29-3

magical cook · Posted: Tue Aug 28, 2007 11:18 am Post subject: 29-3

Can you pls help with your explanation? thanks!

agps · Posted: Wed Aug 29, 2007 8:46 am Post subject:

I think it's B.

1) if n=0 Very Happy

x^0=1 and 1/(x^0) = 1/1=1, then 0=0, true for any value of X not sufficient

2) if n not 0 x=1 (not 100% sure) sufficient

what is the OA?

beny · Posted: Wed Aug 29, 2007 9:18 am Post subject:

x^n - x^(-n) = 0

This can be solved to:
x^(2n) = 1

There are several possibilities:

1.) x is not equal to 1, n = 0
2.) x = 1, n can equal anything
3.) x = -1, n can equal anything

Statement 1:
x is an integer... it can still equal anything if n = 0. Not sufficient.

Statement 2:
n is not equal to 0... it can still equal -1 or 1. Not sufficient.

Answer is E.

agps · Posted: Wed Aug 29, 2007 10:18 am Post subject:

yup, i think beny is right.
i knew there was something missing from statement 2. thanks beny.

magical cook · Posted: Wed Aug 29, 2007 12:07 pm Post subject:

Thanks it is E.

I understood the logic why it's E but I am not too sure this is how to get x^(2n) = 1 ..

from x^n - x^(-n) = 0

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

1 - x^-2n = 0

x^(2n) = 1

Can you tell me if I'm wrong as to the progress to get to x^(2n) = 1
??

Thanks!

agps · Posted: Wed Aug 29, 2007 12:35 pm Post subject:

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

so x^n+x^-n=0 -> x^n+(1/x)^n = 0 ->[(x^n*x^n)+1]/x^n = = ->(x^n)^2+1/x^n = 0 ->(x^2n)-1/x^n = 0 -> x^2n=1

magical cook · Posted: Wed Aug 29, 2007 1:06 pm Post subject:

Thank you!

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

sorry, how come 1 turns to -1 ....

beny · Posted: Wed Aug 29, 2007 2:08 pm Post subject:

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

magical cook · Posted: Wed Aug 29, 2007 2:49 pm Post subject:

I see now - thank you Very Happy

agps · Posted: Thu Aug 30, 2007 1:31 am Post subject:

sorry i typed it wrong for most of it and missed it.