5.7 Applications of Exponential and Logarithmic Functions

Vanessa Coffelt

5.7 Applications of Exponential and Logarithmic Functions

As we mentioned in Sections 5.2 and 5.3, exponential and logarithmic functions are used to model a wide variety of behaviors in the real world. In the examples that follow, note that while the applications are drawn from many different disciplines, the mathematics remains essentially the same. Due to the applied nature of the problems we will examine in this section, we will often express our final answers as decimal approximations (after finding exact answers first, of course!)

Equation 5.1 Simple Interest

The amount of interest $I$ accrued at an annual rate $r$ on an investment^[1] $P$ after $t$ years is

$I = Prt$

The amount in the account after $t$ years, $A(t)$ is given by

$A(t) = P + I = P + Prt = P(1+rt)$

5.7.1 Applications of Exponential Functions

Perhaps the most well-known application of exponential functions comes from the financial world. Suppose you have 100 dollars to invest at your local bank and they are offering a whopping $5 \, \%$ annual percentage interest rate. This means that after one year, the bank will pay you $5 \%$ of that 100 dollars, or 100(0.05) = 5 dollars in interest, so you now have 105 dollars. This is in accordance with the formula for simple interest which you have undoubtedly run across at some point before.

Suppose, however, that six months into the year, you hear of a better deal at a rival bank.^[2] Naturally, you withdraw your money and try to invest it at the higher rate there. As six months is one half of a year, that initial 100 dollars yields $100(0.05)\left(\frac{1}{2}\right) = 2.50$ dollars in interest.

You take your 102.50 dollars off to the competitor and find out that those restrictions which may apply actually $\underline{do}$ apply, so you return to your bank and re-deposit the 102.50 dollars for the remaining six months of the year.

For the first six months of the year, interest was earned on the original principal of 100 dollars, but for the second six months, interest was earned on 102.50 dollars, that is, you earned interest on your interest. This is the basic concept behind compound interest.

In the previous discussion, we would say that the interest was compounded twice per year, or semiannually.^[3] If more money can be earned by earning interest on interest already earned, one wonders what happens if the interest is compounded more often, say every three months – 4 times a year, or `quarterly.’

In this case, the money is in the account for three months, or $\frac{1}{4}$ of a year, at a time. After the first quarter, we have

$A = P(1+rt) = 100 \left(1 + 0.05 \cdot \frac{1}{4} \right) = 101.25 \text{ dollars}$

We now invest the 101.25 dollars for the next three months and find that at the end of the second quarter, we have

$A = 101.25 \left(1 + 0.05 \cdot \frac{1}{4} \right)\approx 102.51 \text{ dollars}$

Continuing in this manner, the balance at the end of the third quarter is 103.79 dollars, and, at last, we obtain 105.08 dollars. The extra two cents hardly seems worth it, but we see that we do in fact get more money the more often we compound.

In order to develop a formula for this phenomenon, we need to do some abstract calculations. Suppose we wish to invest our principal $P$ at an annual rate $r$ and compound the interest $n$ times per year. This means the money sits in the account $\frac{1}{n}^{\text{\tiny th}}$ of a year between compoundings. Let $A_{k}$ denote the amount in the account after the $k^{\text{\tiny th}}$ compounding.

Then $A_{1} = P\left(1 + r\left(\frac{1}{n}\right)\right)$ which simplifies to $A_{1} = P \left(1 + \frac{r}{n}\right)$ . After the second compounding, we use $A_{1}$ as our new principal and get $A_{2} = A_{1} \left(1 + \frac{r}{n}\right) = \left[P \left(1 + \frac{r}{n}\right)\right]\left(1 + \frac{r}{n}\right) = P \left(1 + \frac{r}{n}\right)^2$ . Continuing in this fashion, we get $A_{3} =P \left(1 + \frac{r}{n}\right)^3$ , $A_{4} =P \left(1 + \frac{r}{n}\right)^4$ , and so on, so that $A_{k} = P \left(1 + \frac{r}{n}\right)^k$ .

As we compound the interest $n$ times per year, after $t$ years, we have $nt$ compoundings. We have just derived the general formula for compound interest below.

Equation 5.2 Compounded Interest

If an initial principal $P$ is invested at an annual rate $r$ and the interest is compounded $n$ times per year, the amount in the account after $t$ years, $A(t)$ is given by

$A(t) = P \left(1 + \frac{r}{n}\right)^{nt}$

If we take $P = 100$ , $r = 0.05$ , and $n = 4$ , Equation 5.2 becomes $A(t) = 100\left(1+ \frac{0.05}{4}\right)^{4t}$ which reduces to $A(t) = 100(1.0125)^{4t}$ .

To check this new formula against our previous calculations, we find $A\left(\frac{1}{4}\right) = 100(1.0125)^{4 \left(\frac{1}{4}\right)} = 101.25$ , $A\left(\frac{1}{2}\right) \approx 102.51$ dollars, $A\left(\frac{3}{4}\right) \approx 103.79$ dollars, and $A(1) \approx 105.08$ dollars.

Example 5.7.1

Example 5.7.1.1

Suppose 2000 dollars is invested in an account which offers $7.125 \%$ compounded monthly.

Express the amount $A(t)$ in the account as a function of the term of the investment $t$ in years.

Solution:

Express the amount $A(t)$ in the account as a function of the term of the investment $t$ in years.

Substituting $P = 2000$ , $r = 0.07125$ , and $n = 12$ (interest is compounded monthly) into Equation 5.2 yields

$\begin{array}{rcl} A(t) & = & 2000\left(1 + \frac{0.07125}{12}\right)^{12t} \\[6pt] & = & 2000 (1.0059375)^{12t} \end{array}$

Example 5.7.1.2

Suppose 2000 dollars is invested in an account which offers $7.125 \%$ compounded monthly.

