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(1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)

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(1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1) [#permalink]

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\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=\)

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)
[Reveal] Spoiler: OA
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1) [#permalink]

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New post 29 Dec 2012, 06:01
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1) [#permalink]

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New post 29 Dec 2012, 06:14
Expert's post
Walkabout wrote:
\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=\)

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)


whenevr you see a fraction raised to a negative power think reciprocal

so we have : 2^3 * 4^2 * 16^1

Consolidate: 2^3 * 2^4 * 2^4

Clearly as you can see we have 2^11 only B fits

otherwise take the reciprocal of 2^11 ----> 1/2^-11

B is the best
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1) [#permalink]

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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1) [#permalink]

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New post 21 Mar 2014, 00:21
Answer = B


Keep the base as 2 & modify the powers accordingly

\((\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}\)

= \((\frac{1}{2})^{-3}(\frac{1}{2})^{-4}(\frac{1}{2})^{-4}\)

With the base the same, add the powers

-3-4-4 = -11

\((\frac{1}{2})^{-11}\)
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1) [#permalink]

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New post 03 Dec 2014, 07:20
Resolved it as 16 x 2^3 x 4^2
After 2:20sec :(
Simplified it as 1/2
B
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1) [#permalink]

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New post 01 Apr 2016, 21:28
Hello from the GMAT Club BumpBot!

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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)   [#permalink] 01 Apr 2016, 21:28
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