Question

The air in a factory is being filtered so that the quantity
of a pollutant, P (in mg/liter), is decreasing according to
the function P = P(naught)e^−kt, where t is time in hours. If
10% of the pollution is removed in the first five hours:
(a) How long is it before the pollution is reduced by 50%?
(b) Explain why the quantity of pollutant might decrease in this way.

Answer 1

First you need to figure out what k is.
\[\large P = P_0e^{−kt}\]
Since the pollution goes down by 10% in 5 hours this can be changed to:
\[\large 0.90P = Pe^{−k(5)}\]Reduce:
\[\large 0.90=e^{−k(5)}\]\[\large ln(0.90)=ln(e^{−k(5)})\]\[\large ln(0.90)=-5k\]\[\large k=\frac{ln(0.90)}{-5}=0.0210721031316\]Now that you know k you can solve when the pollution will be decreased to 50%:\[\large 0.50=e^{-0.0210721031316t}\]
Just solve that for t using the same method we used to solve for k.

For the second part of the question you just need to understand that this is a decay function because e is raised to a -k value.