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by Canman » Thu Jul 17, 2008 9:24 am
This is hard to type out given all the exponents, but I'll try. It will be more clear if you write it out after

-- (2^64)^(2^64) = 2^p
-- (2^(64*2^64) = 2^p
-- p = 64*2^64
-- p = 2^6*2^64
-- p = 2^70

Ans C
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by sudhir3127 » Thu Jul 17, 2008 9:33 am
i will go with C .
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by Chitts » Thu Jul 17, 2008 9:42 am
answer is C
Regards,
Chitts
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by anksanks » Thu Jul 17, 2008 10:18 am
I wasn't sure so asked my roomie for this...
Thanks to him for the answer...
Solve this kind using logs

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

taking log both sides,

d log d = p log 2
inserting d,

2^64 ( log 2^64) = p log 2
=>2^64 (64 log 2) = p log 2 (Since log a^b = b log a)
=>(2^ 64) * (2^6) * log 2 = p log 2 (We can cancel log 2 from both ends now)
=> 2 ^(64+6) = p (Since (a^x) * (a^y) = a^(x+y) )
=> p = 2^70
Shape me with your bounds,
Wrap me all around,
Define this adorable thought,
And break the silence.
(c) Anks
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by mberkowitz » Mon Jul 21, 2008 11:15 am
anksanks,

you said,

2^64 ( log 2^64) = p log 2
=>2^64 (64 log 2) = p log 2 (Since log a^b = b log a)
=>(2^ 64) * (2^6) * log 2 = p log 2 (We can cancel log 2 from both ends now)
=> 2 ^(64+6) = p (Since (a^x) * (a^y) = a^(x+y) )
=> p = 2^70

how does =>2^64 (64 log 2) make 2^ 64) * (2^6) * log 2?

thanks,
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