If n is an integer and (x^n)-(x^-n)=0, what is the value of x?
1)x is an integer
2)n is not equal to 0.
Pls explain why the ans cant be 'b'. Correct ans given is 'e'.
integers
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first simplify the equation.
(x^n) - (1/x^n)=0
Statement I.
x is an integer. Insufficient we dont know anything about n.
Statement II.
n is not equal to 0.
Plug in numbers.
x=1
n=2
1^2 - (1/1^2) = 0
1-1=0
0=0
x=2
n=2
2^2 - (1/2^2) = 3/4 is not equal to 0
hence insufficient.
Even after combining both the statements we cannot get anything different from what is shown above, hence answer is E.
(x^n) - (1/x^n)=0
Statement I.
x is an integer. Insufficient we dont know anything about n.
Statement II.
n is not equal to 0.
Plug in numbers.
x=1
n=2
1^2 - (1/1^2) = 0
1-1=0
0=0
x=2
n=2
2^2 - (1/2^2) = 3/4 is not equal to 0
hence insufficient.
Even after combining both the statements we cannot get anything different from what is shown above, hence answer is E.
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Hi parallel_chase,
taking lcm of the above eq1
{(x^n)(x^n) -1 }/(x^n) = 0
Therefore x^2n - 1 = 0
x^2n = 1 ----------------------------------- eq2
Wen we try to plug in values into this eq2 .. then we can say that x has 2 be 1 right ?
Thanks,
Vignesh
Can we simplify this as below ?parallel_chase Posted: Mon Aug 04, 2008 12:59 am Post subject:
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first simplify the equation.
(x^n) - (1/x^n)=0 ------------------------------ eq1
taking lcm of the above eq1
{(x^n)(x^n) -1 }/(x^n) = 0
Therefore x^2n - 1 = 0
x^2n = 1 ----------------------------------- eq2
Wen we try to plug in values into this eq2 .. then we can say that x has 2 be 1 right ?
Thanks,
Vignesh
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Can we simplify this as below ?
taking lcm of the above eq1
{(x^n)(x^n) -1 }/(x^n) = 0
Therefore x^2n - 1 = 0
x^2n = 1 ----------------------------------- eq2
Wen we try to plug in values into this eq2 .. then we can say that x has 2 be 1 right ?
What you are doing is right.
But the statement will still not be sufficient.
X=1
n=2
1^4 =1
1=1
LHS=RHS
x=2
n=1
2^2=1
4=1
4 is not equal to 1
Hence insufficient.
Always analyze different possibilities for the same variable. Especially in DS.
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Hey parallel_chase,
I stil have a doubt
I dont think it works the way u have explained.
statement B says N is not equal to zero.
So for the LHS to be equal to RHS the value of X can take the value "1" only. Since only 1^n is 1.
Do u understand what i am trying to say ?
Regards,
Vignesh
I stil have a doubt
I dont think it works the way u have explained.
statement B says N is not equal to zero.
So for the LHS to be equal to RHS the value of X can take the value "1" only. Since only 1^n is 1.
Do u understand what i am trying to say ?
Regards,
Vignesh
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You are right I misinterpreted the question, i was just trying to prove the statement instead we need to find the value of x.Vignesh.4384 wrote:Hey parallel_chase,
I stil have a doubt
I dont think it works the way u have explained.
statement B says N is not equal to zero.
So for the LHS to be equal to RHS the value of X can take the value "1" only. Since only 1^n is 1.
Do u understand what i am trying to say ?
Regards,
Vignesh
Anyways,
Even if we simplify the statement
x^2n =1
we still get two values i.e. x=1 or -1
Since the power is even for x^2n , for any of the two values of x i.e. 1 or -1, the result will always be 1.
Hence Insufficient.
Even combining both the statements together, we cannot conclude anything hence E is the answer.
Thanks.