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

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GMAT Date: 10-25-2013
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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, 10:52
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45% (medium)

Question Stats:

68% (02:14) correct 32% (02:51) wrong based on 242 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 55609
Re: If 4^4x = 1600, what is the value of (4^x–1)^2?  [#permalink]

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18 Oct 2013, 10:59
5
3
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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02 Nov 2013, 23:29
6
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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20 Oct 2013, 06:00
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.
Math Expert
Joined: 02 Sep 2009
Posts: 55609
Re: If 4^4x = 1600, what is the value of (4^x–1)^2?  [#permalink]

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20 Oct 2013, 06:06
1
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, 03:35
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 $$4^{4x} = 1600$$ --> $$4^{2x} = 40$$ - ?
Senior Manager
Joined: 08 Apr 2012
Posts: 344
Re: If 4^4x = 1600, what is the value of (4^x–1)^2?  [#permalink]

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03 Nov 2013, 01:30
Great!!!
Thanks a lot.
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Joined: 16 Feb 2013
Posts: 6
Re: If 4^4x = 1600, what is the value of (4^x–1)^2?  [#permalink]

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20 Feb 2014, 16:43
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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Joined: 02 Sep 2009
Posts: 55609
Re: If 4^4x = 1600, what is the value of (4^x–1)^2?  [#permalink]

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21 Feb 2014, 00:18
1
1
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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Updated on: 02 Sep 2014, 21:01
3
One more approach:

Refer Step I to Step VI
Attachment:

power.jpg [ 47.28 KiB | Viewed 4901 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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Originally posted by PareshGmat on 24 Feb 2014, 01:28.
Last edited by PareshGmat on 02 Sep 2014, 21:01, 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, 09:32
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!
Math Expert
Joined: 02 Sep 2009
Posts: 55609
Re: If 4^4x = 1600, what is the value of (4^x–1)^2?  [#permalink]

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27 Feb 2014, 11:08
2
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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Re: If 4^4x = 1600, what is the value of (4^x–1)^2?  [#permalink]

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04 Feb 2019, 23:21
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

What is wrong in this approach?? Can you please share your thoughts on this.

If 4^(4x) = 1600

$$2^{8x} = 2^6 * 5^2$$
8x = 6
x= 3/4

Now [4^(x–1)]^2

[2^2(x–1)]^2

[ 2 ^(2x -2)] ^ 2 (substituted the value of x= 3/4 here)

[2^-1/2]^2

1/2

Here i was able to guess D, because the denominator was 2 and somehow that 5 has come to the numerator as it is the root of 25.
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Posts: 55609
Re: If 4^4x = 1600, what is the value of (4^x–1)^2?  [#permalink]

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05 Feb 2019, 00:06
1
KanishkM 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

What is wrong in this approach?? Can you please share your thoughts on this.

If 4^(4x) = 1600

$$2^{8x} = 2^6 * 5^2$$
8x = 6
x= 3/4

Now [4^(x–1)]^2

[2^2(x–1)]^2

[ 2 ^(2x -2)] ^ 2 (substituted the value of x= 3/4 here)

[2^-1/2]^2

1/2

Here i was able to guess D, because the denominator was 2 and somehow that 5 has come to the numerator as it is the root of 25.

The red part is not correct.
If you could get 8x = 6 from here $$2^{8x} = 2^6 * 5^2$$, then you'd get 1=5^2, which is obviously not correct. The point is that you cannot equate the powers there. Some power of 2 (2^(8x)) to be equal to 2^6 * 5^2, x must be some irrational number.
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Re: If 4^4x = 1600, what is the value of (4^x–1)^2?  [#permalink]

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05 Feb 2019, 00:14
Bunuel wrote:
KanishkM wrote:
Hi Bunuel

What is wrong in this approach?? Can you please share your thoughts on this.

If 4^(4x) = 1600

$$2^{8x} = 2^6 * 5^2$$
8x = 6
x= 3/4

.

The red part is not correct.
If you could get 8x = 6 from here $$2^{8x} = 2^6 * 5^2$$, then you'd get 1=5^2, which is obviously not correct. The point is that you cannot equate the powers there. Some power of 2 (2^(8x)) to be equal to 2^6 * 5^2, x must be some irrational number.

Thank you Bunuel, I think i got your point

So basically if i had to get a value of x, i need to find a value which will completely consume RHS

Which now i realize, will be a cumbersome task.
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Re: If 4^4x = 1600, what is the value of (4^x–1)^2?   [#permalink] 05 Feb 2019, 00:14
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