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Big-Oh algebra with logarithms that I don't get?

  1. May 10, 2012 #1
    My textbook says O(3log2 n) can be written as O(nlog2 3). Why is that?

    Thank you.
     
  2. jcsd
  3. May 10, 2012 #2

    I like Serena

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    Welcome to PF, Nishiura_high! :smile:

    One of the log rules is that ##\log a^b = b \log a##.

    So:
    $$\log_2(3^{\log_2 n}) = \log_2 n \cdot \log_2 3$$
    and also:
    $$\log_2(n^{\log_2 3}) = \log_2 3 \cdot \log_2 n$$
     
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