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If y^2 = 2, what is the value of 2^(x-y)(x+y)/2^(x-1)(x+1)  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 78% (01:19) correct 22% (01:50) wrong based on 168 sessions

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If y^2 = 2, what is the value of $$\frac{2^{(x-y)(x+y)}}{2^{(x-1)(x+1)}}$$?

(A) -2
(B) -1
(C) 1/2
(D) 2
(E) 4

Any idea how to solve?

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Originally posted by enigma123 on 26 Feb 2012, 15:40.
Last edited by Bunuel on 26 Feb 2012, 16:05, edited 1 time in total.
Edited the question
Math Expert V
Joined: 02 Sep 2009
Posts: 57238
Re: Value question  [#permalink]

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enigma123 wrote:
If $$y^2$$ = 2, what is the value of $$\frac{2 (x-y) (x+y)}{2(x-1) (x+1)}$$
(A) -2
(B) -1
(C) $$\frac{1}{2}$$
(D) 2
(E) 4

Any idea how to solve?

From the given OA I think that the question should be as follows:

If y^2 = 2, what is the value of $$\frac{2^{(x-y)(x+y)}}{2^{(x-1)(x+1)}}$$?

Apply $$(a-b)(a+b)=a^2-b^2$$: $$\frac{2^{(x^2-y^2)}}{2^{(x^2-1)}}=2^{(x^2-y^2-(x^2-1))}=2^{(-y^2+1)}=2^{(-2+1)}=2^{(-1)}=\frac{1}{2}$$.

Answer: C.

Similar question to practice: gmatprep-special-products-question-107238.html

Hope it helps.
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Re: If y^2 = 2, what is the value of 2^(x-y)(x+y)/2^(x-1)(x+1)  [#permalink]

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enigma123 wrote:
If y^2 = 2, what is the value of $$\frac{2^{(x-y)(x+y)}}{2^{(x-1)(x+1)}}$$?

(A) -2
(B) -1
(C) 1/2
(D) 2
(E) 4

Let’s first separately simplify the numerator and denominator.

2^[(x-y)(x+y)] = 2^(x^2-y^2)

2^[(x-1)(x+1)] = 2^(x^2-1)

Now we have:

[2^(x^2-y^2)]/[2^(x^2-1)] = 2^[(x^2-y^2) - (x^2-1)] = 2^(-y^2 +1)

Since y^2 = 2, -y^2 = -2, so we have:

2^(-2 +1) = 2^-1 = 1/2.

Answer: C
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Re: If y^2 = 2, what is the value of 2^(x-y)(x+y)/2^(x-1)(x+1)  [#permalink]

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_________________ Re: If y^2 = 2, what is the value of 2^(x-y)(x+y)/2^(x-1)(x+1)   [#permalink] 24 Dec 2018, 11:40
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