Since the tortoise travels 3/5 the remaining distance, the remaining distance at the end of the day is:
5/5 - 3/5 = 2/5
2/5 of what it was at the beginning of the day.
So, the function can be modeled by an exponential with a "growth" factor of 2/5, as follows:
r(n) = 1000 × (2/5)ⁿ
where, r is the number of remaining feet after n days of travel.
To find the remaining distance after 4 days of travel:
r(4) = 1000 × (2/5)⁴
r(4) = 25.6 ft
