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2^(4-1)^2/2^(3-2)

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2^(4-1)^2/2^(3-2) [#permalink]

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New post 07 Nov 2010, 12:21
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\(\frac{2^{(4-1)^2}}{2^{(3-2)}}=\)

A. 2^8
B. 2^7
C. 2^6
D. 2^5
E. 2^4

I'm overlooking something incredibly basic here. I know it. It's the first question I got on the GMAT prep math and I was shocked to see I got it WRONG. I've looked over it time and again, but can't find how to get the answer they are saying. They insist it's A, but I can take one look at that and see they are asking 2^6 over 2^2. Which would be 2^4. What am I missing?
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Oct 2013, 04:05, edited 1 time in total.
Edited the question.

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Re: I'm missing something basic here, but no idea what [#permalink]

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New post 07 Nov 2010, 12:36
The parenthesis leads me to believe the question requires you to first square the result of (4-1). This would simplify the question to:

\(\frac{2^9}{2^2} = 2^7\)

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Re: I'm missing something basic here, but no idea what [#permalink]

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New post 07 Nov 2010, 12:44
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LJ wrote:
The parenthesis leads me to believe the question requires you to first square the result of (4-1). This would simplify the question to:

\(\frac{2^9}{2^2} = 2^7\)


*blink blink* There has got to be a better way to display that...

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Re: I'm missing something basic here, but no idea what [#permalink]

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New post 07 Nov 2010, 12:56
Pollux wrote:
LJ wrote:
The parenthesis leads me to believe the question requires you to first square the result of (4-1). This would simplify the question to:

\(\frac{2^9}{2^2} = 2^7\)


*blink blink* There has got to be a better way to display that...


Can you post a screenshot of what it looked like on the test screen?

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Re: I'm missing something basic here, but no idea what [#permalink]

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New post 07 Nov 2010, 13:11
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Pollux wrote:
I'm overlooking something incredibly basic here. I know it. It's the first question I got on the GMAT prep math and I was shocked to see I got it WRONG. I've looked over it time and again, but can't find how to get the answer they are saying. They insist it's A, but I can take one look at that and see they are asking 2^6 over 2^2. Which would be 2^4. What am I missing?


2^(4-1)^2 over 2^(3-1)

A) 2^8
B) 2^7
C) 2^6
D) 2^5
E) 2^4


Writing mathematical symbols in posts: writing-mathematical-symbols-in-posts-72468.html

Also please check the questions when posting. Original question is \(\frac{2^{(4-1)^2}}{2^{(3-2)}}=?\)

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 \((a^m)^n\).

According to above:

\(\frac{2^{(4-1)^2}}{2^{(3-2)}}=\frac{2^{3^2}}{2}=\frac{2^{(3^2)}}{2}=\frac{2^9}{2}=2^8\).

Answer: A.

For more check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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Re: I'm missing something basic here, but no idea what [#permalink]

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New post 07 Nov 2010, 13:25
Bunuel replied before I posted mine. It all makes sense now. Top down. Top down. Top down. Good to know!
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Re: I'm missing something basic here, but no idea what [#permalink]

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New post 07 Nov 2010, 20:26
i am new to this forum...please provide me complete information.

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Re: I'm missing something basic here, but no idea what [#permalink]

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New post 07 Nov 2010, 20:30
Bunuel corrected the question and explained the answer! I will never be able to beat that! :)

I guess what Pollux missed is that he thought 2^(3^2) = 2^6 but in fact it is 2^9.

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Re: I'm missing something basic here, but no idea what [#permalink]

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New post 08 Nov 2010, 16:53
Bunuel wrote:
Pollux wrote:
I'm overlooking something incredibly basic here. I know it. It's the first question I got on the GMAT prep math and I was shocked to see I got it WRONG. I've looked over it time and again, but can't find how to get the answer they are saying. They insist it's A, but I can take one look at that and see they are asking 2^6 over 2^2. Which would be 2^4. What am I missing?


2^(4-1)^2 over 2^(3-1)

A) 2^8
B) 2^7
C) 2^6
D) 2^5
E) 2^4


Writing mathematical symbols in posts: writing-mathematical-symbols-in-posts-72468.html

Also please check the questions when posting. Original question is \(\frac{2^{(4-1)^2}}{2^{(3-2)}}=?\)

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 \((a^m)^n\).

According to above:

\(\frac{2^{(4-1)^2}}{2^{(3-2)}}=\frac{2^{3^2}}{2}=\frac{2^{(3^2)}}{2}=\frac{2^9}{2}=2^8\).

Answer: A.

For more check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.



I am just curious as I got when the OP posted they said it was \(\frac{2^{(4-1)^2}}{2^{(3-1)}}=?\)

and when you answered it you changed the denominator's exponent from (3-1) to (3-2), was it just a typo by the OP? I"m confused because I got 2^7
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Re: I'm missing something basic here, but no idea what [#permalink]

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New post 08 Nov 2010, 17:51
Answer: B
Explanation:
\(\frac{{2^{(4-1)^2}}}{{2^{(3-1)}}} = \frac{{2^{3^2}}}{{2^2}} = \frac{{2^9}}{{2^2}} = 2^{(9-2)} = 2^7\)

Remember that \(2^{3^2}\) is not same to \((2^3)^2\) because the formulas are \(2^{x^y}\) and \((2^x)^y = 2^{xy}\)are different from each other and if we solve \(2^{3^2}\) we get \(2^9\) (here solve from top to down) and by solving \((2^3)^2\)we get \(2^6=64\) or \(8^2=64\)

Please! check your Official Answer because the answer can't be \(2^8\) by solving with the forum timer, I got it wrong as it says that OA is A, which according to rule can't be (just explained above).
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Re: I'm missing something basic here, but no idea what [#permalink]

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New post 10 Oct 2013, 13:24
AtifS wrote:
Answer: B
Explanation:
\(\frac{{2^{(4-1)^2}}}{{2^{(3-1)}}} = \frac{{2^{3^2}}}{{2^2}} = \frac{{2^9}}{{2^2}} = 2^{(9-2)} = 2^7\)

Remember that \(2^{3^2}\) is not same to \((2^3)^2\) because the formulas are \(2^{x^y}\) and \((2^x)^y = 2^{xy}\)are different from each other and if we solve \(2^{3^2}\) we get \(2^9\) (here solve from top to down) and by solving \((2^3)^2\)we get \(2^6=64\) or \(8^2=64\)

Please! check your Official Answer because the answer can't be \(2^8\) by solving with the forum timer, I got it wrong as it says that OA is A, which according to rule can't be (just explained above).


Hi,
The question given by original poster was incorrect; it has been edited and corrected.
Thanks.
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Re: 2^(4-1)^2/2^(3-2) [#permalink]

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New post 22 Oct 2013, 05:16
waltiebikkiebal wrote:
Small question, could there be an indicator of below 600 level questions?
This is obviously a 400-500 level question, and it is a bit misleading for some of us to waste precious time on easy questions, when someone wants to practice questions of a higher level.

Thanks.


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Re: 2^(4-1)^2/2^(3-2) [#permalink]

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New post 05 Feb 2014, 01:31
I think this question is testing that you know PEMDAS. If you follow that you will get the right answer ..

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Re: 2^(4-1)^2/2^(3-2) [#permalink]

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New post 18 Feb 2014, 02:26
Imp math rule; Please read below:
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Re: 2^(4-1)^2/2^(3-2) [#permalink]

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Re: 2^(4-1)^2/2^(3-2)   [#permalink] 13 Jan 2016, 10:21
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