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If 2^2n + 2^2n + 2^2n + 2^2n = 4^24, then n =

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If 2^2n + 2^2n + 2^2n + 2^2n = 4^24, then n =  [#permalink]

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New post Updated on: 18 Oct 2018, 02:11
3
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

74% (01:04) correct 26% (01:22) wrong based on 134 sessions

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Originally posted by carcass on 11 Sep 2012, 17:55.
Last edited by Bunuel on 18 Oct 2018, 02:11, edited 2 times in total.
Edited the question.
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Re: If 2^2n + 2^2n + 2^2n + 2^2n = 4^24, then n =  [#permalink]

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New post 11 Sep 2012, 20:30
2^2n + 2^2n + 2^2n + 2^2n = 4^24

=> 4 x 2^2n = 4^24 = 2^48

=> 2^2 x 2^2n = 2^48

=> 2^(2n+2) = 2^48

=> 2n+2 = 48=> n =23

So. Answer will be D.

Hope it helps!
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Re: If 2^2n + 2^2n + 2^2n + 2^2n = 4^24, then n =  [#permalink]

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New post 11 Sep 2012, 22:38
3
carcass wrote:
If 2^2n + 2^2n + 2^2n + 2^2n = 4^24, then n =

1) 3

2) 6

3)12

4)23

5)24

A tough question.

:)


Not really! The takeaway from this question is: when you have addition/subtraction between terms with exponents, you need to think about taking common. Also, in all exponent questions, consider making the base of every term same, if possible.

\(2^{2n} + 2^{2n} + 2^{2n} + 2^{2n} = 4^{24}\)

Taking \(2^{2n}\) common,

\(2^{2n}(1 + 1 + 1 + 1) = 2^{48}\)

\(2^{2n}*4 = 2^{48}\)
\(2^{2n}*2^2 = 2^{48}\)
\(2^{2n+2} = 2^{48}\)

Now simply put 2n + 2 = 48
n = 23
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Re: If 2^2n + 2^2n + 2^2n + 2^2n = 4^24, then n =  [#permalink]

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New post 26 Feb 2014, 01:32
4. 2^2n = 4^24

4. 4^n = 4^24

4^(n+1) = 4^24

Equating powers, n+1=24; n=23 Answer = D
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Re: If 2^2n + 2^2n + 2^2n + 2^2n = 4^24, then n =  [#permalink]

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New post 08 Jan 2019, 19:51
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Re: If 2^2n + 2^2n + 2^2n + 2^2n = 4^24, then n =   [#permalink] 08 Jan 2019, 19:51
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