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If 2^k = 3, then 2^(3k+2) =

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If 2^k = 3, then 2^(3k+2) =  [#permalink]

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New post 01 Apr 2015, 03:25
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A
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D
E

Difficulty:

  25% (medium)

Question Stats:

71% (00:56) correct 29% (01:24) wrong based on 248 sessions

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If 2^k = 3, then 2^(3k+2) =  [#permalink]

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New post 01 Apr 2015, 03:38
Bunuel wrote:
If 2^k = 3, then 2^(3k+2) =

A. 29
B. 54
C. 81
D. 83
E. 108


Kudos for a correct solution.


2^k=3
2^3k=3^3
2^3k=27

2^(3k+2)
=2^3k*2^2
=27*4
=108

Answer: E
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Re: If 2^k = 3, then 2^(3k+2) =  [#permalink]

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New post 01 Apr 2015, 20:00
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Answer = E. 108

\(2^{3k+2} = 2^{3k} * 2^2 = (2^{k})^{3} * 4 = 3^3 * 4 = 108\)
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Re: If 2^k = 3, then 2^(3k+2) =  [#permalink]

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New post 06 Apr 2015, 05:04
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Re: If 2^k = 3, then 2^(3k+2) =  [#permalink]

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New post 08 Feb 2017, 18:52
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Bunuel wrote:
If 2^k = 3, then 2^(3k+2) =

A. 29
B. 54
C. 81
D. 83
E. 108


We need to determine the value of (2^3k)(2^2) or (2^3k)(4).

Since 2^k = 3, 2^3k = (2^k)^3 = 3^3 = 27.

Therefore, (2^3k)(4) = 27 x 4 = 108.

Answer: E
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If 2^k = 3, then 2^(3k+2) =  [#permalink]

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New post 08 Feb 2017, 19:22
Bunuel wrote:
If 2^k = 3, then 2^(3k+2) =

A. 29
B. 54
C. 81
D. 83
E. 108


Kudos for a correct solution.


2^k = 3---I

2^(3k+2)=(2^3k)*(2^2)

Cube I and multiple with 2^2

(2^k)^3==>3^3==>27*4=108
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Re: If 2^k = 3, then 2^(3k+2) =  [#permalink]

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New post 26 May 2017, 14:23
pretty straight forward...
you need to be careful on re-writing everything
2^(3k+2) can be rewritten as (2^k)^3 * 2^2
(2^k)^3 = 3^3 = 27
2^2 = 4

27*4 = 108
answer is E.
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If 2^k = 3, then 2^(3k+2) =  [#permalink]

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New post 05 Oct 2018, 21:37
Bunuel wrote:
If 2^k = 3, then 2^(3k+2) =

A. 29
B. 54
C. 81
D. 83
E. 108


Kudos for a correct solution.


Given: \(2^k = 3\)

Expanding the expression we have at hand, we will get the following

\(2^{3k+2} = 2^{3k}*2^2 = (2^k)^3*4 = 3^3*4 = 27*4 = 108\)

Therefore, the value of the expression \(2^{3k+2}\) is 108 (Option A)
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If 2^k = 3, then 2^(3k+2) = &nbs [#permalink] 05 Oct 2018, 21:37
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