APPLICATIONS OF EXPONENTIAL
AND
LOGARITHMIC FUNCTIONS



DECAY WORD PROBLEMS:



To solve an exponential or logarithmic word problem, convert the narrative to an equation and solve the equation.



Example 3: The radioactive isotope sodium-24 is used as a tracer to measure the rate of flow in an artery or vein. The half-life of sodium-24 is 14.9 hours. Suppose that a hospital buys a 40-g sample of sodium-24 and will reorder when the sample is reduced to 3 g.




Solution and Explanation:



First, what does it mean to say that the half-life of sodium-24 is 14.9 hours? It means that after 14.9 hours only half of the original amount remains. After another 14.9 hours one-half of that one-half amount remains. Another way of saying that is that after 29.8 hours only $\displaystyle \frac{1}{2}$ of $\displaystyle \frac{1}{2}$ or $\displaystyle \frac{1}{4}$ of the original amount remains. Make a table showing the relationship between the number of hours that have passes and the amount of sodium-24 remaining.

\begin{eqnarray*}&& \\
&& \\
0\mbox{ hours } &:&{\quad 40 g} \\
&& \\
14.9\m...
...&{\quad }\displaystyle \frac{1}{2}\left( 20{ g}\right) =10
{ g}
\end{eqnarray*}

\begin{eqnarray*}44.7\mbox{ hours } &:&{\quad }\displaystyle \frac{1}{2}\left( 1...
...\quad }\displaystyle \frac{1}{2}\left( 2.5{ g}\right) =1.25
{ g}
\end{eqnarray*}


Let's us form the equation with base e:


\begin{eqnarray*}&& \\
f\left( t\right) &=&a\cdot e^{bt} \\
&& \\
&&
\end{eqnarray*}


At time 0, the hospital had 40 g. We can say the same thing with the equation

\begin{eqnarray*}&& \\
f\left( 0\right) &=&40. \\
&& \\
&&
\end{eqnarray*}


The equation is now

\begin{eqnarray*}&& \\
f\left( t\right) &=&40\cdot e^{bt}. \\
&& \\
&&
\end{eqnarray*}


After 14.9 hours,there is only $\displaystyle \frac{1}{2}$ of 40=20 g left. Another way of say this is

\begin{eqnarray*}&& \\
f\left( 14.9\right) &=&20 \\
&& \\
f(14.9) &=&40\cdot e^{b\left( 14.9\right) }
\end{eqnarray*}


\begin{eqnarray*}\displaystyle \frac{20}{40} &=&e^{b\left( 14.9\right) } \\
&& \\
\displaystyle \frac{1}{2} &=&e^{14.9b} \\
&& \\
&&
\end{eqnarray*}


Take the natural log of both sides of the equation.

\begin{eqnarray*}&& \\
\ln \left( \displaystyle \frac{1}{2}\right) &=&\ln \left...
...&\\
\ln \left( \displaystyle \frac{1}{2}\right) &=&14.9b\cdot 1
\end{eqnarray*}

\begin{eqnarray*}b &=&\displaystyle \frac{\ln \left( \displaystyle \frac{1}{2}\right) }{14.9} \\
&& \\
b &\approx &-0.046520
\end{eqnarray*}


The equation can be written as

\begin{eqnarray*}f\left( t\right) &=&40\cdot e^{-0.046520t}
\end{eqnarray*}


The decay constant is $b=-0.046520.\bigskip\bigskip $



How much of the sodium-24 will remain after 50 hours? Just replace t in the formula with 50.

\begin{eqnarray*}f\left( 50\right) &=&40\cdot e^{-0.046520\left( 50\right) } \\ ...
...\\
&=&40\cdot \left( 0.097686\right) \\
&& \\
&=&3.907428{ g}
\end{eqnarray*}


How often before the hospital has to reorder sodium-24? or How long will it before the sample is reduced to 3 g? Replace $f\left( t\right) $ with 3 g and solve for t.

\begin{eqnarray*}f\left( t\right) &=&40\cdot e^{-0.046520t} \\
&& \\
3 &=&40\c...
...046520t} \\
&& \\
\displaystyle \frac{3}{40} &=&e^{-0.046520t}
\end{eqnarray*}


Take the natural logarithm of both sides of the equation.

\begin{eqnarray*}\displaystyle \frac{3}{40} &=&e^{-0.046520t} \\
&& \\
\ln \le...
...laystyle \frac{3}{40}\right) &=&\ln \left( e^{-0.046520t}\right)
\end{eqnarray*}

\begin{eqnarray*}\ln \left( \displaystyle \frac{3}{40}\right) &=&-0.046520t\cdot...
...ln \left( \displaystyle \frac{3}{40}\right) &=&-0.046520t\cdot 1
\end{eqnarray*}

\begin{eqnarray*}t &=&\displaystyle \frac{\ln \left( \displaystyle \frac{3}{40}\right) }{-0.046520} \\
&& \\
t &\approx &55.680722\mbox{ hours }
\end{eqnarray*}


How much of the sample will remain after 1 year?


First convert 1 year to hours.

\begin{eqnarray*}1\mbox{ year }\cdot \displaystyle \frac{365\mbox{ days }}{1\mbo...
...le \frac{24\mbox{ hours }}{1\mbox{ day }} &=&8,760\mbox{
hours }
\end{eqnarray*}


Substitute 8,760 for t in the equation $f\left( t\right) =40\cdot
e^{-0.046520t}$.

\begin{eqnarray*}f\left( 8,760\right) &=&40\cdot e^{-0.046520\left( 8,760\right) } \\
&& \\
&=&4.17308604588\times 10^{-176}
\end{eqnarray*}


This is an extremely small number.




If you would like to work another example, click on example


If you would like to test your knowledge by working some problems, click on problem.


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Author: Nancy Marcus

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