Question

In an exponential decay process given by M = M0 e-kt
the original amount M0
has been reduced by a factor 16 in 321 days.
How many days did it take to be reduced by a factor of2? What is the value of k ?

Answer 1

Answer to Question #84208 – Math – Algebra

Question

In an exponential decay process given by $M = M_0 e^{-kt}$ the original amount $M_0$ has been reduced by a factor 16 in 321 days. How many days did it take to be reduced by a factor of 2? What is the value of $k$?

Solution

$$\frac{M_0}{16} = M_0 e^{-321k}$$ $$\frac{1}{16} = e^{-321k}$$ $$\ln \frac{1}{16} = -321k$$ $$k = -\frac{1}{321} \cdot \ln \frac{1}{16} = 0.00864$$ $$M = M_0 e^{-0.00864t}$$ $$\frac{M_0}{2} = M_0 e^{-0.00864t}$$ $$\frac{1}{2} = e^{-0.00864t}$$ $$\ln \frac{1}{2} = -0.00864t$$ $$t = -\frac{1}{0.00864} \cdot \ln \frac{1}{2} = 80.225 \approx 80 \text{ days}.$$

Answer: about 80 days, $k = -\frac{1}{321} \cdot \ln \frac{1}{16} = 0.00864$.

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