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+2or 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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