As we mentioned in Sections ExponentialFunctions and LogarithmicFunctions, 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!)

1 Applications of Exponential Functions

Perhaps the most well-known application of exponential functions comes from the financial world. Suppose you have \( \$ 100\) 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 \(\$100\), or \( \$ 100(0.05) =\$ 5\) in interest, so you now have \(\$105\). 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] (Some restrictions may apply.) Naturally, you withdraw your money and try to invest it at the higher rate there. Since six months is one half of a year, that initial \(\$100\) yields \(\$100(0.05)\left (\frac {1}{2}\right ) = \$ 2.50\) in interest.

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

To your surprise and delight, at the end of the year your statement reads \(\$105.06\), not \(\$105\) as you had expected. ^[3] (Actually, the final balance should be \(\$105.0625\).) Where did those extra six cents come from?

For the first six months of the year, interest was earned on the original principal of \(\$100\), but for the second six months, interest was earned on \(\$102.50\), 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. ^[4] (Using this convention, simple interest after one year is the same as compounding the interest only once.) 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 \(A = P(1+rt) = \$100 \left (1 + 0.05 \cdot \frac {1}{4} \right ) = \$101.25\). We now invest the \(\$101.25\) for the next three months and find that at the end of the second quarter, we have \(A = \$101.25 \left (1 + 0.05 \cdot \frac {1}{4} \right )\approx \$102.51\). Continuing in this manner, the balance at the end of the third quarter is \(\$103.79\), and, at last, we obtain \(\$105.08\). 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 \(\frac {1}{n}\) of a year between compoundings. Let \(A_{k}\) denote the amount in the account after the \(k^{\text {th}}\) 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 \(A_{k} = P \left (1 + \frac {r}{n}\right )^k\).

Since 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.

If we take \(P = 100\), \(r = 0.05\), and \(n = 4\), Formula 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 \(A\left (\frac {1}{4}\right ) = 100(1.0125)^{4 \left (\frac {1}{4}\right )} = 101.25\), \(A\left (\frac {1}{2}\right ) \approx \$102.51\), \(A\left (\frac {3}{4}\right ) \approx \$103.79\), and \(A(1) \approx \$105.08\).

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. ^[8] (See Section ?? to understand this pun.)

Consider an investment of \(\$ 1\) invested at \(100 \%\) interest for \(1\) year compounded \(n\) times a year. Formula 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 \(A\).

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 ^[9] (Or define, depending on your point of view.) that \(\lim _{n \rightarrow \infty } \left (1+\frac {1}{n}\right )^{n} = e\), where \(e\) is the natural base first presented in Section ??. Taking the number of compoundings per year to infinity results in what is called continuously compounded interest.

Using the limit definition of \(e\) along with some limit properties, we can derive a general formula for continuously compounded interest.

Consider the limit \(\lim _{n \rightarrow \infty } P\left (1 + \frac {r}{n}\right )^{nt}\). In order to use the limit definition of ‘\(e\),’ we need to make a substitution to make ‘\(\left (1 + \frac {r}{n}\right )\)’ look like ‘\(\left (1 + \frac {1}{\text {something}}\right )\).’

If we let \(u = \frac {n}{r}\), we get \(n = r \, u\), so \(1 + \frac {r}{n} = 1 + \frac {r}{ru} = 1 + \frac {1}{u}\) and \(nt = (ru)t = urt\). In the context of compound interest, \(r > 0\), so \(n \rightarrow \infty \) implies \(u \rightarrow \infty \) and vice-versa so the limit becomes:

Using Properties of Limits, Theorem ??, we get:

A couple of remarks are in order. First, in the limit \(\lim _{n \rightarrow \infty } \left (1+\frac {1}{n}\right )^{n} = e\), the variable here, \(n\), takes on natural number values, and as such, is a discrete variable, not a continuous one. ^[10] (See Section ?? for a discussion of the difference, if needed!) We’ll revisit limits of discrete variables in Section ??.

