bchow
May 19, 2007, 10:32 AM
n^{log{\small_2}n}^{log{\small_2}n}
={n^{log{\small_2}n}}^{log{\small_2}n}
={{n^{(log{\small_n}2)}}^{-log{\small_2}n}
={ 2^{(log{\small_2}n)}}^{-1}
= \frac{1}{n}
log{\small_2}n =\frac{ log{\small_n}n}{ log{\small_n}2} =\frac{ 1}{ log{\small_n}2} = (log{\small_n}2)^{-1}
={n^{log{\small_2}n}}^{log{\small_2}n}
={{n^{(log{\small_n}2)}}^{-log{\small_2}n}
={ 2^{(log{\small_2}n)}}^{-1}
= \frac{1}{n}
log{\small_2}n =\frac{ log{\small_n}n}{ log{\small_n}2} =\frac{ 1}{ log{\small_n}2} = (log{\small_n}2)^{-1}