exponents

vipulgoyal wrote:62. If 2^(-2x) + 2^(-x) - 6 = 0, then x - (1/x) =
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

0

Judging from the correct answer, I've added some brackets to the expression in order to avoid any ambiguity.

First recognize that 2^(-2x) = [2^(-x)]^2
So, we take the equation 2^(-2x) + 2^(-x) - 6 = 0 and rewrite it as [2^(-x)]^2 + 2^(-x) - 6 = 0

At this point, we'll use a technique called u-substitution to make the equation easier to solve.
If we let u = 2^(-x), then our equation becomes u^2 + u - 6 = 0
Now solve this new equation for u.
Factor to get: (u + 3)(u - 2) = 0
This means that u = -3 or u = 2

Since u = 2^(-x), we can conclude that 2^(-x) = -3 or 2^(-x) = 2

NOTE: 2^(-x) cannot equal -3, since 2^(any value) will always be positive.
So, it must be the case that 2^(-x) = 2
In other words, 2^(-x) = 2^1, which means -x = 1, which means x = -1

If x = -1, then x - (1/x) = (-1) - 1/(-1)
= -1 - (-1)
= 0

Answer: C

Cheers,
Brent