APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

	APPLICATIONS OF EXPONENTIAL
	AND
	LOGARITHMIC FUNCTIONS

DECAY WORD PROBLEMS:

As with any word problem, the trick is convert a narrative statement or question to a mathematical statement.

Example 1: If you start a biology experiment with 5,000,000 cells and 45% of the cells are dying every minute, how long will it take to have less than 1,000 cells?

Solution:

Let's start by calculation some points, graphing the points, and then determining what mathematical model best describes the points. The number of remaining cells at any time depends on the amount of time since the experiment began. All the points will have the form (time since experiment began, number of remaining cells). This sounds complicated, but it is not

For example, at the start of the experiment no time has passed and time=0. At the start of the experiment there are 5,000,000 cells. Therefore the first point is $\left( 0,5,000,000\right) .$

One minute after the experiment starts, 45% of the five million cells die leaving 55% X 5,000,000 = 2,750,000 cells. The corresponding point is $\left( 1,2,750,000\right) .\bigskip\bigskip$

Two minutes after the experiment starts, 45% of the 2,750,000 cells die leaving 55% X 2,750,000 = 1,512,500 cells. The corresponding point is $\left( 2,1,512,500\right) .\bigskip\bigskip$

Three minutes after the experiment starts, 45% of 1,512,500 cells die leaving 55% X 1,512,500 = 831,875 cells. The corresponding point is $\left( 3,831,875\right) .\bigskip\bigskip$

When you plot these points, you note that the curve looks exponential. Therefore, the mathematical model is probably exponential.

A mathematical model is nothing more than a fancy word for an equation. An exponential equation looks something like the following equation:

$\begin{eqnarray*}&& \\ f\left( t\right) &=&a\cdot e^{bt} \\ && \end{eqnarray*}$ where f(t) represents the number of cells remaining t minutes after the experiment started, a represents the number of cells at the start of the experiment $\left( 5,000,000\right)$ ,t represents the numbers of minutes since the experiment began, and b represents the decay constant at time t based on a base of e. (Note that the decay constant changes when the base changes. See example at the end of this section.)

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

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

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

The equation is now modified:

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

We know that there are 2,750,000 cells after 1 minute. Another way of saying this is that $f\left( 1\right) =2,750,000.$ In the above equation, replace $f\left( 1\right)$ with 2,750,000 and replace t with 1.

$\begin{eqnarray*}&& \\ f\left( 1\right) &=&5,000,000\cdot e^{b\left( 1\right) }... ...000,000\cdot e^{b} \\ && \\ && \\ 0.55 &=&e^{b} \\ && \\ && \end{eqnarray*}$

Take the natural logarithm of both sides of the equation:

$\begin{eqnarray*}&&\\ 0.55 &=&e^{b} \\ && \\ \ln \left( 0.55\right) &=&\ln \l... ...\ln \left( 0.55\right) &=&b\cdot \ln \left( e\right) =b\cdot 1=b \end{eqnarray*}$

$\begin{eqnarray*}b &=&\ln \left( 0.55\right) \\ && \\ b &\approx &-0.597837 \\ && \\ \end{eqnarray*}$

The decay factor is -0.597837

The equation describing the number of cells remaining after an experiment has begun is

$\begin{eqnarray*}&& \\ f\left( t\right) &=&5,000,000\cdot e^{-0.597837\cdot t} \\ && \\ && \\ && \end{eqnarray*}$

Let's check it out by seeing if this model will give us 1,512,500 cells after two minutes.

$\begin{eqnarray*}&& \\ f\left( 2\right) &=&5,000,000\cdot e^{-0.597837\left( 2\... ...02291 \\ && \\ &\approx &1,512,500 \\ && \\ && \\ && \\ && \end{eqnarray*}$

The model is $f\left( t\right) =5,000,000\cdot e^{-0.597837\left( t\right) }$

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 and solve for t.

$\begin{eqnarray*}&& \\ f\left( t\right) &=&5,000,000\cdot e^{-0.597837\left( t\... ...1,000}{5,000,000} &=&e^{-0.597837\left( t\right) } \\ && \\ && \end{eqnarray*}$

Take the natural logarithm of both sides of the equation.

$\begin{eqnarray*}&& \\ \displaystyle \frac{1,000}{5,000,000} &=&e^{-0.597837\le... ...,000}\right) &=&\ln \left( e^{-0.597837\left( t\right) }\right) \end{eqnarray*}$

$\begin{eqnarray*}\ln \left( \displaystyle \frac{1,000}{5,000,000}\right) &=&-0.5... ...displaystyle \frac{1,000}{5,000,000}\right) &=&-0.597837t\cdot 1 \end{eqnarray*}$

$\begin{eqnarray*}t &=&\displaystyle \frac{\ln \left( \displaystyle \frac{1,000}{... ...7837} \\ && \\ && \\ t &\approx &14.2466812717 \\ && \\ && \end{eqnarray*}$

It will take about 14.25 minutes for the cell population to drop below a 1,000 count.

Let's check our answer:

In the mathematical model $f\left( t\right) =5,000,000\cdot e^{-0.597837\left( t\right) },$ everywhere there is a t, substitute 14.25. If the answer is less than 1,000, our answer is correct.

$\begin{eqnarray*}&& \\ f\left( t\right) &=&5,000,000\cdot e^{-0.597837\left( t\... ...37\left( 14.25\right) }\\ &&\\ &=&998.0179<1,000 \\ && \\ && \end{eqnarray*}$

Relationship between base and decay constant:

Suppose you have an equation $f\left( t\right) =75\cdot e^{4t}.$ Recall that you can rewrite this equation as

$\begin{eqnarray*}&& \\ f\left( t\right) &=&75\cdot \left( e^{4}\right) ^{t} \\ ... ...right) &=&75\cdot \left( 54.5981500331\right) ^{6} \\ && \\ && \end{eqnarray*}$

Suppose we wanted a base of 7 instead of a base of e. In order words, we wanted our equation to look like $f\left( t\right) =75\cdot 7^{bt}.$ This can re rewritten as $f\left( t\right) =75\cdot \left( 7^{b}\right) ^{6}.$ If we want both equations to represent the same thing, we must find a value for b such that 7^b=54.5981500331 or something very close to this quantity.

$\begin{eqnarray*}7^{b} &=&54.5981500331 \\ && \\ \log \left( 7^{b}\right) &=&\log \left( 54.5981500331\right) \end{eqnarray*}$

$\begin{eqnarray*}b\cdot \log \left( 7\right) &=&\log \left( 54.5981500331\right)... ... \frac{\log \left( 54.5981500331\right) }{\log \left( 7\right) } \end{eqnarray*}$

$\begin{eqnarray*}b &\approx &2.05559336948 \\ && \\ && \\ f\left( t\right) &=&75\cdot 7^{2.05559336948t} \\ && \\ && \\ && \end{eqnarray*}$

Note that $f\left( t\right) =75\cdot 7^{2.05559336948t}$ can be rewritten as $f\left( t\right) =75\cdot \left( 7^{2.05559336948}\right) ^{t}=75\cdot \left( 54.981500333\right) ^{t}.\bigskip\bigskip$

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.

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