big $\Theta$ question dealing with $\log_2{n}$ and $\log_{10}{n}$

Question

Show that $\log_{10}{n} = \Theta(log_2{n})$. I know that I have to show that

1) $\log_{10}{n} = O(\log_2{n})$

show: $\log_{10}{n} \le C * \log_2{n}$

and

2) $\log_2{n} = O(\log_{10}{n})$

show: $\log_2{n} \le C * \log_{10}{n}$

but I'm not quite sure where to begin. Any help would be greatly appreciated.

Answer 1

Hint: $\log_x i$ = $\log_y i$ / $\log_y x$