [#permalink] ### Show Tags 22 Dec 2014, 01:42 1 viktorija wrote: x^n=x^(n+2) for any integer n. Is it true that x>0? (1) x=x^2-2 (2)2x<x^5 $$x^n = x^{n+2}$$ st.1 x=x^2-2 now x^2-2-x=0 x^2-2x+x-2=0 x(x-2)+1(x-2)=0 (x-2)(x+1)=0 x=2 or x=-1 out of these values of x, only x=-1 will satisfy the equation $$x^n = x^{n+2}$$. hence we will reject x=2. thus only possible value of x=-1. which answer the original question as false. Thus st.1 alone is sufficient. st.2 x^5>2x x(x^4-2)>0 this equation is satisfied by x=-1, x=2,3,4... now out of these values of x, only x=-1, satisfy the equation $$x^n = x^{n+2}$$. hence x=-1 . which answer the original question as false. thus st.2 alone is sufficient. hence answer must be D EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13087 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0? [#permalink] ### Show Tags 22 Dec 2014, 14:40 1 2 Hi viktorija, manpreetsingh86's explanation is spot on, so I won't rehash that here. The information given in the prompt involves a really specific Number Property. Knowing this Number Property would make solving the rest of the question considerably easier: We're told that x^n=x^(n+2) for ANY integer n. This equation offers a SERIOUS restriction; for it to hold true for ANY integer value of n, the value of X can only be one of 3 things: -1, 0 or 1. Taking THAT knowledge into Fact 1 and Fact 2 would allow you to work through them quickly and avoid any excess math. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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21 Jan 2015, 05:00
2
Here x can only be 0 OR 1 OR -1(if n=even)
Statement 1 tells that x=2 or x=-1.
So x=-1. Sufficient.

Statement 2 can holds only when x=-1 and not when x=0 OR x=1.
So x=-1. Sufficient.

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x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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08 Jan 2016, 00:29
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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10 Jan 2016, 18:16
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.

x^n = x^(n+2) for any integer n. Is it true that x > 0?

(1) x = x^2 -2
(2) 2x < x^5

When you modify the original condition and the question, it becomes x=-1,0,1. There is 1 variable(x), which should match with the number equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely make D the answer.
For 1), from x^2-x-2=0, (x-2)(x+1)=0, x=2,-1, only x=-1 is possible, which is no and sufficien.t
For 2), what satisfies 2x<x^5 is only x=-1, which is no and sufficient. Therefore, the answer is D.