How much is in the account after $5$ years?

Solution:

How much is in the account after $5$ years?

To find the amount in the account after $5$ years, we compute

$\begin{array}{rcl} A(5) &=& 2000 (1.0059375)^{12(5)} \\[6pt] & \approx & 2852.92 \end{array}$

After $5$ years, we have approximately 2852.92 dollars.

Example 5.7.1.3

Suppose 2000 dollars is invested in an account which offers $7.125 \%$ compounded monthly.

How long will it take for the initial investment to double?

Solution:

How long will it take for the initial investment to double?

Our initial investment is 2000 dollars, so to find the time it takes this to double, we need to find $t$ when $A(t) = 4000$ . That is, we need to solve

$\begin{array}{rcl} 2000 (1.0059375)^{12t} &= &4000 \\[6pt] (1.0059375)^{12t} &= & 2 \end{array}$

Taking natural logs as in Section 5.5, we get $t = \frac{\ln(2)}{12 \ln(1.0059375)} \approx 9.75$ .

Hence, it takes approximately 9 years 9 months for the investment to double.

Example 5.7.1.4

Suppose 2000 dollars is invested in an account which offers $7.125 \%$ compounded monthly.

Compute and interpret the average rate of change^[4] of the amount in the account:

from the end of the fourth year to the end of the fifth year
from the end of the thirty-fourth year to the end of the thirty-fifth year

Solution:

Find and interpret the average rate of change of the amount in the account.

Recall to find the average rate of change of $A$ over an interval $[a,b]$ , we compute $\frac{A(b)-A(a)}{b-a}$ .

Find and interpret the average rate of change of the amount in the account from the end of the fourth year to the end of the fifth year.
The average rate of change of $A$ from the end of the fourth year to the end of the fifth year is $\frac{A(5)-A(4)}{5-4} \approx 195.63$ .
This means that the value of the investment is increasing at a rate of approximately 195.63 dollars per year between the end of the fourth and fifth years.
Find and interpret the average rate of change of the amount in the account from the end of the thirty-fourth year to the end of the thirty-fifth year.
Likewise, the average rate of change of $A$ from the end of the thirty-fourth year to the end of the thirty-fifth year is $\frac{A(35)-A(34)}{35-34} \approx 1648.21$ , so the value of the investment is increasing at a rate of approximately 1648.21 dollars per year during this time.

So, not only is it true that the longer you wait, the more money you have, but also the longer you wait, the faster the money increases.^[5]

We have observed that the more times you compound the interest per year, the more money you will earn in a year. Let’s push this notion to the limit.^[6]

Consider an investment of 1 dollar invested at $100 \%$ interest for $1$ year compounded $n$ times a year. Equation 5.2 tells us that the amount of money in the account after $1$ year is $A = \left(1+\frac{1}{n}\right)^{n}$ . Below is a table of values relating $n$ and $A$ .

$\begin{array}{|r||r|} \hline n & A \\ \hline 1 & 2 \\ \hline 2 & 2.25 \\ \hline 4 & \approx 2.4414 \\ \hline 12 & \approx 2.6130 \\ \hline 360 & \approx 2.7145 \\ \hline 1000 & \approx 2.7169 \\ \hline 10000 & \approx 2.7181 \\ \hline 100000 & \approx 2.7182 \\ \hline \end{array}$

As promised, the more compoundings per year, the more money there is in the account, but we also observe that the increase in money is greatly diminishing.

We are witnessing a mathematical `tug of war’. While we are compounding more times per year, and hence getting interest on our interest more often, the amount of time between compoundings is getting smaller and smaller, so there is less time to build up additional interest.

With Calculus, we can show^[7] that as $n \rightarrow \infty$ , $A = \left(1+\frac{1}{n}\right)^{n} \rightarrow e$ , where $e$ is the natural base first presented in Section 5.2. Taking the number of compoundings per year to infinity results in what is called continuously compounded interest.

Theorem 5.10

Investing 1 dollar at $100 \%$ interest compounded continuously for one year returns $e$ dollars.

Using this definition of $e$ and a little Calculus, we can take Equation 5.2 and produce a formula for continuously compounded interest.

Equation 5.3 Continuously Compounded Interest

If an initial principal $P$ is invested at an annual rate $r$ and the interest is compounded continuously, the amount in the account after $t$ years, $A(t)$ is given by

$A(t) = P e^{rt}$

If we take the scenario of Example 5.7.1 and compare monthly compounding to continuous compounding over $35$ years, we find that monthly compounding yields $A(35) = 2000 (1.0059375)^{12(35)}$ which is about 24,035.28 dollars, whereas continuously compounding gives $A(35) = 2000e^{0.07125 (35)}$ which is about 24,213.18 dollars, a difference of less than $1 \%$ .

Equations 5.2 and 5.3 both use exponential functions to describe the growth of an investment. It turns out, the same principles which govern compound interest are also used to model short term growth of populations. As with many concepts in this text, these notions are best formalized using the language of Calculus. Nevertheless, we do our best here.

In Biology, The Law of Uninhibited Growth states as its premise that the instantaneous rate at which a population increases at any time is directly proportional to the population at that time.^[8] In other words, the more organisms there are at a given moment, the faster they reproduce. Formulating the law as stated results in a differential equation, which requires Calculus to solve. Solving said differential equation gives us the formula below.

