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# Given 2^(4x) = 1600, what is the value of [2^(x-1)]^4

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Joined: 29 Nov 2012
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Given 2^(4x) = 1600, what is the value of [2^(x-1)]^4  [#permalink]

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Updated on: 17 Jun 2013, 06:18
2
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Difficulty:

45% (medium)

Question Stats:

71% (02:12) correct 29% (04:06) wrong based on 194 sessions

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Given $$2^{4x} = 1600$$, what is the value of $$\frac{[2^{(x-1)}]^4}{[2^x]^2}$$

A. 2
B. 5/2
C. 5
D. 25/4
E. 24

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Originally posted by fozzzy on 17 Jun 2013, 06:07.
Last edited by Bunuel on 17 Jun 2013, 06:18, edited 1 time in total.
Edited the question.
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Posts: 49417
Re: Given 2^4x = 1600, what is the value of [2^(x-1)]^4 / [2^x]^  [#permalink]

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17 Jun 2013, 06:17
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4
Given $$2^{4x} = 1600$$, what is the value of $$\frac{[2^{(x-1)}]^4}{[2^x]^2}$$

A. 2
B. 5/2
C. 5
D. 25/4
E. 24

$$\frac{[2^{(x-1)}]^4}{[2^x]^2}=\frac{2^{4(x-1)}}{2^{2x}}=2^{4x-4-2x}=2^{2x-4}=\frac{2^{2x}}{16}$$.

Since $$2^{4x} = 1600$$, then $$2^{2x}=\sqrt{1600}=40$$.

Therefore, $$\frac{2^{2x}}{16}=\frac{40}{16}=\frac{5}{2}$$.

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Re: Given 2^(4x) = 1600, what is the value of [2^(x-1)]^4  [#permalink]

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21 Mar 2017, 06:17
fozzzy wrote:
Given $$2^{4x} = 1600$$, what is the value of $$\frac{[2^{(x-1)}]^4}{[2^x]^2}$$

A. 2
B. 5/2
C. 5
D. 25/4
E. 24

We can simplify the question:

2^(4x - 4)/2^(2x)

2^(4x - 4 - 2x)

2^(2x - 4)

2^(2x)/2^4

We know that 2^4 = 16, and notice that 2^(2x) = √2^(4x). Thus

√2^(4x) = √1600

2^(2x) = 40

Thus, 2^(2x)/2^4 = 40/16 = 5/2.

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Re: Given 2^(4x) = 1600, what is the value of [2^(x-1)]^4  [#permalink]

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12 Apr 2018, 07:15
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Re: Given 2^(4x) = 1600, what is the value of [2^(x-1)]^4 &nbs [#permalink] 12 Apr 2018, 07:15
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# Given 2^(4x) = 1600, what is the value of [2^(x-1)]^4

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