The sum of n terms of a geometric series is given by [tex]S_{n}=a_{1}\dfrac{r^{n}-1}{r-1}[/tex]
Substituting the given numbers, you have [tex]2730=2\dfrac{4^{n}-1}{4-1}\\\\2730\dfrac{3}{2}=4^{n}-1\\\\4096=4^{n}\\\\\dfrac{\log(4096)}{\log(4)}=n=6[/tex]
