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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]

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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]

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New post 03 Mar 2013, 16:22
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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]

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New post 03 Mar 2013, 22:53
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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]

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Re: If a and n are positive numbers, does 2a^{2x}=n?   [#permalink] 12 May 2014, 22:12
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