Question 979123
That seems good, depending on what source you use for natural logarithms.


{{{y=p*e^(0.008t)}}}
y, value for price after time t in months;
p, initial value when t is 0 months;


Finding time to double the price:
{{{2=1*e^(0.008t)}}}
{{{ln(2)=0.008t}}}
{{{t=ln(2)/0.008}}}
{{{highlight(t=86.64)}}}  but you might say 86.6 months.


Eight years is {{{8*12=96*months}}}.
y after 96 months,
{{{y=2.25*e^(0.008*96)}}}
{{{highlight(y=4.85)}}}