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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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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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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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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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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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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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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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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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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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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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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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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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22 Oct 2013, 05:16
Expert's post
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.

Check the tags please:
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Re: 2^(4-1)^2/2^(3-2) [#permalink]

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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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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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