 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 1) and 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. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior Manager Joined: 02 Apr 2014 Posts: 473 GMAT 1: 700 Q50 V34 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0? [#permalink] ### Show Tags 06 Nov 2017, 20:27 Bunuel wrote: x^n = x^(n+2) for any integer n. Is it true that x > 0? For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or -1. (1) x = x^2 - 2 --> x^2 - x - 2 = 0 --> (x - 2)(x + 1) = 0 --> x = 2 or x = -1. Since we know that x can only be 0, 1, or -1, then x = -1 < 0. Sufficient. (2) 2x < x^5 --> from the possible values of x (0, 1, -1), only -1 satisfies 2x < x^5, thus x = -1 < 0. Sufficient. Answer: D. Hi Bunuel I have one question for x must be 0,1,-1 for all integers n , if x = 0, and if n= 0, then 0 ^ 0 = 0 ^ 2 , value of 0 ^ 0 = 0 or 0 ^ 0 = 1 or it is out of scope of gmat? I came across similiar question https://gmatclub.com/forum/what-is-the- ... 51809.html where 0 ^ 0 is out of scope, if this were true, x can't take value of 0 right? only 1 or -1 Please help Thanks Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0? [#permalink] ### Show Tags 06 Nov 2017, 20:53 hellosanthosh2k2 wrote: Bunuel wrote: x^n = x^(n+2) for any integer n. Is it true that x > 0? For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or -1. (1) x = x^2 - 2 --> x^2 - x - 2 = 0 --> (x - 2)(x + 1) = 0 --> x = 2 or x = -1. Since we know that x can only be 0, 1, or -1, then x = -1 < 0. Sufficient. (2) 2x < x^5 --> from the possible values of x (0, 1, -1), only -1 satisfies 2x < x^5, thus x = -1 < 0. Sufficient. Answer: D. Hi Bunuel I have one question for x must be 0,1,-1 for all integers n , if x = 0, and if n= 0, then 0 ^ 0 = 0 ^ 2 , value of 0 ^ 0 = 0 or 0 ^ 0 = 1 or it is out of scope of gmat? I came across similiar question https://gmatclub.com/forum/what-is-the- ... 51809.html where 0 ^ 0 is out of scope, if this were true, x can't take value of 0 right? only 1 or -1 Please help Thanks 0^0, in some sources equals to 1 (not 0), some mathematicians say it's undefined. But you won't need this for the GMAT because the case of 0^0 is not tested on the GMAT. For this problem, x = n = 0 or x = n + 2 = 0 is technically out because we are given that x^n = x^(n+2) for ANY integer n because for these cases x^n or x^(n + 2) will become undefined and could not be equal to other part. Apart from these exception x = 0 is perfectly OK. _________________ Senior Manager Joined: 02 Apr 2014 Posts: 473 GMAT 1: 700 Q50 V34 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0? [#permalink] ### Show Tags 06 Nov 2017, 22:04 Bunuel wrote: hellosanthosh2k2 wrote: Bunuel wrote: x^n = x^(n+2) for any integer n. Is it true that x > 0? For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or -1. (1) x = x^2 - 2 --> x^2 - x - 2 = 0 --> (x - 2)(x + 1) = 0 --> x = 2 or x = -1. Since we know that x can only be 0, 1, or -1, then x = -1 < 0. Sufficient. (2) 2x < x^5 --> from the possible values of x (0, 1, -1), only -1 satisfies 2x < x^5, thus x = -1 < 0. Sufficient. Answer: D. Hi Bunuel I have one question for x must be 0,1,-1 for all integers n , if x = 0, and if n= 0, then 0 ^ 0 = 0 ^ 2 , value of 0 ^ 0 = 0 or 0 ^ 0 = 1 or it is out of scope of gmat? I came across similiar question https://gmatclub.com/forum/what-is-the- ... 51809.html where 0 ^ 0 is out of scope, if this were true, x can't take value of 0 right? only 1 or -1 Please help Thanks 0^0, in some sources equals to 1 (not 0), some mathematicians say it's undefined. But you won't need this for the GMAT because the case of 0^0 is not tested on the GMAT. For this problem, x = n = 0 or x = n + 2 = 0 is technically out because we are given that x^n = x^(n+2) for ANY integer n because for these cases x^n or x^(n + 2) will become undefined and could not be equal to other part. Apart from these exception x = 0 is perfectly OK. Thanks Bunuel for explanation Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6639 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0? [#permalink] ### Show Tags 09 Nov 2017, 07:58 viktorija wrote: x^n = x^(n+2) for any integer n. Is it true that x > 0? (1) x = x^2 - 2 (2) 2x < x^5 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. $$x^n = x^{n+2}$$ $$x^{n+2} - x^n = 0$$ $$x^n(x^2-1) = 0$$ $$x^n(x+1)(x-1) = 0$$ $$x -1$$ or $$x = 0$$ or $$x = 1$$ The original condition states $$x -1$$ or $$x = 0$$ or $$x = 1$$. Condition 1) $$x = x^2 - 2$$ $$x^2 - x - 2 = 0$$ $$(x+1)(x-2) = 0$$ $$x = -1$$ or $$x = 2$$ From the original condition, we have x = -1. Then answer is "No" This is sufficient by CMT(Common Mistake Type) 1, which states "No is also an answer", t Condition 2) $$2x < x^5$$ $$x^5 - 2x > 0$$ $$x(x^4 - 2 ) > 0$$ Only $$x = -1$$ from the original condition satisfies this condition. Then answer is "No" This is sufficient by CMT(Common Mistake Type) 1, which states "No is also an answer", t The answer is D. 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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07 Mar 2018, 04:01
viktorija wrote:
x^n = x^(n+2) for any integer n. Is it true that x > 0?

(1) x = x^2 - 2
(2) 2x < x^5

hard problem
from x^n<x^(n+2)
we see that x =0 or 1 or -1

put these number into condition 1 and 2. we see that for both, only x=-1 is good.
so, each condition is ok

D.
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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02 Dec 2018, 14:27
This came up in a GMAT Focus test for me and I selected (B).

Plugging-in numbers and being cognizant of limitations given would have helped me solve this. I think other people fall into this trap as well judging by the difficulty.
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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05 Dec 2018, 11:54
I do not know why no body has solved in this way-

$$x^n$$=$$x^{(n+2)}$$
$$x^n$$=$$x^n$$*$$x^2$$

so
$$x^2$$=1
x=-1 or +1

Statement 1: sufficient
put -1 and +1 in the expression
only -1 satisfy the equation
so x=-1

Statement 1: sufficient
put -1 and +1 in the expression
only -1 satisfy the equation
so x=-1

so both the statements gives the solution hence answer is D
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0? &nbs [#permalink] 05 Dec 2018, 11:54
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