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2^64)^(2^64) = 2^P What does P equal to? This looks

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2^64)^(2^64) = 2^P What does P equal to? This looks [#permalink] New post 17 Feb 2006, 22:45
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(2^64)^(2^64) = 2^P
What does P equal to?


This looks like a really simple problem. Too bad I remember absolutely nothing from my math days...I do remember a rule that lets you move the power to the front of the equation? Or am I mistaken on that?
Can anyone refresh that concept with me if I remembered it correctly?

Thanks!

P.S. By the way, I don't have the OA...
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 [#permalink] New post 17 Feb 2006, 22:54
lol, oops I plugged in some numbers and finally figured it out. The rule was that you could multiply the first power with the base of the second power, in this case you could multiply the first 64 with the second (2^64) on the left side of the equation.....
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Re: PS - two to the powers of... [#permalink] New post 18 Feb 2006, 07:42
cattalk wrote:
(2^64)^(2^64) = 2^P, What does P equal to?

This looks like a really simple problem. Too bad I remember absolutely nothing from my math days...I do remember a rule that lets you move the power to the front of the equation? Or am I mistaken on that?
Can anyone refresh that concept with me if I remembered it correctly?

Thanks! P.S. By the way, I don't have the OA...


(2^3)^(2^3) = 2^P
(2x2x2)^(2x2x2) = 2^P
(2)^(2x2x2) x (2)^(2x2x2) x (2)^(2x2x2) = 2^P
(2)^(8) x (2)^(8) x (2)^(8) = 2^P
(2)^(8+8+8) = 2^P
(2)^(24) = 2^P

therefore, p = 24 = 3 (2^3)

similarly (2^64)^(2^64) = 2^P
p = 64 x (2^64). i will check, if any.
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 [#permalink] New post 18 Feb 2006, 07:58
(x^y)^z= x^(y*z)

Hence:
(2^64)^(2^64)
=2^(64*(2^64))
=2^((2^6)*(2^64))
=2^(2^(6+64))
=2^(2^70)

P=2^70
  [#permalink] 18 Feb 2006, 07:58
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