91Ó°ÊÓ

Under certain water conditions, the free chlorine (hypochlorous acid, HOCl) in a swimming pool decomposes according to the law of uninhibited decay. After shocking his pool, Ben tested the water and found the amount of free chlorine to be 2.5 parts per million (ppm). Twenty-four hours later, Ben tested the water again and found the amount of free chlorine to be 2.2 ppm.

The amount of material present at a time t is given by the function $A\left(t\right)=A_0{e}^{kt}$

where $k<0$represents the decay rate.

Substitute role="math" localid="1647398653281" $A_0=2.5,A\left(t\right)=2.2,t=1$in the function role="math" localid="1647398711504" $A\left(t\right)=A_0{e}^{kt}$

$$\begin{gathered} A\left(t\right)=A_0{e}^{kt} \\ 2.2=2.5e^{k1} \\ \frac{2.2}{2.5}=e^k \\ \ln \left(\frac{2.2}{2.5}\right)=\ln \left(e^k\right) \\ \ln \left(\frac{2.2}{2.5}\right)=k \\ k=-0.1278 \end{gathered}$$

Reading after 3 days can be calculated by substituting role="math" localid="1647398851938" $A_0=2.5,2,t=3,k=-0.1278$in the function role="math" localid="1647398900359" $A\left(t\right)=A_0{e}^{kt}$

$$\begin{gathered} A\left(t\right)=A_0{e}^{kt} \\ A\left(t\right)=2.5e^{-0.1278×3} \\ A\left(t\right)=1.7 \end{gathered}$$

Substitute $A_0=2.5,2,A\left(t\right)=1.0,k=-0.1278$in the function $A\left(t\right)=A_0{e}^{kt}$

localid="1647399148322" $$\begin{gathered} A\left(t\right)=A_0{e}^{kt} \\ 1.0=2.5e^{-0.1278×t} \\ \frac{1.0}{2.5}=e^{-0.1278×t} \\ 0.4=e^{-0.1278×t} \\ \ln \left(0.4\right)=\ln \left(e^{-0.1278×t}\right) \\ \ln \left(0.4\right)=-0.1278×t \\ t=\frac{\ln \left(0.4\right)}{-0.1278} \\ t=7.17 \end{gathered}$$