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

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

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21 Dec 2014, 20:41
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x^n = x^(n+2) for any integer n. Is it true that x > 0?

(1) x = x^2 - 2
(2) 2x < x^5
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x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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21 Jan 2015, 02:59
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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.

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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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22 Dec 2014, 02:42
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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.

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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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22 Dec 2014, 15:40
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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.

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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, 06:00
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, 01:29
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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, 19:16
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 $79 for 1 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: 468 Location: India Schools: XLRI"20 GMAT 1: 700 Q50 V34 GPA: 3.5 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0? [#permalink] ### Show Tags 06 Nov 2017, 21:27 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: 58381 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0? [#permalink] ### Show Tags 06 Nov 2017, 21:53 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: 468 Location: India Schools: XLRI"20 GMAT 1: 700 Q50 V34 GPA: 3.5 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0? [#permalink] ### Show Tags 06 Nov 2017, 23:04 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: 8011 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, 08:58 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$79 for 1 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, 05:01
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, 15:27
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, 12:54
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?  [#permalink]

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29 Dec 2018, 05:04
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.

I know this is a silly doubt, but this is how I solved it.

x^n = x^(n+2)
So, x^n / x^(n+2) =1
x^(n - n - 2) = 1
x^-2 = 1
1/x^2 = 1
x^2 = 1
So, x = 1 or x = -1.

With this approach, I am not getting x = 0.
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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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25 Apr 2019, 14:30
Hi Bunuel,

I have gone through your solution but need couple of clarifications in my approach.

I have always committed the same mistake in previous questions so wanted to set the record straight.

So here is what i did
We are given that :
$$x^{n}=x^{n+2}$$
dividing the above equation by$$x^{n}$$ we have
$$x^{2}$$ =1

so x can be +1 or -1

Logically i know that even 0 satisfies the above equation , how and why did i not get algebraically get x= 0 as one of its values .

Second
we have
x= $$x^{2}$$-2
$$x^{2}$$-x= 2
x(x-1)=2
Is this step correct that we say
either x=2 or
(x-1)=2 => x= 3. I know x=3 does not satisfy the equation, but then what is the error i am committing.

Logically i know that either x=2 or x=-1. and since from the stem we have $$x^{2}$$=1 we can say therefor that x=-1

Now in second statement:
we have 2x<$$x^{5}$$
Case (1)
Lets say x>0
then 2<$$x^{4}$$ but since we have $$x^{2}$$=1 then $$x^{4}$$=1 and hence this is incorrect.
Case (2)
Lets say x<0
then 2>$$x^{4}$$ but since we have $$x^{2}$$=1 then $$x^{4}$$=1 and hence this is correct.

Need help in above errors and method .

Thanks
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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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25 Apr 2019, 23:47
Sir Bunuel will even give you more stuff but here is my take
As i said earlier you can’t divide by X^n since x can be Zero. Also you missed the third value
I suggest it should be this way
Simplify the question first
x^n=x^(n+2) => x^n=x^n•x^2 => x^n-x^n•x^2=0 => x^n(1-x^2)=0 , x^n =0 , x=0 ,x^2=1 => x=1 or -1
So you see now the we got the third value
Probus wrote:
Hi Bunuel,

I have gone through your solution but need couple of clarifications in my approach.

I have always committed the same mistake in previous questions so wanted to set the record straight.

So here is what i did
We are given that :
$$x^{n}=x^{n+2}$$
dividing the above equation by$$x^{n}$$ we have
$$x^{2}$$ =1

so x can be +1 or -1

Logically i know that even 0 satisfies the above equation , how and why did i not get algebraically get x= 0 as one of its values .

Second
we have
x= $$x^{2}$$-2
$$x^{2}$$-x= 2
x(x-1)=2
Is this step correct that we say
either x=2 or
(x-1)=2 => x= 3. I know x=3 does not satisfy the equation, but then what is the error i am committing.

Logically i know that either x=2 or x=-1. and since from the stem we have $$x^{2}$$=1 we can say therefor that x=-1

Now in second statement:
we have 2x<$$x^{5}$$
Case (1)
Lets say x>0
then 2<$$x^{4}$$ but since we have $$x^{2}$$=1 then $$x^{4}$$=1 and hence this is incorrect.
Case (2)
Lets say x<0
then 2>$$x^{4}$$ but since we have $$x^{2}$$=1 then $$x^{4}$$=1 and hence this is correct.

Need help in above errors and method .

Thanks

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

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15 Aug 2019, 07:50
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.

(2) 2x < x^5 --> from the possible values of x (0, 1, -1), only -1 satisfies 2x < x^5, thus x = -1 < 0. Sufficient.

Hi Bunuel

For Statement 2 --

1) I tried to solve this algebraically. Just wondering, where is the problem in my procedure ?

We are given in S2 2x < x^5

So moving 2x to the RHS of the equation

-- x^5 - 2x > 0

-- x (x^4 - 2) >0

-- hence there are two possibilities for x with this equation

possibility 1 : x >0
or
possibility 2 : x^4 - 2 >0 or x^4 > 2 or x can be positive or negative

Given we have possibility 1 (Where x has to be positive) or possibility 2(x can be positive or negative) -- hence i marked this as insufficient

I clearly am going wrong somewhere -- just wondering, where is the error in this strategy ?

Thank you !!
x^n = x^(n+2) for any integer n. Is it true that x > 0?   [#permalink] 15 Aug 2019, 07:50
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