Second, when we use the limit definition of ‘\(e\)’ in the above argument, we used \(\lim _{u \rightarrow \infty } \left (1+\frac {1}{u}\right )^{u} = e\). Like \(n\), the variable \(u\) is a discrete variable (not necessarily natural numbers, but discrete nonetheless). Since both \(n\) and \(u\) are tending to infinity, this change in dummy variable doesn’t affect the limit. ^[11] (Both \(n\) and \(u\) are taking discrete steps to infinity - one is doing so at just a slower pace.)

Last, we note that the Real Numbers Power rule applies since \(e \approx 2.718 > 0\). We codify this result in the following theorem.

It is worth noting that if we take the scenario of Example 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 \(\$ 24,\!035.28\), whereas continuously compounding gives \(A(35) = 2000e^{0.07125 (35)}\) which is about \(\$ 24,\!213.18\) - a difference of less than \(1 \%\).

Formulas 2 and ?? 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. 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.

It is worth taking some time to compare Formulas ?? and 4. In Formula ??, we use \(P\) to denote the initial investment; in Formula 4, we use \(N_0\) to denote the initial population. In Formula ??, \(r\) denotes the annual interest rate, and so it shouldn’t be too surprising that the \(k\) in Formula 4 corresponds to a growth rate as well. While Formulas ?? and 4 look entirely different, they both represent the same mathematical concept.

Whereas Formulas ?? and 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 ?? in Section ??. There, the value of a car decreased from its purchase price of \(\$25,\!000\) 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 Formula 4 with the exception that the rate constant \(k\) is negative.

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 ?? of Section ??.

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, ^[14] (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.) 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 since the physics behind warming and cooling is the same, we combine both cases in the equation below.

If we re-examine the situation in Example ?? with \(T_0 = 160\), \(T_{a} = 70\), and \(k = 0.1\), we get, according to Formula 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 Formula 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.

If we had taken the time to graph \(y=T(t)\) in Example 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 suggests 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 \(350^{\circ }\text {F}\). ^[16] (at which point it would be more toast than roast.)

The function \(T\) in this situation is sometimes called a limited growth model, since 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.

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. ^[18] (Which can be just as damaging as diseases.)

If we take the time to analyze the graph of \(y=N(t)\) in Example 5, we can see graphically how logistic growth combines features of uninhibited and limited growth.

We can see graphically that there is an inflection point in the graph. In this case, the inflection point is called the point of diminishing returns. Even though the function is still increasing through the inflection point (more people are hearing the rumor), 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. ^[21] (We’ll explore this in the Exercises.) (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. These regions are indicated graphically in the desmos interactive below.

2 Applications of Logarithms

Just as many physical phenomena can be modeled by exponential functions, the same is true of logarithmic functions. In Exercises ??, ?? and ?? of Section ??, 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.

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 ?? in Section ??.

Another place logarithms are used is in data analysis. Suppose, for instance, we wish to model the spread of influenza A (H1N1), the so-called ‘Swine Flu’. Below is data taken from the World Health Organization ( WHO) where \(t\) represents the number of days since April 28, 2009, and \(N\) represents the number of confirmed cases of H1N1 virus worldwide.

Using desmos, we make a scatter plot of the data treating \(t\) as the independent variable and \(N\) as the dependent variable. Which function family could produce data that trends like these?

Thinking back Section ??, we have desmos run a quadratic regression. We find \(N(t) \approx 16.713 t^2 +149.68t -233.15\) with \(R^2 = 0.992\), indicating a pretty good fit.

However, is there any underlying scientific principle which would account for these data to be quadratic? Are there other models which fit the data better?

To answer these questions, scientists often use logarithms in an attempt to ‘linearize’ non-liner data sets such as the one before us. To see how this could work, suppose we guessed the relationship between \(N\) and \(t\) is something from Section ??, \(N(t) = a t^{p}\).

By taking the natural logs of both sides and using the Product and Power Rules, in turn, we find that \(\ln (N(t)) = \ln (a t^p) = \ln (a) + \ln (t^p) = \ln (a) + p \ln (t) = p \ln (t) + \ln (a)\). If we let \(x = \ln (t)\) and \(y= \ln (N(t))\), the model takes the form \(y = p x + \ln (a)\) which is a linear model with slope \(p\) and \(y\)-intercept \(\ln (a)\). So, instead of plotting \(N(t)\) versus \(t\), we plot \(y=\ln (N(t))\) versus \(x=\ln (t)\).

