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If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
18 Oct 2013, 09:52
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Question Stats:

69% (02:35) correct

31% (02:13) wrong

based on 113 sessions
If 4^(4x) = 1600, what is the value of [4^(x–1)]^2?

A. 40

B. 20

C. 10

D. 5/2

E. 5/4

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Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
18 Oct 2013, 09:59
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Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
20 Oct 2013, 05:00

Bunuel wrote:

Yash12345 wrote:

If 4^(4x) = 1600, what is the value of [4^(x–1)]^2? A. 40 B. 20 C. 10 D. 5/2 E. 5/4

4^{4x} = 1600 -->

4^{2x} = 40 -

4^{(x-1)^2}=4^{2(x-1)}=4^{2x-2}=\frac{4^{2x}}{4^2}=\frac{40}{16}=\frac{5}{2} .

Answer: D.

I am getting the answer as 5/4.. I just can figure out what have i missed?

4^4 . 4^x = 4^3 . 5^2

4^2x = 25

now, 4^2x/4^2 = 25/16 =5/4.

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Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
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Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
02 Nov 2013, 02:35

Bunuel wrote:

Yash12345 wrote:

If 4^(4x) = 1600, what is the value of [4^(x–1)]^2? A. 40 B. 20 C. 10 D. 5/2 E. 5/4

4^{4x} = 1600 -->

4^{2x} = 40 -

4^{(x-1)^2}=4^{2(x-1)}=4^{2x-2}=\frac{4^{2x}}{4^2}=\frac{40}{16}=\frac{5}{2} .

Answer: D.

Is there another aproach?What if I don't realize the

4^{4x} = 1600 -->

4^{2x} = 40 - ?

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Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
02 Nov 2013, 22:29
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ronr34 wrote:

Bunuel wrote:

Yash12345 wrote:

If 4^(4x) = 1600, what is the value of [4^(x–1)]^2? A. 40 B. 20 C. 10 D. 5/2 E. 5/4

4^{4x} = 1600 -->

4^{2x} = 40 -

4^{(x-1)^2}=4^{2(x-1)}=4^{2x-2}=\frac{4^{2x}}{4^2}=\frac{40}{16}=\frac{5}{2} .

Answer: D.

Is there another aproach?What if I don't realize the

4^{4x} = 1600 -->

4^{2x} = 40 - ?

I think what Bunuel did is the easiest approach. However, if you are worried that this might not strike you, start with the unknown entity.

4^{(x-1)^2} = [\frac{4^x}{4^1}]^2 = [\frac{4^{2x}}{4^2}] and let

t = [\frac{4^{2x}}{4^2}] Now, given that

4^{4x} = 1600. Thus,

t^2 = [\frac{4^{4x}}{4^4}] =

[\frac{1600}{16*16}] = [\frac{100}{16}] and

t = \frac{10}{4} =\frac{5}{2} _________________

All that is equal and not-Deep Dive In-equality Hit and Trial for Integral Solutions

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Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
03 Nov 2013, 00:30

Great!!! Thanks a lot.

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Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
20 Feb 2014, 15:43

Bunuel wrote:

Yash12345 wrote:

If 4^(4x) = 1600, what is the value of [4^(x–1)]^2? A. 40 B. 20 C. 10 D. 5/2 E. 5/4

4^{4x} = 1600 -->

4^{2x} = 40 -

4^{(x-1)^2}=4^{2(x-1)}=4^{2x-2}=\frac{4^{2x}}{4^2}=\frac{40}{16}=\frac{5}{2} .

Answer: D.

Hi Bunuel, why isn't (x-1)^2 is not treated like (a-b)^2 formula?

Thanks

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Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
20 Feb 2014, 23:18
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streamingline wrote:

Bunuel wrote:

Yash12345 wrote:

If 4^(4x) = 1600, what is the value of [4^(x–1)]^2? A. 40 B. 20 C. 10 D. 5/2 E. 5/4

4^{4x} = 1600 -->

4^{2x} = 40 -

4^{(x-1)^2}=4^{2(x-1)}=4^{2x-2}=\frac{4^{2x}}{4^2}=\frac{40}{16}=\frac{5}{2} .

Answer: D.

Hi Bunuel, why isn't (x-1)^2 is not treated like (a-b)^2 formula?

Thanks

Actually parenthesis were missing there. Edited, it should read:

(4^{(x-1)})^2=4^{2(x-1)}=4^{2x-2}=\frac{4^{2x}}{4^2}=\frac{40}{16}=\frac{5}{2} .

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:

a^m^n=a^{(m^n)} and not

(a^m)^n , which on the other hand equals to

a^{mn} .

So:

(a^m)^n=a^{mn} ;

a^m^n=a^{(m^n)} and not

(a^m)^n .

Hope it helps.

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If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
24 Feb 2014, 00:28
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One more approach:

Refer Step I to Step VI

Attachment:

power.jpg [ 47.28 KiB | Viewed 576 times ]
4^{4x} = 1600 Dividing both sides by

4^4 \frac{4^{4x}}{4^4} = \frac{1600}{4^4} 4^{4x-4} = \frac{100}{16} 4^{(x-1)^4} = \frac{10^2}{4^2} Square root both sides

4^{(x-1)^2} = \frac{10}{4} = \frac{5}{2} Answer = D

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Last edited by

PareshGmat on 02 Sep 2014, 20:01, edited 1 time in total.

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Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
27 Feb 2014, 08:32

Bunuel wrote:

Yash12345 wrote:

If 4^(4x) = 1600, what is the value of [4^(x–1)]^2? A. 40 B. 20 C. 10 D. 5/2 E. 5/4

4^{4x} = 1600 -->

4^{2x} = 40 -

(4^{(x-1)})^2=4^{2(x-1)}=4^{2x-2}=\frac{4^{2x}}{4^2}=\frac{40}{16}=\frac{5}{2} .

Answer: D.

Hi Bruno,

thank you for posting all these answers. They are a great tool!!

Quick question though. I just want to confirm the steps of

4^{4x} = 1600 TO

4^{2x} = 40 -

Do you just squareroot the two sides?

\sqrt{4^{4x}} = \sqrt{1600} So the base, 4, doesn't change, only the ^4x gets rooted to ^2x. Is that right?

Thank you again!

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Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink ]
27 Feb 2014, 10:08
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Re: If 4^4x = 1600, what is the value of (4^x–1)^2?
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27 Feb 2014, 10:08