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If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x =

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Bunuel
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If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   04 Dec 2014, 07:34
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Tough and Tricky questions: Exponents.



If \(x - \frac{1}{2^6} - \frac{1}{2^7} - \frac{1}{2^8} = \frac{2}{2^9}\), then x =

A) 1/2
B) 1/2^3
C) 1/2^4
D) 1/2^5
E) 1/2^9

Kudos for a correct solution.

Source: Chili Hot GMAT
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D
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PareshGmat
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Re: If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   04 Dec 2014, 19:39
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\(x - \frac{1}{2^6} - \frac{1}{2^7} - \frac{1}{2^8} = \frac{2}{2^9}\)

\(x = \frac{1}{2^6} + \frac{1}{2^7} + \frac{1}{2^8} + \frac{2}{2^9}\)

\(x = \frac{1*2*2*2}{2^9} + \frac{1*2*2}{2^9} + \frac{1*2}{2^9} + \frac{2}{2^9}\)

\(x = \frac{8+4+2+2}{2^9} = \frac{16}{2^9} = \frac{1}{2^5}\)

Answer = D
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Re: If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   04 Dec 2014, 07:44
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x= 2/2^9+1/2^6+1/2^7+1/2^8
Take LCM i.e. 2^9

x=(2+2^3+2^2+2)/2^9
x=(2+8+4+2)/2^9
x=16/2^9
x=2^4/2^9
x=1/2^5

Ans - D
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Re: If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   05 Dec 2014, 07:29
Expert Reply
Bunuel wrote:

Tough and Tricky questions: Exponents.



If \(x - \frac{1}{2^6} - \frac{1}{2^7} - \frac{1}{2^8} = \frac{2}{2^9}\), then x =

A) 1/2
B) 1/2^3
C) 1/2^4
D) 1/2^5
E) 1/2^9

Kudos for a correct solution.

Source: Chili Hot GMAT


The correct answer is D.
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Re: If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   05 Dec 2014, 11:40
Choice D

When we start, notice that the RHS equals 2/2^9 = 1/2^8. Then add the factor of 1/2^8 to both sides yielding 2/2^8 on the RHS.
Keep doing this until you reach x-1/2^6 = 1/2^6, yielding x = 1/2^5.

D
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Re: If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   08 Mar 2015, 13:14
From this
\(x = \frac{1}{2^6} + \frac{1}{2^7} + \frac{1}{2^8} + \frac{2}{2^9}\)


To this
\(x = \frac{1*2*2*2}{2^9} + \frac{1*2*2}{2^9} + \frac{1*2}{2^9} + \frac{2}{2^9}\)

\(x = \frac{8+4+2+2}{2^9} = \frac{16}{2^9} = \frac{1}{2^5}\)



How?
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Re: If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   08 Mar 2015, 13:39
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erikvm wrote:
From this
\(x = \frac{1}{2^6} + \frac{1}{2^7} + \frac{1}{2^8} + \frac{2}{2^9}\)


To this
\(x = \frac{1*2*2*2}{2^9} + \frac{1*2*2}{2^9} + \frac{1*2}{2^9} + \frac{2}{2^9}\)

\(x = \frac{8+4+2+2}{2^9} = \frac{16}{2^9} = \frac{1}{2^5}\)



How?


This is also basic (finding the common denominator):

\(x = \frac{1}{2^6} + \frac{1}{2^7} + \frac{1}{2^8} + \frac{2}{2^9}=\frac{2^3}{2^9} + \frac{2^2}{2^9} + \frac{2}{2^9} + \frac{2}{2^9}=\frac{2^3+2^2+2+2}{2^9}\).
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Re: If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   08 Feb 2017, 19:49
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Bunuel wrote:

Tough and Tricky questions: Exponents.



If \(x - \frac{1}{2^6} - \frac{1}{2^7} - \frac{1}{2^8} = \frac{2}{2^9}\), then x =

A) 1/2
B) 1/2^3
C) 1/2^4
D) 1/2^5
E) 1/2^9


We can multiply the given equation by 2^9 and we have:

x(2^9) - 2^3 - 2^2 - 2^1 = 2

x(2^9) = 2 + 2^1 + 2^2 + 2^3

x(2^9) = 16

x = 16/2^9

x = 2^4/2^9

x = 1/2^5

Answer: D
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Re: If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   18 Aug 2019, 05:33
x= (2/2^9) +(1/2)^8 + (1/2)^7 + (1/2)^6
x =( 1/2^8) + ( 1/2^8)+ (1/2)^7 + (1/2)^6
x= (2/2^8) + (1/2)^7 + (1/2)^6
x = (1/2^7) +(1/2)^7 + (1/2)^6
x = (2/2^7) + (1/2)^6
x= (1/2)^6 + (1/2)^6
x = (2/2^6)
x = 1/2^5

Hence answer is D
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If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   18 Aug 2019, 07:23
Bunuel wrote:

Tough and Tricky questions: Exponents.



If \(x - \frac{1}{2^6} - \frac{1}{2^7} - \frac{1}{2^8} = \frac{2}{2^9}\), then x =

A) 1/2
B) 1/2^3
C) 1/2^4
D) 1/2^5
E) 1/2^9

Kudos for a correct solution.

Source: Chili Hot GMAT


Given: \(x - \frac{1}{2^6} - \frac{1}{2^7} - \frac{1}{2^8} = \frac{2}{2^9}\)

Asked: x = ?

\(x - \frac{1}{2^6} - \frac{1}{2^7} - \frac{1}{2^8} = \frac{2}{2^9}\)

\(x = \frac{1}{2^6} + \frac{1}{2^7} +\frac{1}{2^8} + \frac{2}{2^9}\)

\(x = \frac{(2-1)}{2^6} + \frac{(2-1)}{2^7} +\frac{(2-1)}{2^8} + \frac{1}{2^8}\)

\(x = \frac{1}{2^5}-\frac{1}{2^6} + \frac{1}{2^6}-\frac{1}{2^7} +\frac{1}{2^7}-\frac{1}{2^8} + \frac{1}{2^8}\)

\(x = \frac{1}{2^5}\)

IMO D
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Re: If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   05 Jun 2020, 01:21
we can multiply 2^9 on both sides-
\(\\
2^9 x - 2^9/2^6 - 2^9/2^7 - 2^9/2^8 = (2^9*2/2^9)\)
\(2^9 x - 2^3 - 2^2 - 2 = 2 \) (taking 2 on right side)
\(2^9 x - 2^3 - 2^2 = 4\)
\(2^2 ( 2^7 x - 2 - 1 ) = 2^2\)
\(2^7 x -3 = 1\) (taking 2^2on right side)
\(2^7 x= 4\)
\(x = 2^2 / 2^7\)
\(x = 2^-5\)
\(x = (1/2)^5\)

ANSWER (D)
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If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink] New post   05 Jul 2020, 13:03
\(x − \frac{1}{2^6} − \frac{1}{2^7} − \frac{1}{2^8} = \frac{2}{2^9}\)
\(x=\frac{1}{2^6} + \frac{1}{2^7} + \frac{1}{2^8} + \frac{2}{2^9}\)
\(x=\frac{8}{2^9} + \frac{4}{2^9} + \frac{2}{2^9} + \frac{2}{2^9}\)
\(x=\frac{16}{2^9}\)
\(x=\frac{1}{2^5}\)
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If x − 1/2^6 − 1/2^7 − 1/2^8 = 2/2^9, then x = [#permalink]
05 Jul 2020, 13:03
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