Using desmos, we find a linear regression for this data set.

We see \(r=0.991\), which is very close to \(1\) indicating a very good fit. The slope of the regression line is \(m \approx 1.512\) which corresponds to our exponent \(p\). The \(y\)-intercept \(b \approx 4.513\) corresponds to \(\ln (a)\), so that \(a \approx 91.201\). Hence, we get the model \(N = 91.201 t^{1.512}\), graphed below along with the original data set.

Of interest here is that desmos has its own built-in power regression model. If the ‘log mode’ square is checked, the graphing utility returns the same model we obtained using our linearization (since the routine which determines the coefficients uses logarithms as well.) ^[25] (If, however, we uncheck that box, we get a different power function model, \(N(t) = 62.318 t^{1.675}\) which chooses \(a\) and \(p\) directly to minimize the total squared error. See here for more details.)

At this point, the quadratic model fits the data better, ostensibly because we have three parameters we can adjust in the formula \(N(t) = at^2+bt+c\) to minimize our error as opposed to just two parameters in the formula \(N(t) = a t^{p}\). Neither model, however, is based on any underlying scientific principle.

If we think about this situation from a scientific perspective, it does seem to make sense that, at least in the early stages of the outbreak, the more people who have the flu, the faster it will spread. This suggests we fit the data to an uninhibited growth model.

As written, Formula 4 gives uninhibited growth as \(N(t) = N_0e^{kt}\). Here, for simplicity’s sake, we relabel \(N_0 = a\) and \(e^{k} = b\) so that we are looking for parameters \(a\) and \(b\) so that \(N(t) = a \cdot b^{t}\).

If we assume \(N(t) = a \cdot b^{t}\) then, taking logs as before, we get \(\ln (N(t)) = t \ln (b) + \ln (a)\). If we let \(y= \ln (N(t))\), then, once again, we get a linear model this time with slope \(\ln (b)\) and \(y\)-intercept \(\ln (a)\). We present the results of the regression below. While there is a strong correlation, \(r = 0.962\), the plot doesn’t instill the greatest of confidence in this model.

From the slope of the model, we have \(m = \ln (b) \approx 0.202\) so \(b \approx 1.223\). From the \(y\)-intercept of the model, we get \(B = \ln (a) \approx 5.596\) so \(a \approx 269.35\), so that our model is \(N(t) = 269.35(1.223)^{t}\). Using the built-in exponential regression, seen below, (again, with ‘log mode’ checked) returns the model \(N(t) = 269.41 (1.223)^{t}\), the discrepancy between \(269.35\) and \(269.41\) stemming ostensibly from round-off error.

The exponential model didn’t fit the data as well as the quadratic or power function model, but it stands to reason that, perhaps, the spread of the flu is not unlike that of the spread of a rumor and that a logistic model can be used to model the data. Again, for simplicity, we abbreviate the model given inFormula 7 from \(N(t) =\frac {L}{1 + Ce^{-kLt}}\) to \(N(t) = \frac {L}{1 + Ce^{Kt}}\).

Running the data, a logistic function appears to be an excellent fit, both judging by the graph as well as the coefficient of determination, \(R^2 \approx 0.995\). Moreover, the underlying principles which lead to the formulation of this model seem reasonable enough.

While the quadratic model also fits extremely well, our logistic model takes into account that only a finite number of people will ever get the flu (according to our model, \(L=10,\!739\)), whereas the quadratic model predicts no limit to the number of cases. As we have stated several times before in the text, mathematical models, regardless of their sophistication, are just that: models, and they all have their limitations. ^[26] (Speaking of limitations, as of June 3, 2009, there were 19,273 confirmed cases of influenza A (H1N1). This is well above our prediction of 10,739. Each time a new report is issued, the data set increases and the model must be recalculated. We leave this recalculation to the reader.)