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Best answer
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let put n = 2m

now

1) n = 2m

2) (log n)100

(log 2m)100  

=m100  (if take base 2)

now u can see that 1st one is exponential time and 2nd one is pollynomial time 

so n > (log n)100

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''logn'' raise to any constant power will be always smaller than ''n'' asymptotically....you can take log on both sides and check it. However, ''logn'' raised to the power ''logn'' itself will be greater than n.
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1 votes
n     ?    (log n)^100

Taking log on both sides,

log n      log(log n)^100

log n      100*loglog(n)

now put any big value of n

let n = 2^1024

log 2^1024      100*loglog(2^1024)

1024 * 1             100*log(1024)

1024                 100*10

1024          >       1000

so n > (log n)^100.

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