Base of Logs

$\log_{10}(n) = \Theta(\log_2(n))$

$\begin{eqnarraystar} f(n) &=& \log_{10}(n) \ 10^{f(n)} &=& n \ \log_2(10^{f(n)... ...f(n) \log_2(10) &=& \log_2(n) \ f(n) &=& \log_2(n)/\log_2(10)\end{eqnarraystar}$

So for $c_1 = (1/2)/\log_2(10)$ , $c_2 = 2/\log_2(10)$ , n₀ > 0,

$0 \le c_1 \log_2(n) \le f(n) \le c_2 \log_2(n)$ for all $n \ge n_0$ .

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