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If N = 10^10

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by niks_01.27 » Sun Sep 16, 2007 1:10 pm
Substitute the value of both N's in;

N^N = 10^k.

You will get k = 10^102.
regards
niks...
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by optimisticsam » Sun Sep 16, 2007 1:23 pm
Little fuzzy on this one still - anyone care to explain a little further?
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by kajcha » Sun Sep 16, 2007 1:49 pm
(x^y)^z can be written as (x)^(y*z).

In this case N = 10^100

N^N = (10^100)^(10^100) = 10^k

= 10^(100*(10^100)) = 10^k

k = 100*(10^100) => (10^2)*(10^100) = 10^102
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by imago » Sun Sep 16, 2007 3:33 pm
kajcha wrote:(x^y)^z can be written as (x)^(y*z).

In this case N = 10^100

N^N = (10^100)^(10^100) = 10^k

= 10^(100*(10^100)) = 10^k

k = 100*(10^100) => (10^2)*(10^100) = 10^102

Thanks Kajcha is very clear!
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by optimisticsam » Sun Sep 16, 2007 3:58 pm
Ah! Got it. Thanks!

I was forgetting that (x^a)(x^b) = x^(a+b)

Thanks for clarifying! Very helpful!
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