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 2: Hospitals utilize the radioactive substance iodine-131 in the diagnosis of conditions of the thyroid gland. The half-life of iodine-131 is eight days.




Solution and Explanation:



First, what does it mean to say that the half-life of iodine-131 is eight days ?


It means that after eight days only half of the original amount remains. After another eight days one-half of that amount remains. Another way of saying this is that after 16 days 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 days that have passes and the remaining amount of the iodine-131.


\begin{eqnarray*}&& \\
0 \mbox{ days } &:&{\quad }2{ g} \\
&& \\
8 \mbox{ day...
...{\quad }\displaystyle \frac{1}{2}\left( 1{ g}\right) =0.50
{ g}
\end{eqnarray*}


\begin{eqnarray*}24 \mbox{ days } &:&{\quad }\displaystyle \frac{1}{2}\left( 0.5...
...d }\displaystyle \frac{1}{2}\left( 0.125{ g}\right)
=0.0625{ 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 2 g. We can say the same thing with the equation

\begin{eqnarray*}&& \\
f\left( 0\right) &=&2. \\
&& \\
&&
\end{eqnarray*}


The equation is now

\begin{eqnarray*}&& \\
f\left( t\right) &=&2\cdot e^{bt}. \\
&& \\
&&
\end{eqnarray*}


After 8 days, there is only 1 g left. Another way to say this is

\begin{eqnarray*}&& \\
f\left( 8\right) &=&2\cdot e^{b\left( 8\right) } \\
&& \\
f(8) &=&1
\end{eqnarray*}

\begin{eqnarray*}1 &=&2\cdot e^{b\left( 8\right) } \\
&& \\
\displaystyle \frac{1}{2} &=&e^{b\left( 8\right) } \\
&& \\
&&
\end{eqnarray*}


Take the natural logarithm of both sides of the equation what you started The model now can be written as

\begin{eqnarray*}&&\\
\displaystyle \frac{1}{2}&=&e^{b\left( 8\right) } \\
&&\...
...le \frac{1}{2}\right) &=&\ln \left( e^{b\left( 8\right) }\right)
\end{eqnarray*}

\begin{eqnarray*}\ln \left( \displaystyle \frac{1}{2}\right) &=&8b\cdot \ln \lef...
...
&&\\
\ln \left( \displaystyle \frac{1}{2}\right) &=&8b\cdot 1
\end{eqnarray*}

\begin{eqnarray*}b&=&\displaystyle \frac{\ln \left( \displaystyle \frac{1}{2}\ri...
...\
&&\\
f\left( t\right)&=&2\cdot e^{-0.086643\left( t\right) }
\end{eqnarray*}


The decay constant is $b=-0.086643.\bigskip\bigskip\bigskip $



How much of the iodine-131 will remain after 20 days? Just replace t in the equation with 20.

\begin{eqnarray*}&& \\
f\left( t\right) &=&2\cdot e^{-0.086643\left( t\right) }...
... \\
f\left( 20\right) &=&2\cdot e^{-0.086643\left( 20\right) }
\end{eqnarray*}

\begin{eqnarray*}&=&2\cdot \left( 0.176778\right) \\
&& \\
&=&0.353556{ } \\
&& \\
&& \\
&&
\end{eqnarray*}



After 20 days, there will be 0.353556 g left.


How long will it be until only 0.01 g remains? Replace $f\left( t\right) $ with 0.01 g and solve for t.

\begin{eqnarray*}&& \\
f\left( t\right) &=&2\cdot e^{-0.086643\left( t\right) }...
...c{0.01}{2} &=&e^{-0.086643\left( t\right) } \\
&& \\
&& \\
&&
\end{eqnarray*}



Take the natural logarithm of both sides of the equation.

\begin{eqnarray*}&& \\
\ln \left( \displaystyle \frac{0.01}{2}\right) &=&\ln \l...
...ystyle \frac{0.01}{2}\right) &=&-0.086643\left( t\right) \cdot 1
\end{eqnarray*}

\begin{eqnarray*}t &=&\displaystyle \frac{\ln \left( \displaystyle \frac{0.01}{2...
...t) }{-0.086643} \\
&& \\
t &\approx &61.151130{ \mbox{ days }}
\end{eqnarray*}



It would take a little over 61 days for the sample to be reduced to 0.01 g. To be more specific, it would take

\begin{eqnarray*}&& \\
&& \\
61.151130 \mbox{ days } &=&61{ }\mbox{ days }+0.1...
...
\mbox { day }} &=&61{ }\mbox{ days }+3.627120{ }\mbox{ hours }
\end{eqnarray*}

\begin{eqnarray*}61{ }\mbox{ days }+3.627120{ }\mbox{ hours } &=&61{ }\mbox{ day...
...\mbox{ hours }{ }37.6272{ }\mbox{ minutes } \\
&& \\
&& \\
&&
\end{eqnarray*}



It would take 61 days, 3 hours, and about 37 minutes for the 2 g sample to be reduced to 0.01 g.



Let's us check this answer

\begin{eqnarray*}&& \\
f\left( 61.151130\right) &=&2\cdot e^{-0.086643\left( 61.151130\right) } \\
&& \\
&=&0.01 \\
&& \\
&&
\end{eqnarray*}




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.


If you would like to go back to the table of contents, click on contents. 

[Exponential Rules] [Logarithms]

[Algebra] [Trigonometry ] [Complex Variables]

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Author: Nancy Marcus

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