seems simple, still cant solve

[AJWILL]

if 2^(-2x) + 2^(-x)-6=0. what is x-1/x?

[eagleeye]

if 2^(-2x) + 2^(-x)-6=0. what is x-1/x?

Couple of ways to do this:
First way: observation-

2^(-2x) + 2^(-x) - 6=0
=> 2^(-2x) + 2^(-x) = 6 = 2*3
=> 2^(-x) * (2^(-x) + 1) = 2 * 3 = 2 * (2 + 1)
This works if 2^(-x) = 2 => -x = 1 => x= -1
Hence x-1/x = - 1 + 1 = 0

Second way: Solving a quadratic after substituting:
Let 2^(-x) = a , where a>0.

Then 2^(-2x) + 2^(-x) - 6=0
becomes a^2 + a = 6
=> a^2+a-6 = 0
=> (a-2)(a+3) = 0
=> a =2 or a = -3. Since a>0, a can't be -3. Hence a = 2
=> 2^(-x) = 2 = 2^1. Since bases are equal, -x = 1 => x = -1.

Then x - 1/x = -1 - (1/-1) = -1 + 1 = 0

Let me know if this helps

[snipfer]

the variable change is the systematic way to go.

But what is the logic behind a > 0 ?

Thanks!