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let \(2^{-x} = a\) so the question becomes \(a^2 + a - 6 = 0\). Solving this we get \((a-2)(a+3) = 0\). Therefore \(a = 2\) or \(a =-3\). But as "\(a\)" is a positive number (\(2\) raised to some power will always be positive), hence \(a = -3\) is not possible So we get \(2^{-x} = 2\) or \(x = -1\)

equation becomes a^2+a-6=0...comes out to be a=-3(not possible,since 2^-x is not equal to -3).Thus a=-2, 2^-x=-2 x=-1 and finally the answer comes out to be 0

Re: If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? [#permalink]

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24 Sep 2017, 21:39

Plugin values of X 2ˆ(-2x)+2ˆ(-x)−6=0 We need a value of x which results the LHS of the equation to be zero. Trial and error we can take x=(-1) I checked from given answer choices. 2ˆ(-2*-1)+2(-*-1)-6=0 2ˆ2+2-6 4+2-6=0 x-1/X=-1-1/(-1)=-1+1=0 Answer c

We can let y = 2^(-x) and rewrite the equation as:

(2^(-x))^2 + 2^(-x) - 6 = 0

y^2 + y - 6 = 0

(y + 3)(y - 2) = 0

y = -3 or y = 2

Since y = 2^(-x), we can say 2^(-x) = -3 or 2^(-x) = 2. However, since 2 is positive, 2^(-x) will be positive regardless of the value of x. So, 2^(-x) can’t be -3, and thus it must be 2. Let’s solve the equation 2^(-x) = 2:

2^(-x) = 2^1

-x = 1

x = -1

Thus, x - 1/x = -1 - 1/(-1) = -1 + 1 = 0.

Answer: C
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