Equation 5.4 Uninhibited Growth

If a population increases according to The Law of Uninhibited Growth, the number of organisms at time $t$ , $N(t)$ is given by the formula

$N(t) = N_{0}e^{kt},$

where $N(0) = N_{0}$ (read ` $N$ nought’) is the initial number of organisms and $k>0$ is the constant of proportionality which satisfies the equation:

$\left( \textit{ instantaneous rate of change of } N(t) \textit{ at time } t\right) = k \, N(t)$

It is worth taking some time to compare Equations 5.3 and 5.4. In Equation 5.3, we use $P$ to denote the initial investment; in Equation 5.4, we use $N_{0}$ to denote the initial population. In Equation 5.3, $r$ denotes the annual interest rate, and so it shouldn’t be too surprising that the $k$ in Equation 5.4 corresponds to a growth rate as well. While Equations 5.3 and 5.4 look entirely different, they both represent the same mathematical concept.

Example 5.7.2

In order to perform arthrosclerosis research, epithelial cells are harvested from discarded umbilical tissue and grown in the laboratory. A technician observes that a culture of twelve thousand cells grows to five million cells in one week. Assuming that the cells follow The Law of Uninhibited Growth, find a formula for the number of cells, in thousands, after $t$ days, $N(t)$ .

Solution:

We begin with $N(t) = N_{0}e^{kt}$ . Recall $N(t)$ gives the number of cells in thousands, so $N_{0} = 12$ and $N(t) = 12e^{kt}$ .

Next, we need to determine the growth rate $k$ . We know that after one week, the number of cells has grown to five million. As $t$ measures days and the units of $N(t)$ are in thousands, this translates mathematically to $N(7) = 5000$ or $12e^{7k} = 5000$ .

Solving, we get $k = \frac{1}{7} \ln\left(\frac{1250}{3}\right)$ , so $N(t) = 12e^{ \frac{t}{7} \ln\left(\frac{1250}{3}\right)}$ .

Of course, in practice, we would approximate $k$ to some desired accuracy, say $k \approx 0.8618$ , which we can interpret as an $86.18 \%$ daily growth rate for the cells.

Whereas Equations 5.3 and 5.4 model the growth of quantities, we can use equations like them to describe the decline of quantities.

One example we’ve seen already is Example 5.2.2 in Section 5.2. There, the value of a car decreased from its purchase price of 25,000 dollars to nothing at all.

Another real world phenomenon which follows suit is radioactive decay. There are elements which are unstable and emit energy spontaneously. In doing so, the amount of the element itself diminishes. The assumption behind this model is that the rate of decay of an element at a particular time is directly proportional to the amount of the element present at that time. In other words, the more of the element there is, the faster the element decays.

This is precisely the same kind of hypothesis which drives The Law of Uninhibited Growth, and as such, the equation governing radioactive decay is hauntingly similar to Equation 5.4 with the exception that the rate constant $k$ is negative.

Equation 5.5 Radioactive Decay

The amount of a radioactive element at time $t$ , $A(t)$ is given by the formula

$A(t) = A_{0}e^{kt},$

where $A(0) = A_{0}$ is the initial amount of the element and $k<0$ is the constant of proportionality which satisfies the equation

$\left( \textit{ instantaneous rate of change of } A(t) \textit{ at time } t\right) = k \, A(t)$

Example 5.7.3

Iodine-131 is a commonly used radioactive isotope and is used to help detect how well the thyroid is functioning. Suppose the decay of Iodine-131 follows the model given in Equation 5.5, and that the half-life^[9] of Iodine-131 is approximately $8$ days. If $5$ grams of Iodine-131 is present initially, find a function which gives the amount of Iodine-131, $A$ , in grams, $t$ days later.

Solution:

As we start with $5$ grams initially, Equation5.5 gives $A(t) = 5e^{kt}$ .

Because the half-life is $8$ days, it takes $8$ days for half of the Iodine-131 to decay, leaving half of it behind. Mathematically, this translates to $A(8) = 2.5$ , or

$\begin{array}{rcl} 5e^{8k} &=& 2.5 \\[4pt] k &= & \frac{1}{8} \ln\left(\frac{1}{2}\right) \\[4pt] &=& -\frac{\ln(2)}{8} \\[4pt] & \approx & -0.08664 \end{array}$

which we can interpret as a loss of material at a rate of $8.664 \%$ daily.

Hence, our final answer is $A(t) = 5 e^{-\frac{t\ln(2)}{8}} \approx 5 e^{-0.08664t}$ .

We now turn our attention to some more mathematically sophisticated models. One such model is Newton’s Law of Cooling, which we first encountered in Example 5.2.3 of Section 5.2.

In that example we had a cup of coffee cooling from $160^{\circ}\text{F}$ to room temperature $70^{\circ}\text{F}$ according to the formula $T(t) = 70 + 90 e^{-0.1 t}$ , where $t$ was measured in minutes. In that situation, we knew the physical limit of the temperature of the coffee was room temperature,^[10] and the differential equation which gives rise to our formula for $T(t)$ takes this into account.

Whereas the radioactive decay model had a rate of decay at time $t$ directly proportional to the amount of the element which remained at time $t$ , Newton’s Law of Cooling states that the rate of cooling of the coffee at a given time $t$ is directly proportional to how much of a temperature gap exists between the coffee at time $t$ and room temperature, not the temperature of the coffee itself. In other words, the coffee cools faster when it is first served, and as its temperature nears room temperature, the coffee cools ever more slowly.

Of course, if we take an item from the refrigerator and let it sit out in the kitchen, the object’s temperature will rise to room temperature, and as the physics behind warming and cooling is the same, we combine both cases in the equation below.

