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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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Each moment of time ought to be put to proper use, either in business, in improving the mind, in the innocent and necessary relaxations and entertainments of life, or in the care of the moral and religious part of our nature.

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 _________________

"Popular opinion is the greatest lie in the world"-Thomas Carlyle

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. _________________

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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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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