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x^(n) – x^(-n) = 0

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Source: — Data Sufficiency |
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by cramya » Tue Nov 04, 2008 2:06 am
Stmt I)

x is an integer

since n value is not given if n was 0 then x could be any integer

Stmt II

n <>0

Gives us no definite value of x even wihtout knowing x is an intger or not we can have x=-1 or x=1


Stmt I and Ii put together still not sufficient

I go for E

It could be 1 or -1 (both statements put together are not sufficient)

Hope I dint miss something!
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by cramya » Tue Nov 04, 2008 2:07 am
By just simplified the question stem to

x ^ n = 1 / x ^ n in evaluating the stmts for sufficiency
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by jsl » Tue Nov 04, 2008 2:19 am
good question...

I simplified question stem to:
( x^2n - 1 )/(x^n) = 0
x^2n = 1

Given x^2n = 1

There are only 2 ways of getting x^2n to equal 1. That is if x = 1 and n = 0 or x = 0 and n=1 (I THINK!). I wish I could remember the exponent rules surrounding 0's and 1's.

I think...
1^0 = 1
2^0 = 1
3^0 = 1, etc....

and...
0^1 = 1
0^2 = 1
0^3 = 1, etc....

Therefore, I'm guessing B...
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by scoobydooby » Tue Nov 04, 2008 5:35 am
jsl wrote:good question...

I simplified question stem to:
( x^2n - 1 )/(x^n) = 0
x^2n = 1

Given x^2n = 1

There are only 2 ways of getting x^2n to equal 1. That is if x = 1 and n = 0 or x = 0 and n=1 (I THINK!). I wish I could remember the exponent rules surrounding 0's and 1's.

I think...
1^0 = 1
2^0 = 1
3^0 = 1, etc....

and...
0^1 = 1
0^2 = 1
0^3 = 1, etc....

Therefore, I'm guessing B...
jsl, 0^1, 0^ (anything) would be 0
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by jsl » Tue Nov 04, 2008 5:50 am
scoobydooby wrote:jsl, 0^1, 0^ (anything) would be 0
Thanks for correcting me! I really need to revise this stuff!

So what's the OA?
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by scoobydooby » Tue Nov 04, 2008 6:05 am
i would think E

if x^n-x^(-n)=0 then x^n=x^(-n) or x^n=1/(x^n)
x cannot be 0 or 1/0 would be undefined, so possibilities are n=0, x=1 or x=-1(provided n is even)

1) x is an integer, it may be that x=1 or -1, we dont know n either. not sufficient

2) n is not 0, so x can be 1 or -1 depending on whether n is even or odd. not sufficient

combining x is an integer, n not zero, we still do not know if n is even or odd, x=1 or -1. therefore E
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by gmat009 » Tue Nov 04, 2008 10:06 am
IMO B
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by logitech » Tue Nov 04, 2008 10:22 am
jsl wrote:good question...

I simplified question stem to:
( x^2n - 1 )/(x^n) = 0
x^2n = 1

Given x^2n = 1

There are only 2 ways of getting x^2n to equal 1. That is if x = 1 and n = 0 or x = 0 and n=1 (I THINK!). I wish I could remember the exponent rules surrounding 0's and 1's.

I think...
1^0 = 1
2^0 = 1
3^0 = 1, etc....

and...
0^1 = 1
0^2 = 1
0^3 = 1, etc....

Therefore, I'm guessing B...
JSL, thanks for the explanation.

( x^2n - 1 )/(x^n) = 0
x^2n = 1

And we have to keep in mind that X^n can not be equal to ZERO

So this leaves us:

either x = -1 or + 1 and n is any integer but ZERO

OR X is ANY INTEGER and n is ZERO

1) Insuff
2) Suff , because it tells us n &#8800; 0 and eliminates of of the choices.

by the way, if I remember correctly

0^1 = 1
0^2 = 0
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by ska7945 » Tue Nov 04, 2008 2:16 pm
oa e
let's beat GMAT.
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by elmiko » Tue Nov 04, 2008 3:36 pm
According to a check on my TI-84 0^any number = 0

I think B is insufficient because also because it only tells us that n is not equal to 0.

Therefore given x^2n = 1
1^any number = 1
however x could also be -1 because n could be 2, 4, 6, etc.

therefore 2 is insuff also.

I go with E also.
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by cramya » Tue Nov 04, 2008 4:06 pm
One samll caveat: 0 ^ 0 would be undefined if I am not mistaken
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by logitech » Tue Nov 04, 2008 4:12 pm
cramya wrote:One samll caveat: 0 ^ 0 would be undefined if I am not mistaken
Exponentiation: x0 = 1, except that the case x = 0 may be left undefined in some contexts; see Zero to the zero power. For all positive real x, 0x = 0.

https://en.wikipedia.org/wiki/Zero
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by Fab » Thu Nov 06, 2008 3:11 pm
I'll go with E.
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