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If a and n are positive numbers, does 2a^{2x}=n?

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If a and n are positive numbers, does 2a^{2x}=n? [#permalink] New post 03 Mar 2013, 14:44
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If a and n are positive numbers, does 2a^{2x}=n?

(1) a^x+\frac{1}{a^x}=\sqrt{n+2}

(2) x > 0
[Reveal] Spoiler: OA

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Re: If a and n are positive numbers, does 2a^{2x}=n? [#permalink] New post 03 Mar 2013, 15:22
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PraPon wrote:
If a and n are positive numbers, does 2a^{2x}=n?

(1) a^x+\frac{1}{a^x}=\sqrt{n+2}

(2) x > 0


(1) a^x+\frac{1}{a^x}=\sqrt{n+2} ALONE when squarred : ( a^{x} + a^{-x} )^2 = n+2

so, n = (a^{x} + a^{-x})^{2} - 2

n = a^{2x}+a^{-2x}+2a^{x}*a^{-x}-2 (a^{x}*a^{-x}=1)

n = a^{2x}+a^{-2x}

back to the question :

2a^{2x}-n = 2a^{2x}-a^{2x}-a^{-2x}
2a^{2x}-n = a^{2x}-a^{-2x}

Hence, A insufficent

(2) x > 0 ALONE

Clearly insufficent

both (1) and (2) INSUFF

Thus, Answer is E ..
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Re: If a and n are positive numbers, does 2a^{2x}=n? [#permalink] New post 03 Mar 2013, 21:53
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PraPon wrote:
If a and n are positive numbers, does 2a^{2x}=n?

(1) a^x+\frac{1}{a^x}=\sqrt{n+2}

(2) x > 0


We know that (a^x+a^-x)^2 = a^2x+a^-2x+2.

From F.S 1, we have (n+2) =a^2x+a^-2x+2

or n =a^2x+a^-2x. Thus, the question stem is asking whether 2a^2x= n?

If it has to be true, then a^2x+a^-2x = 2a^2x.

or a^(4x) = 1. Now, for x=0, we get a YES. But depending on the values of a and x, this will change. Thus, insufficient.

From F.S 2, we only have x>0. Clearly Insufficient.

Combining both, we know that x is not equal to zero. However, if a=1, we can still get a^4x = 1. Insufficient.

E.
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Re: If a and n are positive numbers, does 2a^{2x}=n?   [#permalink] 03 Mar 2013, 21:53
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