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# If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ?

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Joined: 02 Sep 2009
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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ?  [#permalink]

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09 Sep 2017, 08:10
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Question Stats:

76% (01:31) correct 24% (01:57) wrong based on 157 sessions

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If $$2^{(-2x)} + 2^{(-x)} - 6 = 0$$, $$x - \frac{1}{x} =$$ ?

(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

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Joined: 02 Nov 2015
Posts: 167
GMAT 1: 640 Q49 V29
Re: If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ?  [#permalink]

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09 Sep 2017, 08:29
It's an E.
The question breaks down to
1/(4^x)+1/(2^x)-6=0.

Let 1/2^x=a.
Then it becomes a^2+a-6=0.
On solving we get a=2 and -3.
Thus 1/(2^x)=2
Which gives x=-1.
So x-1/x equals 1+1=2.

My formatting may not be very good but have tried to keep it lucid.

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

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09 Sep 2017, 08:44
Since we have been asked to find the value of $$x - \frac{1}{x}$$, this can be rewritten as $$\frac{x^2 - 1}{x}$$

The equation $$2^{(-2x)} + 2^{(-x)} - 6 = 0$$ => $$\frac{1}{{2^{(x)^2}}} + \frac{1}{{2^{(x)}}} - 6 = 0$$

Substituting $$\frac{1}{2^x}$$ be b

Therefore, $$b^2 + b - 6 = 0$$
Solving for b, b=-3 or 2

If b=2 | $$\frac{1}{2^x} = 2^{-x} = 2$$ So, x = -1

The value for the expression $$\frac{x^2 - 1}{x} = \frac{0}{-1} = 0$$(Option C)
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Location: India
GPA: 3.12
If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ?  [#permalink]

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09 Sep 2017, 08:47
kumarparitosh123 wrote:
It's an E.
The question breaks down to
1/(4^x)+1/(2^x)-6=0.

Let 1/2^x=a.
Then it becomes a^2+a-6=0.
On solving we get a=2 and -3.
Thus 1/(2^x)=2
Which gives x=-1.
So x-1/x equals 1+1=2.

My formatting may not be very good but have tried to keep it lucid.

Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app

It must be C, kumarparitosh123

As x=-1, and we need to find the value of x-1/x

The expression $$x-\frac{1}{x} = -1 -(\frac{1}{-1}) = -1 + 1 = 0$$
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Re: If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ?  [#permalink]

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09 Sep 2017, 08:49
pushpitkc wrote:
kumarparitosh123 wrote:
It's an E.
The question breaks down to
1/(4^x)+1/(2^x)-6=0.

Let 1/2^x=a.
Then it becomes a^2+a-6=0.
On solving we get a=2 and -3.
Thus 1/(2^x)=2
Which gives x=-1.
So x-1/x equals 1+1=2.

My formatting may not be very good but have tried to keep it lucid.

Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app

It must be C, kumarparitosh123

As x=-1, and we need to find the value of x-1/x

The expression $$x-\frac{1}{x} = -1 -(\frac{1}{-1}) = -1 + 1 = 0$$

Yes Pushpitkc
My Bad. I don't know in which state of mind did I solve..

Thanks for pointing it out.
A lesson for me not to rush much..?

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

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09 Sep 2017, 08:51
Bunuel wrote:
If $$2^{(-2x)} + 2^{(-x)} - 6 = 0$$, $$x - \frac{1}{x} =$$ ?

(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

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 $$x = -1$$

therefore $$x- \frac{1}{x} = -1 +1 = 0$$

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

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09 Sep 2017, 12:55
Bunuel wrote:
If $$2^{(-2x)} + 2^{(-x)} - 6 = 0$$, $$x - \frac{1}{x} =$$ ?

(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

2^-x=a

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

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09 Sep 2017, 21:06
Bunuel wrote:
If $$2^{(-2x)} + 2^{(-x)} - 6 = 0$$, $$x - \frac{1}{x} =$$ ?

(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

$$2^{(-2x)} + 2^{(-x)} - 6 = 0$$
=> $$1/2^{(2x)} + 1/2^{(x)} - 6 = 0$$

Let 2^x = y
=> $$1/y^2 + 1/y - 6 = 0$$
=> 6y^2 - 6y -1 = 0
=> 6y^2 - 3y + 2y -1 = 0
=> 3y (2y-1) + 2y-1 = 0
=> (3y +1) (2y-1) = 0
=> y = -1/3, 1/2

y = 2^-1
x = -1

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

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

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26 Sep 2017, 16:28
Bunuel wrote:
If $$2^{(-2x)} + 2^{(-x)} - 6 = 0$$, $$x - \frac{1}{x} =$$ ?

(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

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.

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Re: If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? &nbs [#permalink] 26 Sep 2017, 16:28
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