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

[Nishiura_high]

My textbook says O(3^{log₂ n}) can be written as O(n^{log₂ 3}). Why is that?

Thank you.

 

[I like Serena]

Welcome to PF, Nishiura_high!

My textbook says O(3^{log₂ n}) can be written as O(n^{log₂ 3}). Why is that?

Thank you.

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