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.

_________________

All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions