GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 16 Oct 2019, 14:59

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

Find Similar Topics 
Manager
Manager
User avatar
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

New post 21 Dec 2014, 20:41
1
33
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

46% (02:25) correct 54% (02:40) wrong based on 454 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
User avatar
V
Joined: 02 Sep 2009
Posts: 58381
x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

Show Tags

New post 21 Jan 2015, 02:59
3
19
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
Senior Manager
User avatar
Joined: 13 Jun 2013
Posts: 266
Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

Show Tags

New post 22 Dec 2014, 02:42
1
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
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15262
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

New post 22 Dec 2014, 15:40
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
_________________
Contact Rich at: Rich.C@empowergmat.com
Image


The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
Manager
Manager
avatar
Joined: 05 Jun 2014
Posts: 60
GMAT 1: 630 Q42 V35
Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

Show Tags

New post 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.

Answer is D.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58381
x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

Show Tags

New post 08 Jan 2016, 01:29

8. Exponents and Roots of Numbers



_________________
Math Revolution GMAT Instructor
User avatar
V
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

New post 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.
_________________
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"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Senior Manager
Senior Manager
avatar
P
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

New post 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
User avatar
V
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

New post 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
Senior Manager
avatar
P
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

New post 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
User avatar
V
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

New post 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"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Director
Director
avatar
P
Joined: 29 Jun 2017
Posts: 930
Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

Show Tags

New post 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.
VP
VP
User avatar
P
Joined: 14 Feb 2017
Posts: 1186
Location: Australia
Concentration: Technology, Strategy
Schools: LBS '22
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23
GMAT 3: 650 Q47 V33
GMAT 4: 650 Q44 V36
WE: Management Consulting (Consulting)
Reviews Badge CAT Tests
Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

Show Tags

New post 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.
_________________
Goal: Q49, V41

+1 Kudos if I have helped you
Manager
Manager
avatar
S
Joined: 05 Oct 2017
Posts: 65
GMAT 1: 560 Q44 V23
Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

Show Tags

New post 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
_________________

It’s not that I’m so smart, it’s just that I stay with problems longer. -- Albert Einstein
Intern
Intern
avatar
B
Joined: 22 Jan 2018
Posts: 46
Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

Show Tags

New post 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.
Senior Manager
Senior Manager
User avatar
P
Joined: 10 Apr 2018
Posts: 266
Location: United States (NC)
Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

Show Tags

New post 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
_________________
Probus

~You Just Can't beat the person who never gives up~ Babe Ruth
Senior Manager
Senior Manager
User avatar
P
Joined: 20 Mar 2018
Posts: 371
Location: Ghana
Concentration: Finance, Real Estate
Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

Show Tags

New post 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


Posted from my mobile device
Manager
Manager
avatar
S
Joined: 15 Dec 2016
Posts: 100
x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

Show Tags

New post 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.

Answer: D.


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 !!
GMAT Club Bot
x^n = x^(n+2) for any integer n. Is it true that x > 0?   [#permalink] 15 Aug 2019, 07:50
Display posts from previous: Sort by

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  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne