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# If 4^4x = 1600, what is the value of (4^x–1)^2?

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

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

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18 Oct 2013, 09:59
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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

$$4^{4x} = 1600$$ --> $$4^{2x} = 40$$ -

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

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

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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

$$4^{4x} = 1600$$ --> $$4^{2x} = 40$$ -

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

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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20 Oct 2013, 05:06
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Expert's post
chitrasekar2k5 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

$$4^{4x} = 1600$$ --> $$4^{2x} = 40$$ -

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

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.

$$4^{4}*4^x=4^{4+x}$$ not $$4^{4x}$$: $$a^n*a^m=a^{n+m}$$

Check here for more: math-number-theory-88376.html
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Re: If 4^4x = 1600, what is the value of (4^x–1)^2? [#permalink]

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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

$$4^{4x} = 1600$$ --> $$4^{2x} = 40$$ -

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

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]

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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

$$4^{4x} = 1600$$ --> $$4^{2x} = 40$$ -

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

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]

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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]

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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

$$4^{4x} = 1600$$ --> $$4^{2x} = 40$$ -

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

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]

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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

$$4^{4x} = 1600$$ --> $$4^{2x} = 40$$ -

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

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]

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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 2393 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

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

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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]

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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

$$4^{4x} = 1600$$ --> $$4^{2x} = 40$$ -

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

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]

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27 Feb 2014, 10:08
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Expert's post
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

$$4^{4x} = 1600$$ --> $$4^{2x} = 40$$ -

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

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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