(a) Let t=number of times in 24 hours it takes to double the population. After 24 hours the population must grow by a factor of 5 to restore its original numbers, because the fish deplete the population to ⅕ its original population.
2t=5, t=log2(5)=2.3219 approx. So the time period d for doubling is 24/log2(5)=10.3362 hours, approximately, (about 10hr 20min).
(b) P=20×5T, where P is the population percent, and where 0<T≤1, so that when T=0 P=20% and at T=1 P=100%. T=0, 1, 2 represent 8am on the first and successive days, and T=1, 2, 3 represents 8am on the second and successive days. For the second day the formula is P=20×5T-1 where 1<T≤2 and for the third day P=20×5T-2 where 2<T≤3. These equations correspond graphically to 3 curves: 1st curve starts at (0,20) representing 8am on the first day, rising to 100% at (1,100); the 2nd curve starts (1,20) representing a time just after 8am on the second day, rising to 100% at (2,100); the 3rd curve starts at (2,20) representing a time just after 8am on the third day, rising to 100% at (3,100). So each curve steps down each day from 100% to 20% as the fish eat the plankton.