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19% (01:29) correct
80% (00:16) wrong based on 4 sessions
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
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Status: Getting ready for the internship summer
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Re: I'm missing something basic here, but no idea what [#permalink]
 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]
07 Nov 2010, 12:44
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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Manager
Status: Getting ready for the internship summer
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Re: I'm missing something basic here, but no idea what [#permalink]
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]
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.htmlAlso 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.htmlHope it helps.
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Re: I'm missing something basic here, but no idea what [#permalink]
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]
07 Nov 2010, 20:26
i am new to this forum...please provide me complete information.
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Senior Manager
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Re: I'm missing something basic here, but no idea what [#permalink]
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]
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.htmlAlso 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.htmlHope 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]
08 Nov 2010, 17:51
Answer: BExplanation: \frac{{2^{(4-1)^2}}}{{2^{(3-1)}}} = \frac{{2^{3^2}}}{{2^2}} = \frac{{2^9}}{{2^2}} = 2^{(9-2)} = 2^7Remember 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)^2we get 2^6=64 or 8^2=64Please! 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
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08 Nov 2010, 17:51
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