# math calculus

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

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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

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

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