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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} $

ebaines · May 21, 2007, 06:22 AM

For the first set of identities I think you've confused the order of precedence with the exponents. Use parentheses in each expresssion and you'll see that the third expression is incorrect.

The second set looks fine to me.