Equation 5.6 Newton’s Law of Cooling (Warming)

The temperature of an object at time $t$ , $T(t)$ is given by the formula

$T(t) = T_{a} + \left(T_{0} - T_{a}\right) e^{-kt},$

where $T(0) = T_{0}$ is the initial temperature of the object, $T_{a}$ is the ambient temperature^[11] and $k>0$ is the constant of proportionality which satisfies the equation

$\left( \textit{ instantaneous rate of change of } T(t) \textit{ at time } t\right) = k \, \left(T(t) - T_{a}\right)$

If we re-examine the situation in Example 5.2.3 with $T_{0} = 160$ , $T_{a} = 70$ , and $k = 0.1$ , we get, according to Equation 5.6, $T(t) = 70 + (160 - 70)e^{-0.1t}$ which reduces to the original formula given in that example. The rate constant $k = 0.1$ in this case indicates the coffee is cooling at a rate equal to $10 \%$ of the difference between the temperature of the coffee and its surroundings.

Note in Equation 5.6 that the constant $k$ is positive for both the cooling and warming scenarios. What determines if the function $T(t)$ is increasing or decreasing is if $T_{0}$ (the initial temperature of the object) is greater than $T_{a}$ (the ambient temperature) or vice-versa, as we see in our next example.

Example 5.7.4

Example 5.7.4.1

A roast initially at $40^{\circ}\text{F}$ cooked in a $350^{\circ}\text{F}$ oven. After $2$ hours, the temperature of the roast is $125^{\circ}\text{F}$ .

Assuming the temperature of the roast follows Newton’s Law of Warming, write a formula for the temperature of the roast $T(t)$ as a function of its time in the oven, $t$ , in hours.

Solution:

Assuming the temperature of the roast follows Newton’s Law of Warming, write a formula for the temperature of the roast $T(t)$ as a function of its time in the oven, $t$ , in hours.

The initial temperature of the roast is $40^{\circ}\text{F}$ , so $T_{0} = 40$ . The environment in which we are placing the roast is the $350^{\circ}\text{F}$ oven, so $T_{a} = 350$ . Newton’s Law of Warming gives $T(t) = 350 + (40-350)e^{-kt}$ , or after some simplification, $T(t) = 350 - 310e^{-kt}$ .

To determine $k$ , we use the fact that after $2$ hours, the roast is $125^{\circ}\text{F}$ , which means $T(2) = 125$ . This gives rise to the equation $350 - 310e^{-2k} = 125$ which yields $k = -\frac{1}{2} \ln \left( \frac{45}{62} \right) \approx 0.1602$ .

The temperature function is

$T(t) = 350 - 310 e^{\frac{t}{2} \ln \left( \frac{45}{62} \right)} \approx 350- 310 e^{-0.1602 t}.$

Example 5.7.4.2

A roast initially at $40^{\circ}\text{F}$ cooked in a $350^{\circ}\text{F}$ oven. After $2$ hours, the temperature of the roast is $125^{\circ}\text{F}$ .

The roast is done when the internal temperature reaches $165^{\circ}\text{F}$ . When will the roast be done?

Solution:

The roast is done when the internal temperature reaches $165^{\circ}\text{F}$ . When will the roast be done?

To find when the roast is done, we set $T(t) = 165$ . This gives $350- 310 e^{-0.1602 t} = 165$ whose solution is $t = -\frac{1}{0.1602} \ln \left( \frac{37}{62} \right) \approx 3.22$ .

Hence, the roast is done after roughly $3$ hours and $15$ minutes.

If we had taken the time to graph $y=T(t)$ in Example 5.7.4, we would have found the horizontal asymptote to be $y = 350$ , which corresponds to the temperature of the oven. We can also arrive at this conclusion analytically by applying `number sense’.

As $t \rightarrow \infty$ , $-0.1602 t \approx \text{very big } (-)$ so that $e^{-0.1602 t} \approx \text{very small } (+)$ . The larger the value of $t$ , the smaller $e^{-0.1602 t}$ becomes so that $T(t) \approx 350 -\text{very small } (+)$ , which indicates the graph of $y=T(t)$ is approaching its horizontal asymptote $y=350$ from below. Physically, this means the roast will eventually warm up to $350^{\circ}\text{F}$ .

The function $T$ in this situation is sometimes called a limited growth model, because the function $T$ remains bounded as $t \rightarrow \infty$ . If we apply the principles behind Newton’s Law of Cooling to a biological example, it says the growth rate of a population is directly proportional to how much room the population has to grow. In other words, the more room for expansion, the faster the growth rate.

Our final model, the logistic growth model combines The Law of Uninhibited Growth with limited growth and states that the rate of growth of a population varies jointly with the population itself as well as the room the population has to grow.

Equation 5.7 Logistic Growth

If a population behaves according to the assumptions of logistic growth, the number of organisms at time $t$ , $N(t)$ is given by

$N(t) =\dfrac{L}{1 + Ce^{-kLt}},$

where $N(0) = N_{0}$ is the initial population, $L$ is the limiting population,^[12] and $C$ is a measure of how much room there is to grow given by

$C = \dfrac{L}{N_{0}} - 1}$

and $k > 0$ is the constant of proportionality which satisfies the equation

$\left( \textit{ instantaneous rate of change of } N(t) \textit{ at time } t\right) = k \, N(t) \left(L - N(t)\right)$

The logistic function is used not only to model the growth of organisms, but is also often used to model the spread of disease and rumors.^[13]

Example 5.7.5

Example 5.7.5.1

The number of people $N(t)$ , in hundreds, at a local community college who have heard the rumor `Carl’s afraid of Sasquatch’ can be modeled using the logistic equation

$N(t) = \dfrac{84}{1+2799e^{-t}},$

where $t\geq 0$ is the number of days after April 1, 2016.

Compute and interpret $N(0)$ .

Solution:

Compute and interpret $N(0)$ .

We find $N(0) = \frac{84}{1+2799e^{0}} = \frac{84}{2800} = 0.03$ .

