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

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Pollux
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2^(4-1)^2/2^(3-2) [#permalink] New post   Updated on: 05 Feb 2022, 10:13
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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

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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?
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A

Originally posted by Pollux on 07 Nov 2010, 12:21.
Last edited by Bunuel on 05 Feb 2022, 10:13, edited 2 times in total.
Edited the question.
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Bunuel
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Re: 2^(4-1)^2/2^(3-2) [#permalink] 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?


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

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


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 the 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: https://gmatclub.com/forum/math-number-theory-88376.html

Hope it helps.
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Re: 2^(4-1)^2/2^(3-2) [#permalink] New post   05 Jan 2012, 05:48
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Please differentiate between (a)^(b^c) and (a^b)^c

The first expression cannot be simplified further, i.e. you have to raise b to the power c to simplify, whereas the second expression can be simplified to a^bc.

Hope this makes it clear.
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2^(4-1)^2/2^(3-2) [#permalink] New post   30 Sep 2019, 12:42
hi Bunuel Why is the ans not D ... ie 2^5

Reason being 2^ ( 4-1) ^ 2 ie the numerator should it not get solved to be 2^( 3)^2 which solves to be 2^ 3+ 2^ 3 ie is 2^6 and as the denominatior is 2^1 ans should be 2^6-1 ie 2^ 5.
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2^(4-1)^2/2^(3-2) [#permalink] New post   30 Sep 2019, 20:36
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mimajit wrote:
hi Bunuel Why is the ans not D ... ie 2^5

Reason being 2^ ( 4-1) ^ 2 ie the numerator should it not get solved to be 2^( 3)^2 which solves to be F ie is 2^6 and as the denominatior is 2^1 ans should be 2^6-1 ie 2^ 5.



\(2^{3^2}=2^{(3^2)}=2^9\), not \(2^3 + 2^3\).

On the other hand \(2^3 + 2^3=2*2^3=2^4\), not \(2^6\)..

You might find links below useful.

8. Exponents and Roots of Numbers



Check below for more:
ALL YOU NEED FOR QUANT ! ! !
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Re: I thought this was an easy problem [#permalink] New post   25 Jul 2022, 09:12
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Re: I thought this was an easy problem [#permalink]
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