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

=> 4*2^2n = 4*2^(2*23)

=> n=23
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Re: If 2^2n + 2^2n + 2^2n + 2^2n = 4^24, then n = [#permalink]
carcass wrote:
If \(2^{2n} + 2^{2n} + 2^{2n} + 2^{2n} = 4^{24}\), then n =

A. 3
B. 6
C. 12
D. 23
E. 24

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

Or, \(2^{2n} + 2^{2n} + 2^{2n} + 2^{2n} = 2^{48}\)

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

Or, \(2^{2n+2} = 2^{48}\)

So, \(2n + 2 = 48\)

Or, \(2n = 46\), ie \(n = 23\), Answer must be (D)
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Re: If 2^2n + 2^2n + 2^2n + 2^2n = 4^24, then n = [#permalink]
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