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Is it possible to simplify

$\log_3{n} \cdot 2^{\log_3{n}} \cdot n$

I am actually trying to find the Big-O notation for this equation. But if you don't know what it is, is it possible to simplify this? To something close to

$n^{\log{3} / \log{1.5}}$

2 Answers 2

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$\log_3{n} \cdot 2^{\log_3{n}} \cdot n = \log_3{n} \cdot n^{\log_3{2}} \cdot n= \log_3{n} \cdot n^{\log_3{2}+1} =\log_3{n} \cdot n^{\log_3{6}}$

In big O notation, for each $\epsilon >0$, this is $O( n^{\log_3{6}+\epsilon})$.

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    @jiewmeng Also, the epsilon part says that it is not $O( n^{\log_3{6}})$ but it is $O( n^{\alpha})$ for any \alpha > \log_3(6).2011-11-15
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$ \log_3{n} \cdot 2^{\log_3{n}} \cdot n =\log_3{n} \cdot 2^{\log_3{n}} \cdot 3^{\log_3{n}}=\log_3{n} \cdot 6^{\log_3{n} }$