Find the half-life (in hours) of a radioactive substance that is reduced by 30 percent in 40 hours.

Answer 1

a = A e^-kt

.3 = e^-40 k

ln .7 = -.357 = -40 k

k = .008917
so

1/2 = e^-.008917 t if t is half life

ln .5 = - .693 = -.008917 t

t = 77.7 hours

Answer 2

typo in second line
.7 = e^-40 k