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

The equation a = 1/a is satisfied only when a = 1 or -1.

So, to satisfy this equation, \(x^n\) must be 1 or -1.

Statement 1:
If n = 0, \(x^n\) will be 1 for all integral values of x except 0. So x can take any value. We do not know the value of x. Not sufficient.

Statement 2:
If n ≠ 0, \(x^n\) will be 1 if x = 1 and \(x^n\) will be -1 if x = -1 and n is odd. So equation can be satisfied by both x = 1 and x = -1. We get two values for x. Not sufficient.

Using both statements together, we still have 2 values for x. Not sufficient.
Answer (E)
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\(x^n - x^{-n}=0\)

\(x^n - \frac{1}{x^n}=0\)

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

\(x^{2n}=1\) (it can be assumed that \(x \neq 0\) because \(x^{-n}\) would then be undefined)

\(x^{2n}=1\) when:

1) \(x=1\)
2) \(n=0\) and \(x \neq 0\)

Statement 1) it cannot be determined whether \(x=1\), and no information is given on \(n\).

Yes example: \(x=1\), \(n=5\), and \(1^5 - 1^{-5} = 0\)

No example: \(x=5\), \(n=1\), and \(5^1 - 5^{-1} \neq 0\)

Statement 2) no information is given on \(x\). The same yes/no examples can be given to show that statement is insufficient.

Combined) \(n \neq 0\) and \(x\) is an integer, but we used the same example in both statements to demonstrate insufficiency.

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
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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.
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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
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