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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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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 ]
20 Oct 2013, 05:06
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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 \(t = \frac{10}{4} =\frac{5}{2}\)

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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 1513 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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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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hieracity 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 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!

Yes:

\(4^{4x} = 1600\);

\((4^{2x})^2 = 40^2\);

\(4^{2x} =40\).

Hope it's clear.

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