As $N(t)$ measures the number of people who have heard the rumor in hundreds, $N(0)$ corresponds to $3$ people. $t=0$ corresponds to April 1, 2016, thus we may conclude that on that day, $3$ people have heard the rumor.

Example 5.7.5.2

The number of people $N(t)$ , in hundreds, at a local community college who have heard the rumor `Carl’s afraid of Sasquatch’ can be modeled using the logistic equation

$N(t) = \dfrac{84}{1+2799e^{-t}},$

where $t\geq 0$ is the number of days after April 1, 2016.

Write and interpret the end behavior of $N(t)$ .

Solution:

Write and interpret the end behavior of $N(t)$ .

We could simply note that $N(t)$ is written in the form of Equation 5.7, and identify $L = 84$ . However, to see better why the answer is 84, we proceed analytically.

As the domain of $N$ is restricted to $t \geq 0$ , the only end behavior of significance is $t \rightarrow \infty$ . As we’ve seen before,^[14] as $t \rightarrow \infty$ , we have $1997 e^{-t} \rightarrow 0^{+}$ and so $N(t) \approx \frac{84}{1 + \text{ very small } (+)} \approx 84$ .

Hence, as $t \rightarrow \infty$ , $N(t) \rightarrow 84$ . This means that as time goes by, the number of people who will have heard the rumor approaches 8400.

Example 5.7.5.3

The number of people $N(t)$ , in hundreds, at a local community college who have heard the rumor `Carl’s afraid of Sasquatch’ can be modeled using the logistic equation

$N(t) = \dfrac{84}{1+2799e^{-t}},$

where $t\geq 0$ is the number of days after April 1, 2016.

How long until $4200$ people have heard the rumor?

Solution:

How long until $4200$ people have heard the rumor?

To compute how long it takes until $4200$ people have heard the rumor, we set $N(t) = 42$ . Solving

$\begin{array}{rcl} \frac{84}{1+2799e^{-t}} &=& 42 \\[4pt] t &=& \ln(2799) \\ &\approx& 7.937 \end{array}$

so it takes around 8 days until 4200 people have heard the rumor.

Example 5.7.5.4

The number of people $N(t)$ , in hundreds, at a local community college who have heard the rumor `Carl’s afraid of Sasquatch’ can be modeled using the logistic equation

$N(t) = \dfrac{84}{1+2799e^{-t}},$

where $t\geq 0$ is the number of days after April 1, 2016.

Check your answers to 2 and 3 using technology.

Solution:

Graphing $\textcolor{red}{y=N(t)}$ below, we see $y=84$ is the horizontal asymptote of the graph, confirming our answer to number , and the graph intersects the line $\textcolor{blue}{y=42}$ at $t \approx 7.937 \approx \ln(2799)$ , which confirms our answer to number 3.

Rendered by QuickLaTeX.com

If we take the time to analyze the graph of $y=N(t)$ in Example 5.7.5, we can see graphically how logistic growth combines features of uninhibited and limited growth.

The curve is concave up, rising steeply, then at some point, becomes concave down and begins to level off.^[15] The point at which this happens is called an inflection point or is sometimes called the `point of diminishing returns’. Even though the function is still increasing through the inflection point, the rate at which it does so begins to decrease.

With Calculus, one can show the point of diminishing returns always occurs at half the limiting population. (In our case, when $N(t)=42$ .) So with that in mind, we present two portions of the graph of $y=N(x)$ , one on the interval $[0,8]$ , the other on $[8,15]$ . The former looks strikingly like uninhibited growth while the latter like limited growth.

Two graphs side by side. The first graph is of N(t) described in Example 5.7.5 from 0 to 8 inclusively. The points (0,0.03) and (8, 43.322). The curve of N(t) curves upward between the two points. The second graph is for N(t) from 8 to 15 inclusively. The point (8, 43.322) is given and a horizontal dashed line above the 80 mark that the curve approaches. — Graphs of N(t)

5.7.2 Applications of Logarithms

Just as many physical phenomena can be modeled by exponential functions, the same is true of logarithmic functions. In Exercises 81, 82, and 83 of Section 5.3, we showed that logarithms are useful in measuring the intensities of earthquakes (the Richter scale), sound (decibels) and acids and bases (pH). We now present yet a different use of the a basic logarithm function, password strength.

Example 5.7.6

Example 5.7.6.1

The information entropy $H$ , in bits, of a randomly generated password consisting of $L$ characters is given by $H = L \log_{2}(N)$ , where $N$ is the number of possible symbols for each character in the password. In general, the higher the entropy, the stronger the password.

If a $7$ character case-sensitive^[16] password is comprised of letters and numbers only, find the associated information entropy.

Solution:

If a $7$ character case-sensitive password is comprised of letters and numbers only, find the associated information entropy.

There are $26$ letters in the alphabet, $52$ if upper and lower case letters are counted as different. There are $10$ digits ( $0$ through $9$ ) for a total of $N=62$ symbols. The password is to be $7$ characters long, so $L = 7$ . Thus,

$\begin{array}{rcl} H & = & 7 \log_{2}(62) \\[4pt] & = & \frac{7 \ln(62)}{\ln(2)} \\[4pt] & \approx & 41.68 \end{array}$

Example 5.7.6.2

The information entropy $H$ , in bits, of a randomly generated password consisting of $L$ characters is given by $H = L \log_{2}(N)$ , where $N$ is the number of possible symbols for each character in the password. In general, the higher the entropy, the stronger the password.

How many possible symbol options per character is required to produce a 7 character password with an information entropy of 50 bits?

Solution:

How many possible symbol options per character is required to produce a $7$ character password with an information entropy of $50$ bits?

