APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

	APPLICATIONS OF EXPONENTIAL
	AND
	LOGARITHMIC FUNCTIONS

DECAY WORD PROBLEMS:

Problem 1: If you start a biology experiment with 5,000,000 cells and half the cells are dying every 10 minutes, how long will it take to have less than 1,000 cells?

Answer: 123 minutes.

Solution:

At time 0, there are 5,000,000 cells. At time 10 minutes, there are 0.50 x 5,000,000 = 2,500,000 cells remaining. At time 20 minutes, there are 0.50 x 2,500,000 = 1,250,000 cells remaining. At time 30 minutes, there are 0.50 x 1,250,000 = 625,000 cells remaining. When you plot the data, the curve looks exponential. Therefore, the mathematical model is probably exponential The model looks something like

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

where $f\left( t\right)$ represents the number of cells remaining after t minutes of observation, a represents the number of cells at the start of the experiment (5,000,000) , t represents the numbers of minutes since the experiment began, and b represents the decay constant based on a base of e.

We know that a=5,000,000 because we started with five million cells. However, you can verify it in the equation $f\left( t\right) =a\cdot e^{bt}.$ Let t=0 in the equation.

$\begin{eqnarray*}f\left( 0\right) &=&5,000,000 \\ &&\\ f(0) &=&a\cdot e^{b\cdot 0}=a\cdot e^{0} \\ && \\ 5,000,000 &=&a\cdot 1=a \end{eqnarray*}$

The equation is now modified:

$\begin{eqnarray*}f\left( t\right) &=&5,000,000\cdot e^{bt} \end{eqnarray*}$

We know that there are 2,500,000 cells after 10 minutes. Another way of saying this is that $f\left( 10\right) =2,500,000.$ In the above equation, replace $f\left( 10\right)$ with 2,500,000 and replace t with 10.

$\begin{eqnarray*}f\left( 10\right) &=&5,000,000\cdot e^{b\left( 10\right) } \\ && \\ f(10) &=&2,500,000 \end{eqnarray*}$

$\begin{eqnarray*}2,500,000 &=&5,000,000\cdot e^{b\left( 10\right) } \\ && \\ && \\ \displaystyle \frac{1}{2} &=&e^{b\left( 10\right) } \end{eqnarray*}$

Take the natural logarithm of both sides of the equation:

$\begin{eqnarray*}\displaystyle \frac{1}{2} &=&e^{b\left( 10\right) } \\ && \\ ... ...e \frac{1}{2}\right) &=&\ln \left( e^{b\left( 10\right) }\right) \end{eqnarray*}$

$\begin{eqnarray*}\ln \left( \displaystyle \frac{1}{2}\right) &=&10b\cdot \ln \le... ...&& \\ \ln \left( \displaystyle \frac{1}{2}\right) &=&10b\cdot 1 \end{eqnarray*}$

$\begin{eqnarray*}b &=&\displaystyle \frac{\ln \left( \displaystyle \frac{1}{2}\right) }{10} \\ && \\ && \\ b &\approx &-0.069314718 \end{eqnarray*}$

The equation describing the number of cells remaining after a certain number of minutes is

$\begin{eqnarray*}f\left( t\right) &=&5,000,000\cdot e^{-0.069314718\cdot t} \end{eqnarray*}$

Let's check it out by seeing if this model will give us 1,250,000 cells after twenty minutes.

$\begin{eqnarray*}f\left( 20\right) &=&5,000,000\cdot e^{-0.069314718\left( 20\right) } \\ && \\ &=&1,250,000.0014 \\ && \\ &\approx &1,250,000 \end{eqnarray*}$

The model is $f\left( t\right) =5,000,000\cdot e^{-0.069314718\left( t\right) }$

How long will it take the sample to decay to below 1,000 cells? Just substitute 1,000 for $f\left( t\right)$ in the equation.

$\begin{eqnarray*}f\left( t\right) &=&5,000,000\cdot e^{-0.069314718\left( t\righ... ...tyle \frac{1,000}{5,000,000} &=&e^{-0.069314718\left( t\right) } \end{eqnarray*}$

Take the natural logarithm of both sides of the equation.

$\begin{eqnarray*}\displaystyle \frac{1,000}{5,000,000} &=&e^{-0.069314718\left( ... ...00}\right) &=&\ln \left( e^{-0.069314718\left( t\right) }\right) \end{eqnarray*}$

$\begin{eqnarray*}\ln \left( \displaystyle \frac{1,000}{5,000,000}\right) &=&-0.0... ...splaystyle \frac{1,000}{5,000,000}\right) &=&-0.06931478t\cdot 1 \end{eqnarray*}$

$\begin{eqnarray*}t &=&\displaystyle \frac{\ln \left( \displaystyle \frac{1,000}{... ...ight) }{-0.06931478} \\ && \\ && \\ t &\approx &122.877013985 \end{eqnarray*}$

It will take about 123 minutes for the cell population to drop below a 1,000 count.

If you would like to review problem 2, click on problem 2.

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