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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, 05:18
2
9
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Difficulty:

45% (medium)

Question Stats:

72% (02:27) correct 28% (02:49) wrong based on 152 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

Originally posted by fozzzy on 17 Jun 2013, 05:07.
Last edited by Bunuel on 17 Jun 2013, 05:18, edited 1 time in total.
Edited the question.
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Joined: 02 Sep 2009
Posts: 52905
Re: Given 2^4x = 1600, what is the value of [2^(x-1)]^4 / [2^x]^  [#permalink]

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17 Jun 2013, 05:17
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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, 05:17
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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Given 2^(4x) = 1600, what is the value of [2^(x-1)]^4  [#permalink]

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04 Feb 2019, 23:20
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

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

which can be written as

2^{4x - 4 -2x}

2^2x-4

2^2x/16 ---------(a)

$$2^{4x} = 1600$$

2^4x = 40^2

2^{4x*1/2} = 40 ^2*1/2

2^2x = 40

Lets put the value in (a)

40/16 = 5/2

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

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05 Feb 2019, 14:44
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

I need to improve my speed on these questions... Took me 5 minutes because I made careless mistakes and had to re-do the question 3 times.

My reasoning:

We can simply both equations to get the values we need.

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

From $$2^{4x} = 1600$$ we can square both sides by $$\frac{1}{2}$$ to get $$2^{2x} = 40$$

Plugging everything back into the question we get

$$\frac{1600}{40*16}$$ = $$\frac{5}{2}$$
Given 2^(4x) = 1600, what is the value of [2^(x-1)]^4   [#permalink] 05 Feb 2019, 14:44
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