Question 613570
N = P*e^(-kt)


1/2P = P*e^(-k*12.4)


1/2 = e^(-12.4k)


ln(1/2) = -12.4k


-0.693147 = -12.4k


-0.693147/(-12.4) = k


0.05589895 = k


k = 0.05589895

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N = P*e^(-kt)


N = P*e^(-0.05589895t)


0.8P = P*e^(-0.05589895t)


0.8 = e^(-0.05589895t)


ln(0.8) = -0.05589895t


-0.22314355 = -0.05589895t


-0.22314355/(-0.05589895) = t


3.9919095 = t


t = 3.9919095


So it will take roughly 3.9919095 years for 20% of the sample to decompose.