Its half-life is 60 days. That is, after 60 days, a given amount of radioactive iodine will have decayed to half the original amount.

f(t)=ae^(kt)

the original amount of radioactive iodine is 45 pounds when t = 0 follows f(0) = 45 that is a = 45

so the formula becomes f(t)=45e^(kt)

since the half-life is 60 days, when t = 60, f(60) = 45/2. Substituting these values into the formula allows us to solve for k:

45e^(60k) = 45/2

e^(60k) = 1/2

60k = ln(1/2)

k = (1/60)*ln(1/2)

k = - 0.01155 approximately

f(t) = 45e^(- 0.01155t) is the function we are looking for

How long will it take for the radioactive iodine to decay to a level of 20% of the original amount released?

20% of 45 is 0.2*45 = 9 pounds

we need to find the time t when the amount is 9 grams, that is when f(t) = 9

45e^(- 0.01155t) = 9

e^(- 0.01155t) = 1/5

- 0.01155t = ln (1/5)

t = - (1/0.01155) * ln(1/5)

t = 139.354 days

So, 45 pounds of radioactive iodine will have decayed to 9 pounds after about 139.4 days.