exponents

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exponents

by vipulgoyal » Tue Jul 02, 2013 8:41 pm
62. If 2^(-2x) + 2^(-x)-6=0, x-1/x=?
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

0

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by Brent@GMATPrepNow » Tue Jul 02, 2013 9:13 pm
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

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by GMATGuruNY » Wed Jul 03, 2013 3:28 am
vipulgoyal wrote:62. If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ?
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

0
2^(-2x) + 2^(-x) = 6.
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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by sayanpaul » Wed Jul 03, 2013 11:14 am
2^(-2x)+2^(-x) = 6
or 4^(-x)+2^(-x) = 6
or 6^(-x) = 6^(1)
or x = -1
therefore, x-(1/x) = -1-(1/-1) = -1+1 = 0
(c)

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by ganeshrkamath » Thu Jul 04, 2013 2:54 am
vipulgoyal wrote:62. If 2^(-2x) + 2^(-x)-6=0, x-1/x=?
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

0
Given : 1/2^(2x) + 1/(2^x) - 6 = 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