We have $L = 7$ and $H=50$ and we need to find $N$ .

Solving the equation $50 = 7 \log_{2}(N)$ gives $N = 2^{50/7} \approx 141.323$ , so we would need $142$ different symbols to choose from.^[17]

Chemical systems known as buffer solutions have the ability to adjust to small changes in acidity to maintain a range of pH values. Buffer solutions have a wide variety of applications from maintaining a healthy fish tank to regulating the pH levels in blood. Our next example shows how the pH in a buffer solution is a little more complicated than the pH we first encountered in Exercise 83 in Section 5.3.

Example 5.7.7

Blood is a buffer solution. When carbon dioxide is absorbed into the bloodstream it produces carbonic acid and lowers the pH. The body compensates by producing bicarbonate, a weak base to partially neutralize the acid. The equation^[18] which models blood pH in this situation is $\text{pH} = 6.1 + \log\left(\frac{800}{x} \right)$ , where $x$ is the partial pressure of carbon dioxide in arterial blood, measured in torr. Find the partial pressure of carbon dioxide in arterial blood if the pH is 7.4.

Solution:

We set $\text{pH} = 7.4$ and get

$\begin{array}{rcl} 7.4 & = & 6.1 + \log\left(\frac{800}{x} \right) \\[4pt] \log\left(\frac{800}{x} \right) & = & 1.3 \\[4pt] x & = & \frac{800}{10^{1.3}} \\ & \approx & 40.09 \end{array}$

Hence, the partial pressure of carbon dioxide in the blood is about $40$ torr.

5.7.3 Section Exercises

For each of the scenarios given in Exercises 1 – 6,

Express the amount, $A$ , in the account as a function of the term of the investment $t$ in years.
To the nearest cent, determine how much is in the account after 5, 10, 30, and 35 years.
To the nearest year, determine how long will it take for the initial investment to double.
Compute and interpret the average rate of change of the amount in the account from the end of the fourth year to the end of the fifth year, and from the end of the thirty-fourth year to the end of the thirty-fifth year. Round your answer to two decimal places.

500 dollars is invested in an account which offers $0.75 \%$ , compounded monthly.
500 dollars is invested in an account which offers $0.75 \%$ , compounded continuously.
1000 dollars is invested in an account which offers $1.25 \%$ , compounded monthly.
1000 dollars is invested in an account which offers $1.25 \%$ , compounded continuously.
5000 dollars is invested in an account which offers $2.125 \%$ , compounded monthly.
5000 dollars is invested in an account which offers $2.125 \%$ , compounded continuously.
Look back at your answers to Exercises 1 – 6. What can be said about the difference between monthly compounding and continuously compounding the interest in those situations? With the help of your classmates, discuss scenarios where the difference between monthly and continuously compounded interest would be more dramatic. Try varying the interest rate, the term of the investment and the principal. Use computations to support your answer.
How much money needs to be invested now to obtain 2000 dollars in 3 years if the interest rate in a savings account is $0.25 \%$ , compounded continuously? Round your answer to the nearest cent.
How much money needs to be invested now to obtain 5000 dollars in 10 years if the interest rate in a CD is $2.25 \%$ , compounded monthly? Round your answer to the nearest cent.
On May, 31, 2009, the Annual Percentage Rate listed at Jeff’s bank for regular savings accounts was compounded monthly. Use Equation 5.2 to answer the following.
1. If $P = 2000$ what is $A(8)$ ?
2. Solve the equation $A(t) = 4000$ for $t$ .
3. What principal $P$ should be invested so that the account balance is 2000 dollars is three years?
Jeff’s bank also offers a 36-month Certificate of Deposit (CD) with an APR of .
1. If $P = 2000$ what is $A(8)$ ?
2. Solve the equation $A(t) = 4000$ for $t$ .
3. What principal $P$ should be invested so that the account balance is 2000 dollars in three years?
4. The Annual Percentage Yield is the simple interest rate that returns the same amount of interest after one year as the compound interest does. With the help of your classmates, compute the APY for this investment.
A finance company offers a promotion on 5000 dollars loans. The borrower does not have to make any payments for the first three years, however interest will continue to be charged to the loan at $29.9 \%$ compounded continuously. What amount will be due at the end of the three year period, assuming no payments are made? If the promotion is extended an additional three years, and no payments are made, what amount would be due?
Use Equation 5.2 to show that the time it takes for an investment to double in value does not depend on the principal $P$ , but rather, depends only on the APR and the number of compoundings per year. Let $n = 12$ and with the help of your classmates compute the doubling time for a variety of rates $r$ . Then look up the Rule of 72 and compare your answers to what that rule says. If you’re really interested^[19] in Financial Mathematics, you could also compare and contrast the Rule of 72 with the Rule of 70 and the Rule of 69.

In Exercises 14 – 18, we list some radioactive isotopes and their associated half-lives. Assume that each decays according to the formula $A(t) = A_{0}e^{kt}$ where $A_{0}$ is the initial amount of the material and $k$ is the decay constant. For each isotope:

Find the decay constant $k$ . Round your answer to four decimal places.
Find a function which gives the amount of isotope $A$ which remains after time $t$ . (Keep the units of $A$ and $t$ the same as the given data.)
Determine how long it takes for $90 \%$ of the material to decay. Round your answer to two decimal places. (HINT: If $90 \%$ of the material decays, how much is left?)

