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605-655 Level|   Algebra|   Exponents|      
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Bunuel
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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 09 Sep 2017, 08:10
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C
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45% (medium)

Question Stats:

73% (02:11) correct 27% (02:40) wrong based on 356 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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C
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kumarparitosh123
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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 09 Sep 2017, 08:29
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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
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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 09 Sep 2017, 08:44
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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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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 09 Sep 2017, 08:47
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kumarparitosh123
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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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 09 Sep 2017, 08:49
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pushpitkc
kumarparitosh123
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..?

Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app
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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 09 Sep 2017, 08:51
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Bunuel
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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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 09 Sep 2017, 12:55
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Bunuel
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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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 09 Sep 2017, 21:06
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Bunuel
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

Answer C
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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 24 Sep 2017, 21:39
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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
x-1/X=-1-1/(-1)=-1+1=0 Answer c
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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 26 Sep 2017, 16:28
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Bunuel
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.

Answer: C
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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 12 Feb 2019, 20:34
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shashankism
Bunuel
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

Answer C

Hello shashankism !

Could you please explain to me how did you solve "the above in red"?

Thank you so much in advance!
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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 30 Nov 2021, 12:01
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Bunuel
If \(2^{(-2x)} + 2^{(-x)} - 6 = 0\), \(x - \frac{1}{x} =\) ?

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

substitute y = -x
=> \(2^{2y} + 2^{y} - 6 = 0\), \(\frac{1}{y} - y =\)?

\(2^{2y} + 2^{y} - 6 = 0\)
<=> \(2^{y} * (2^{y} + 1) - 2 * 3 = 0\)
<=> \(2^{y} * (2^{y} + 1) - 2^{1} * (2^{1} + 1) = 0\)
=> y = 1

\(\frac{1}{y} - y =\)?
=> \(\frac{1}{1} - 1 = 0\)
=> C
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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 20 Dec 2021, 22:11
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2^(−2x)+2^(−x)−6=0

1/2x^2+1/2^x-6=0

To satisfy the given equation 1/2^2x+1/2^x must be equal to 6

To achieve this 'x' must be NEGATIVE, it can't be +ve, because if it's +ve then we will end up in two decimal values whose sum will never be 6 in this case

∴ x<0 and x=-1, we get

1/2^-2+1/2^-1 = 2^2+2 = 6

so, x-1/x = -1+1 = 0(C)
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If 2^(-2x) + 2^(-x) - 6 = 0, x - 1/x = ? 21 Jan 2025, 03:48
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