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D
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If \(2^{2n} + 2^{2n} + 2^{2n} + 2^{2n} = 4^{24}\), then n =
A. 3
B. 6
C. 12
D. 23
E. 24
A. 3
B. 6
C. 12
D. 23
E. 24
11 Sep 2012, 22:38
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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.
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
General Discussion
11 Sep 2012, 20:30
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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!
=> 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!
