how do you solve an equation using Q(t)=Qoe^kt

Question

The half-life of a radioactive substance is 8.93 years. If 6 grams of the substance is present now, how much will be present in 15 years?

Answer 1

1/2 = e^8.93r where k=r= rate of change measured in grams per year

take natural logs

ln.5 = 8.93r

r = (ln.5)/8.93= about-.078= -7.8% per year

plug in below for k=r, with t=15, Qo=6

Q(t)=Qo(e^kt)

Q(15)= 6e^15r

= about 1.87284468 grams

Answer 2

Q(15 years) = (6 grams)*(1/2)^(15/8.93)

Q(15) ≈ 1.8728446796 grams