29-3
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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.
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.
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- Master | Next Rank: 500 Posts
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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!
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!
-
- Master | Next Rank: 500 Posts
- Posts: 484
- Joined: Sun Jul 30, 2006 7:01 pm
- Thanked: 2 times
- Followed by:1 members
-
- Master | Next Rank: 500 Posts
- Posts: 484
- Joined: Sun Jul 30, 2006 7:01 pm
- Thanked: 2 times
- Followed by:1 members