APPLICATIONS OF EXPONENTIAL
AND
LOGARITHMIC FUNCTIONS



DECAY WORD PROBLEMS:




Problem 3: If Jonly, Iowa had a population of 120,000 in 1945 and a population of 65,000 in 1996, how large was the population in 1960 assuming an exponential population model?



Answer: 100,200



Solution:


Since the population is declining, you know that the growth factor is negative and is called the decay factor or the decay constant. Since we are assuming an exponential model, it means that an exponential equation will describe the size of the population at any time between 1945 and 1996. The generic population model is

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



We need to solve for the value of a and the value of b.


We started examining the population size in 1945, so let t=0 in 1945. This means that when t=0, the population size was 120,000. A mathematical way of saying this is , $f\left( 0\right) =120,000.$ Substitute these values in the above equation.

\begin{eqnarray*}&& \\
f\left( t\right) &=&a\cdot e^{bt}\\
&&\\
f\left( 0\right) &=&a\cdot e^{b\left( 0\right) }
\end{eqnarray*}

\begin{eqnarray*}f(0) &=&120,000 \\
&& \\
120,000 &=&a\cdot e^{b\left( 0\right) } \\
&& \\
120,000 &=&a\cdot 1=a
\end{eqnarray*}



The equation can now be rewritten as

\begin{eqnarray*}f\left( t\right) &=&120,000\cdot e^{bt}
\end{eqnarray*}


We know that in 1996 ( t=1996-1945=51), the population was 65,000. A mathematical way of saying this is

\begin{eqnarray*}f\left( 51\right) &=&65,000 \\
&& \\
&&\mbox{ and } \\
&& \\
f\left( 51\right) &=&120,000\cdot e^{b\left( 51\right) }
\end{eqnarray*}

\begin{eqnarray*}&&\mbox{ so } \\
&& \\
65,000 &=&120,000\cdot e^{b\left( 51\right) }
\end{eqnarray*}


Let's solve for b.


\begin{eqnarray*}\frac{65,000}{120,000} &=&e^{b\left( 51\right) }
\end{eqnarray*}



Take the natural logarithm of both sides of the equation.


\begin{eqnarray*}\frac{65,000}{120,000} &=&e^{b\left( 51\right) } \\
&& \\
&& ...
...000}{120,000}\right) &=&\ln \left( e^{b\left( 51\right)
}\right)
\end{eqnarray*}

\begin{eqnarray*}\ln \left( \frac{65,000}{120,000}\right) &=&51b\cdot \ln \left(...
...\
&& \\
\ln \left( \frac{65,000}{120,000}\right) &=&51b\cdot 1
\end{eqnarray*}

\begin{eqnarray*}b &=&\frac{\ln \left( \frac{65,000}{120,000}\right) }{51} \\
&& \\
&& \\
b &\approx &-0.012021656331
\end{eqnarray*}



The equation describing the population size at any years between 1945 and 1996 is


\begin{eqnarray*}f\left( t\right) &=&120,000\cdot e^{-0.012021656331\cdot t}
\end{eqnarray*}


In 1960, t=1960-1945=15. To find the population in 1960, simply substitute 15 for t in the above equation.

\begin{eqnarray*}f\left( 15\right) &=&120,000\cdot e^{-0.012021656331\cdot \left...
...ght) }
\\
&& \\
&=&100,199.870659 \\
&& \\
&\approx &100,200
\end{eqnarray*}



The population in 1960 was about 100,200.





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Author: Nancy Marcus

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