Answer
$3.8$ months
Work Step by Step
If $y=A(1.20^{t}), \qquad$(by part a.)
we want to find t for which $y=2A$.
$2A=A(1.20)^{t} \qquad/\div A$
$2=(1.20)^{t} \qquad/\log(...)$
$\log 2=t\log 1.20$
$t=\displaystyle \frac{\log 2}{\log 1.20}\approx 3.8$ months.
