A defective nuclear reactor is leaking radioactive iodine-131. The rate of decay of iodine-131is given by the function defined by A(t)=A_0 e^(-0.087t), where t is time in days. How long will it take any quantity of iodine-131 to decay to 25% of its initial amount?

Answer 1

just solve for t:

e^(-.087t) = .25
t = 16.00 days

Alternatively, if you express the function as a power of 1/2, then since

1/2 = e^(ln 1/2) = e^(-ln2)
(1/2)^x = e^(-x/ln2) = e^(-x/.693)
e^(-.087t) = (1/2)^(.125t) = (1/2)^(t/8)
So, the half-life is 8 days
25% left means 2 half-lives, or 16 days

Answer 2

Thank you, Steve!