Suppose that a certain material has a half life of 25 years, and there are A(t)=10(1/2)^t/25 grams remaining after t years. find:
i) initial amount
i) Amount after 80 years
Solutions
i) A(0)=10(1/2)^(0/25)
=10(1/2)^0
=10(1)
=10
ii)A(80)=10(1/2)^(80/25)
=10(1/2)^(16/5)
=10(0.1088)
=1.088