Question 1160348
P=Po(1+(.035/365))^365n
3000=2500(1+(.035/365))^365n=1.2
logs both sides
0.07918=365n 0.0000416=0.0152n (round at end)
n=5.209 years

check using continuous compounding, which should be very close
e^(0.035*5.209)(2500)=$3000.03, just a little more, which is correct