Figure

EXAMPLE 8Application: Decomposition of Salt in Water

Salt (NaCl) decomposes in water into sodium (Na $^{+})$ ions and chloride (Cl $^{-})$ ions according to the law of uninhibited decay $\begin{equation*} A( t) =A_{0}e^{kt} \end{equation*}$

where $A=A( t)$ is the amount (in kilograms) of salt present at time $t$ (in hours), $A_{0}$ is the original amount of salt in the solution, and $k$ is a negative number that represents the rate of decomposition.

(a) If initially there are 25 kilograms (kg) of salt and after 10 hours (h) there are 15 kg of salt remaining, how much salt is left after one day?
How long will it take until $\dfrac{1}{2} {kg}$ of salt remains?
Find $\lim\limits_{t\rightarrow \infty }A( t)$ .
(d) Interpret the answer found in (c).

Solving exponential equations is discussed in Section P.5, pp. 45-46.

Solution

Initially, there are 25 kg of salt, so $A(0) =A_{0}=25$ . To find the number $k$ in $A(t)=A_{0}e^{kt}$ , we use the fact that at $t=10$ , then $A(10) =15$ . That is, $\begin{eqnarray*} A( 10) &=&15=25e^{10k}\qquad {\color{#0066A7}{\hbox{$A (t) =A_{0}{e}^{kt}, A_{0}=25; A(10)=15$}}}\\[3pt] e^{10k} &=&\dfrac{3}{5} \\[3pt] 10k &=&\ln \dfrac{3}{5} \\[3pt] k &=&\dfrac{1}{10}\ln 0.6 \end{eqnarray*}$
So, $A(t)=25e^{(\frac{1}{10}\ln 0.6)t}$ . The amount of salt that remains after one day (24 h) is $\begin{equation*} A( 24) =25e^{(\frac{1}{10}\ln 0.6) 24}\approx 7.337 kilograms \end{equation*}$

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We want to find $t$ so that $A(t) =\dfrac{1}{2} kg$ . Then $\begin{eqnarray*} \dfrac{1}{2} &=&25e^{(\frac{1}{10}\ln 0.6)t} \\[3pt] e^{(\frac{1}{10}\ln 0.6)t} &=&\dfrac{1}{50} \\[3pt] \left(\dfrac{1}{10}\ln 0.6\right)t &=&\ln \dfrac{1}{50} \\[3pt] t &\approx &76.582 \end{eqnarray*}$
After approximately $76.6{h}$ , $\dfrac{1}{2}{kg}$ of salt will remain.
Since $\dfrac{1}{10}\ln 0.6\approx -0.051$ , we have $\lim\limits_{t\rightarrow \infty }A( t) =\lim\limits_{t\rightarrow \infty }(25e^{-0.051t})=\lim\limits_{t\rightarrow \infty }\dfrac{25}{e^{0.051t} }=0$
As $t$ becomes unbounded, the amount of salt in the water approaches 0 kg. Eventually, there will be no salt present in the water.