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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?

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?

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 \(a^{mn}\).

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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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?

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 \(a^{mn}\).

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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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). _________________

"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??

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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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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