1. Cobalt 60, used in food irradiation, initial amount 50 grams, half-life of 5.27 years.
2. Phosphorus 32, used in agriculture, initial amount 2 milligrams, half-life 14 days.
3. Chromium 51, used to track red blood cells, initial amount 75 milligrams, half-life 27.7 days.
4. Americium 241, used in smoke detectors, initial amount 0.29 micrograms, half-life 432.7 years.
5. Uranium 235, used for nuclear power, initial amount 1 kg grams, half-life 704 million years.
6. With the help of your classmates, show that the time it takes for $90 \%$ of each isotope listed in Exercises 14 – 18 to decay does not depend on the initial amount of the substance, but rather, on only the decay constant $k$ . Find a formula, in terms of $k$ only, to determine how long it takes for $90 \%$ of a radioactive isotope to decay.
7. In Example 5.2.2 in Section 5.2, the exponential function $V(x) = 25 \left(\frac{4}{5}\right)^{x}$ was used to model the value of a car over time. Use a change of base formula to rewrite the model in the form $V(t) = 25e^{kt}$ .
8. The Gross Domestic Product (GDP) of the US (in billions of dollars) years after the year 2000 can be modeled by:
  $G(t) = 9743.77 e^{0.0514t}$
  1. Find and interpret $G(0)$ .
  2. According to the model, what should have been the GDP in 2007? In 2010? (According to the US Department of Commerce, the 2007 GDP was 14,369.1 billion dollars and the 2010 GDP was 14,657.8 billion dollars.)
9. The diameter of a tumor, in millimeters, days after it is detected is given by:
  $D(t) = 15e^{0.0277t}$
  1. What was the diameter of the tumor when it was originally detected?
  2. How long until the diameter of the tumor doubles?
10. Under optimal conditions, the growth of a certain strain of E. Coli is modeled by the Law of Uninhibited Growth where is the initial number of bacteria and is the elapsed time, measured in minutes. From numerous experiments, it has been determined that the doubling time of this organism is 20 minutes. Suppose 1000 bacteria are present initially.
  1. Find the growth constant $k$ . Round your answer to four decimal places.
  2. Find a function which gives the number of bacteria $N(t)$ after $t$ minutes.
  3. How long until there are 9000 bacteria? Round your answer to the nearest minute.
11. Yeast is often used in biological experiments. A research technician estimates that a sample of yeast suspension contains 2.5 million organisms per cubic centimeter (cc). Two hours later, she estimates the population density to be 6 million organisms per cc. Let be the time elapsed since the first observation, measured in hours. Assume that the yeast growth follows the Law of Uninhibited Growth .
  1. Find the growth constant $k$ . Round your answer to four decimal places.
  2. Find a function which gives the number of yeast (in millions) per cc $N(t)$ after $t$ hours.
  3. What is the doubling time for this strain of yeast?
12. The Law of Uninhibited Growth also applies to situations where an animal is re-introduced into a suitable environment. Such a case is the reintroduction of wolves to Yellowstone National Park. According to the National Park Service, the wolf population in Yellowstone National Park was 52 in 1996 and 118 in 1999. Using these data, find a function of the form $N(t) = N_{0} e^{kt}$ which models the number of wolves $t$ years after 1996. (Use $t = 0$ to represent the year 1996. Also, round your value of $k$ to four decimal places.) According to the model, how many wolves were in Yellowstone in 2002? (The recorded number is 272.)
13. During the early years of a community, it is not uncommon for the population to grow according to the Law of Uninhibited Growth. According to the Painesville Wikipedia entry, in 1860, the Village of Painesville had a population of 2649. In 1920, the population was 7272. Use these two data points to fit a model of the form $N(t) = N_{0} e^{kt}$ were $N(t)$ is the number of Painesville Residents $t$ years after 1860. (Use $t = 0$ to represent the year 1860. Also, round the value of $k$ to four decimal places.) According to this model, what was the population of Painesville in 2010? (The 2010 census gave the population as 19,563) What could be some causes for such a vast discrepancy?
14. The population of Sasquatch in Bigfoot county is modeled by
  $P(t) = \dfrac{120}{1 + 3.167e^{-0.05t}}$
  
  where $P(t)$ is the population of Sasquatch $t$ years after $2010$ .
  1. Find and interpret $P(0)$ .
  2. Find the population of Sasquatch in Bigfoot county in 2013 rounded to the nearest Sasquatch.
  3. To the nearest year, when will the population of Sasquatch in Bigfoot county reach 60?
  4. Find and interpret the end behavior of the graph of $y = P(t)$ analytically. Check your answer using a graphing utility.
15. Let .
  1. From Calculus, we know the inflection point of the graph of $y=f(x)$ is $(1,5)$ . This means the function is increasing the fastest at $x=1$ , or, equivalently, the slope at $(1,5)$ is the largest anywhere on the graph. Graph $y=f(x)$ using a graphing utility and convince yourself of the reasonableness of this claim.
  2. Find average rate of change of over each of the intervals below. What do you guess the slope of the curve is at ? Zoom in on the graph near to check your guess.
    - $[0.75, 1]$
    - $[0.9, 1]$
    - $[0.99,1]$
    - $[1, 1.01]$
    - $[1, 1.1]$
    - $[1, 1.25]$
16. The half-life of the radioactive isotope Carbon-14 is about 5730 years.
  1. Use Equation 5.5 to express the amount of Carbon-14 left from an initial $N$ milligrams as a function of time $t$ in years.
  2. What percentage of the original amount of Carbon-14 is left after 20,000 years?
  3. If an old wooden tool is found in a cave and the amount of Carbon-14 present in it is estimated to be only 42\% of the original amount, approximately how old is the tool?
  4. Radiocarbon dating is not as easy as these exercises might lead you to believe. With the help of your classmates, research radiocarbon dating and discuss why our model is somewhat over-simplified.
17. Carbon-14 cannot be used to date inorganic material such as rocks, but there are many other methods of radiometric dating which estimate the age of rocks. One of them, Rubidium-Strontium dating, uses Rubidium-87 which decays to Strontium-87 with a half-life of 50 billion years. Use Equation 5.5 to express the amount of Rubidium-87 left from an initial 2.3 micrograms as a function of time $t$ in billions of years. Research this and other radiometric techniques and discuss the margins of error for various methods with your classmates.
18. Use Equation 5.5 to show that $k = -\dfrac{\ln(2)}{h}$ where $h$ is the half-life of the radioactive isotope.
19. A pork roast^[20] was taken out of a hardwood smoker when its internal temperature had reached F and it was allowed to rest in a F house for 20 minutes after which its internal temperature had dropped to F.
  Assuming that the temperature of the roast follows Newton’s Law of Cooling (Equation 5.6),
  1. Express the temperature $T$ (in $^{\circ}$ F) as a function of time $t$ (in minutes).
  2. Find the time at which the roast would have dropped to $140^{\circ}$ F had it not been eaten.
20. If Fritzy the Fox’s speed is the same as Chewbacca the Bunny’s speed, Fritzy’s pursuit curve is given by
  $y(x) = \frac{1}{4} x^2-\frac{1}{4} \ln(x)-\frac{1}{4}$
  
