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

i will go with C .

answer is C

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

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,