62. If 2^(-2x) + 2^(-x)-6=0, x-1/x=?
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2
0
exponents
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Judging from the correct answer, I've added some brackets to the expression in order to avoid any ambiguity.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
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
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2^(-2x) + 2^(-x) = 6.vipulgoyal wrote:62. If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ?
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2
0
In other words, the sum of two POWERS OF 2 is equal to 6, implying the following:
2² + 2¹ = 6.
For the terms in red to match, their exponents must be equal.
Thus:
-x = 1
x = -1.
Thus:
1 - 1/x = -1 - (1/-1) = 0.
The correct answer is C.
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Given : 1/2^(2x) + 1/(2^x) - 6 = 0vipulgoyal wrote:62. If 2^(-2x) + 2^(-x)-6=0, x-1/x=?
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2
0
Take y = 1/(2^x)
We get y^2 + y - 6 = 0
(y-2)(y+3) = 0
y = 2
1/2^x = 2 (we discard the other value since it's negative and powers cannot be negative)
x = -1
x + 1/x = 0