  Graph this path for $x > 0$ using a graphing utility. Describe the behavior of $y$ as $x \rightarrow 0^{+}$ and interpret this physically.
21. The current $i$ measured in amps in a certain electronic circuit with a constant impressed voltage of 120 volts is given by $i(t) = 2 - 2e^{-10t}$ where $t \geq 0$ is the number of seconds after the circuit is switched on. Determine the value of $i$ as $t \rightarrow \infty$ . (This is called the steady state current.)
22. If the voltage in the circuit in Exercise 34 above is switched off after 30 seconds, the current is given by the piecewise-defined function
  $i(t) = \left\{ \begin{array}{rcl} 2 - 2e^{-10t} & \text{if } & 0 \leq t < 30 \\ [6pt] \left(2 - 2e^{-300}\right) e^{-10t+300} & \text{if } & t \geq 30 \end{array} \right.$
  
  With the help of a graphing utility, graph $y = i(t)$ and discuss with your classmates the physical significance of the two parts of the graph $0 \leq t < 30$ and $t \geq 30$ .
23. In Exercise 26 in Section 2.1, we stated that the cable of a suspension bridge formed a parabola but that a free hanging cable did not. A free hanging cable forms a catenary and its basic shape is given by $y = \frac{1}{2}\left(e^{x} + e^{-x}\right)$ . Use a graphing utility to graph this function. What are its domain and range? What is its end behavior? Is it invertible? How do you think it is related to the function given in Exercise 40 in Section 5.5 and the one given in the answer to Exercise 33 in Section 5.6?When flipped upside down, the catenary makes an arch. The Gateway Arch in St. Louis, Missouri has the shape
  $y = 757.7 - \frac{127.7}{2}\left(e^{\frac{x}{127.7}} + e^{-\frac{x}{127.7}}\right)$
  
  where $x$ and $y$ are measured in feet and $-315 \leq x \leq 315$ . Find the highest point on the arch.

Section 5.7 Exercise Answers can be found in the Appendix … Coming soon

Called the principal ↵
Some restrictions may apply. ↵
Using this convention, simple interest after one year is the same as compounding the interest only once per year. ↵
See Definition 1.11 in Section 1.3.1. ↵
In fact, the rate of increase of the amount in the account is exponential as well. This is the quality that really defines exponential functions and we refer the reader to a course in Calculus. ↵
Once you've had a semester of Calculus, you'll be able to fully appreciate this very lame pun. ↵
Or define, depending on your point of view. ↵
The average rate of change of a function over an interval was first introduced in Section 1.3.1. The notion of instantaneous rate of change was introduced in the remarks following Example 1.3.12 and revisited in Example 3.2.3. ↵
The time it takes for half of the substance to decay. ↵
The Second Law of Thermodynamics states that heat can spontaneously flow from a hotter object to a colder one, but not the other way around. Thus, the coffee could not continue to release heat into the air so as to cool below room temperature. ↵
That is, the temperature of the surroundings. ↵
That is, as $t \rightarrow \infty$ , $N(t) \rightarrow L$ ↵
Which can be just as damaging as diseases. ↵
See, for example, Example 5.2.3. ↵
We introduced the notion of concavity in Section 4.2. ↵
That is, upper and lower case letters are treated as different characters. ↵
As there are only $94$ distinct ASCII keyboard characters, to achieve this strength, the number of characters in the password should be increased. ↵
Derived from the Henderson-Hasselbalch Equation. See Exercise 41 in Section 5.4. Hasselbalch himself was studying carbon dioxide dissolving in blood - a process called metabolic acidosis. ↵
Awesome pun! ↵
This roast was enjoyed by Jeff and his family on June 10, 2009. This is real data, folks! ↵

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Functions, Trigonometry, and Systems of Equations Copyright © by Vanessa Coffelt is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Equation 5.1 Simple Interest

5.7.1 Applications of Exponential Functions

Equation 5.2 Compounded Interest

Example 5.7.1

Solution:

Solution:

Solution:

Solution:

Equation 5.3 Continuously Compounded Interest

Equation 5.4 Uninhibited Growth

Example 5.7.2

Solution:

Equation 5.5 Radioactive Decay

Example 5.7.3

Solution:

Equation 5.6 Newton’s Law of Cooling (Warming)

Example 5.7.4

Solution:

Solution:

Equation 5.7 Logistic Growth

Example 5.7.5

Solution:

Solution:

Solution:

Solution:

5.7.2 Applications of Logarithms

Example 5.7.6

Solution:

Solution:

Example 5.7.7

Solution:

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