Exponents of exponents

[Magellan]

How to solve a question like this one:

D = 2^32 and 2^P = D^D

What is the value of P?

Basically, you get to 2^P = 2^32^2^32 --> How to proceed?

[akshatsingh]

is p = 2^37 ?

if yes,

D = 2^32 and 2^P = D^D

D^D = (2^32)^ 2^32
D^D = {(2^(2^5)}^ (2^32)
D^D = (2)^{(2^5)*(2^32)}
D^D = {2^(2^37)} = 2^P

Therefore, P = 2^37

[Magellan]

I don't think it is the answer.

IMO, this transition is not correct:
D^D = {(2^(2^5)}^ (2^32)
D^D = (2)^{(2^5)*(2^32)}


I think the answer is the following:
D^D
= 2^32^2^32
= 2^2^5^2^32
= 2^2^320

P = 2^320

[akshatsingh]

Hmm..I think you are right..

Thats correct..