If n is an integer and x^n – x^(-n) = 0, what is the value of x ?

x^n - x^-n = 0 is possible only in either of the two ways
Case 1. n =0 (in this case value of x doesnt matter)
Case 2. x= -1 or +1 (in this case n can be any integer, since no power can change the total expression)

Now coming to the options (1) is insufficient from our deductions in case 2. x can be -1 or + 1

(2) is also insufficient since in case 1 and 2 we found that n can be of any integer


Together, (1) & (2) are insufficient because we still have x = -1 or +1 and n= any integer >0

Answer is E

x^n-1/x^n=0
x^2n-1/x^n =0
1. x is an integer
possible values
x=2 x=1 x=-1
n=0 n=1 n=1
Hence insufficient
2. From statement 2
still 2 possible values of x
x=1 x=-1
n=1 n=1
Hence insufficient
Combining, we will get same
x=1 x=-1
n=1 n=1
Hence insufficient
Answer E

If n is an integer and x^n – x^-n = 0, what is the value of x ?
x^n – x^-n = 0
x^{2n} = 1
If n=0; Any x will satisfy x^{2n} = 1
If n≠ 0; x = 1 or -1

(1) x is an integer.
If n=0; Any x will satisfy x^{2n} = 1
If n≠ 0; x = 1 or -1
NOT SUFFICIENT

(2) n ≠ 0
x = 1 or -1
NOT SUFFICIENT

(1) + (2)
(1) x is an integer.
(2) n ≠ 0
x = 1 or -1
NOT SUFFICIENT

IMO E

If n is an integer and \(x^{n} – x^{-n} = 0\), what is the value of x ?

\(x^{n} – \frac{1}{x^{n}} = 0\)

\(x^{n} = \frac{1}{x^{n}}\)

This is true when:
1) \(x^{n}\) = -1, 1
2) \(n = 0\)

Lets take a look at our statements:

(1) x is an integer.

x can be -1 or 1; x can be any value if n = 0. Insufficient.

(2) n ≠ 0

if n ≠ 0, x can still be -1 or 1.

(1+2) x is an integer AND n ≠ 0

x can be -1 or 1. Insufficient.

Answer is E.

Given: n is an integer and \(x^n – x^{-n} = 0\)
In other words: \(x^n = x^{-n} \)

----------ASIDE----------------
Important: Many students will incorrectly conclude that, since \(x^n = x^{-n}\), it must be the case that \(n = -n \). However, this is true only if \(x \neq 0\), \(x \neq 1\) and \(x \neq -1\)
To illustrate this point, consider the equation \(1^2 = 1^5\)
Even though the two powers have the same base (1), we certainly can't conclude that \(2 = 5\)

Likewise, if \(0^m = 0^n\), we can't then conclude that \(m=n\)
-----------------------------

So, if \(x^n = x^{-n}\), then there are 4 possible cases:
case a: \(x \neq 0\), \(x \neq 1\) and \(x \neq -1\), and \(n = -n \) (which means \(n = 0\))
case b: \(x = 0\), and \(n\) can have any positive value
case c: \(x = 1\), and \(n\) can have any value
case d: \(x = -1\), and \(n\) is an even integer

Target question: What is the value of x?

Statement 1: x is an integer
There are several values of x (and n) that satisfy statement 1. Here are two:
Case a: x = 1 and n = 1. In this case, the answer to the target question is x = 1
Case b: x = -1 and n = 2. In this case, the answer to the target question is x = -1
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: n ≠ 0
PRO TIP: Before I start testing various values of x and n, I first check to see whether I can reuse any of the values I tested for statement 1....it turns out that both pairs of values I used in statement 1 also satisfy statement 2. That is....
Case a: x = 1 and n = 1. In this case, the answer to the target question is x = 1
Case b: x = -1 and n = 2. In this case, the answer to the target question is x = -1
Since we can’t answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Since I was able to use the same counter-examples to show that each statement ALONE is not sufficient, the same counter-examples will satisfy the two statements COMBINED.
In other words,
Case a: x = 1 and n = 1. In this case, the answer to the target question is x = 1
Case b: x = -1 and n = 2. In this case, the answer to the target question is x = -1
Since we still can’t answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent

If n is an integer and x^n – x^(-n) = 0, what is the value of x ?

(1) x is an integer.
(2) n ≠ 0

x^n – x^(-n) = 0
no matter what value of n we use, the entire equation will always be in the form of (x-1/x)(some other value)=0

Why?
if n=3, we get (x-1/x)(x^2 + 1/(x^2) + 1) = 0. (x^2 + 1/(x^2) + 1) can never be equal to zero. Extending the same logic in all cases

if x = 1/x then x = either +1 or -1

statement 1 - x is an integer
no use to me since both +1 and -1 are integers

statement 2, n ≠ 0
no use to me as no matter what value of n I put, x will always be either +1 or -1

Hence both statements are insufficient