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# d^d = 2^p d = 2^64 p =

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d^d = 2^p d = 2^64 p = [#permalink]

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05 Feb 2006, 15:15
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d^d = 2^p

d = 2^64

p = ?
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05 Feb 2006, 15:23
Is it 2^70?

(a^b)^c = a^(bc)

d^d = 2^(64*2^64) = 2^(2^6*2^64) = 2^(2^70)

=> p = 2^70
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05 Feb 2006, 15:50
2^64^((2)^64) is 2^(64*64*2)=2^8192
it's a bit big...
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05 Feb 2006, 16:02
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thearch wrote:
2^64^((2)^64) is 2^(64*64*2)=2^8192
it's a bit big...

thearch,

(a^b)^c = a^(bc)

a = 2
b = 64
c = 2^64
For simplification just ignore the base a and focus on b*c

b*c = 64*2^64 = 2^6*2^64 = 2^(6+64) = 2^70

HTH
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05 Feb 2006, 16:38
TeHCM wrote:

What if the question asked (a^b)^c^c?

Isn't it the same?

(a^b)^c^c = a^(b*c^c)

but ((a^b)^c)^c) = a^(b*c*c) = a^(b*c^2)

This is what I remember from high school algebra, I will verify and update the post if this is wrong.
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05 Feb 2006, 21:05

therefore, p=2^70

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05 Feb 2006, 21:37
kook44 wrote:

therefore, p=2^70

how did u use the power..? i mean carry a ^ 2 like the way we write?
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05 Feb 2006, 22:18
OA is 2^70

I forgot this rule for exponents....tks
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Re: d^d = 2^p d = 2^64 p = [#permalink]

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11 Jul 2016, 03:59
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Re: d^d = 2^p d = 2^64 p =   [#permalink] 11 Jul 2016, 03:59
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# d^d = 2^p d = 2^64 p =

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