Section 3.2: Applications of Exponential Functions

In the previous section, we introduced the concept of an exponential function and learned how exponential graphs behave.  We saw that these functions are much more difficult to compute than the other functions with which we have worked so far.  This naturally raises the question, "Why bother with such functions?"  It is because these functions describe so many man-made and naturally occurring phenomena.  

For example, we use exponential functions in banking and finance to compute compound interest.  In general, exponential models arise whenever quantities grow or shrink by a constant factor, such as in radioactive decay or population growth.  In this section, we will quickly introduce exponential functions in the context of some basic applications.  We will then move on to discuss the application to banking and finance in more detail.  We will return to discuss other applications in the next section.

The Origins of Exponential Functions

We have seen that linear functions arise naturally to model situations where the dependent variable changes by a constant amount each time the independent variable increases by 1 unit.  For example, if C = the total cost of the call and m = the number of minutes that we talk, then C is a function of the length of the call, C = f(m).  If we know that a phone call costs $0.07/ min., and we have to pay a $0.05 connection charge, we can reason that f(m) = 0.07m + 0.05.  

In a similar way, exponential functions are used to model situations where the dependent variable changes by a constant factor each time the independent variable increases by 1 unit.  For example, if you put $200 in the bank at 5% compound interest and do not make any withdrawals, by the end of each year the bank is promising to give you 105% of (i.e., 1.05 times) whatever amount was in the account at the beginning of the year.  That is, if A = the total amount in the bank and t = the number of years you have left it there, then A is a function of years, A = g(t).  If we compute some values for this function, without simplifying the results, we can see a pattern emerge:

You can see that g is given by the exponential formula, g(t) = 200(1.05)^t.  Note: This common financial practice is known as annual compound interest; we will discuss several compound interest formulas in a later section.  Aside: Since you only receive your interest payment at the end of each year, this formula does not hold for fractional values of t; for example, when t = 1.5, your account will still contain the same amount as at t = 1; this means that a more accurate formula would be g(t) = 200(1.05)^ëtû.

A similar example arises when studying populations of cells, animals, or people, as well as looking at simple models for the spread of disease.  For instance, assume that, on average, the world population is growing by 1.3% per year, and that the population in 1999 was 6 billion people.  Then if A = the total population (in billions) and t = the number of years since 1999, then we can reason as before to determine the relationship A = h(t) = 6(1.013)^t.  

Another common example occurs when dealing with radioactive substances.  These substances decrease by a constant fraction each moment.  For example, assume that at a given moment we have a sample of 500 g. of a radioactive substance that decreases by a factor of 1/2 each week.  Note: We usually describe this by saying that it "decays" by 50%.  Letting A = the total amount of the substance and t = the number of weeks since the moment we first weighed the sample, then A is a function of t, A = k(t).  As before, a similar pattern is evident:

This time h is given by the exponential formula, k(t) = 500(1/2)^t.  

You can see a common pattern to all of these exponential functions.  They are all of the form A = Cb^t, where C = the "initial" amount (i.e., when t = 0) and b = the change factor.  Notice how our interpretation of the constant, C, follows immediately from rules of exponents, since A = Cb⁰ = C·1 = C.   Also, you should notice that there is a direct connection between the change factor, b, and the percentage growth or decrease of a given quantity.  When our money grew by 5%, then it increased by a factor of 1.05 = 1 + 0.05.  Likewise, when we said that our radioactive substance decreased by a factor of 1/2 = 0.5 = 1 - 0.50, we could just as easily have said that it decreased by 50%.  In general, if we think of percentages in decimal terms  (i.e., 25% = 0.25, etc.), we have the following:

A quantity grows by r% precisely when it has a change factor of 1 + r.  Similarly, a quantity decreases (or "decays") by r% precisely when it has a change factor of 1 - r. 

We will return to each of these examples to discuss them in greater detail.  In each application, we will write the constant, C, and the change factor, b, in ways that display the detailed characteristics of the application.

Practice working with simple exponential models by completing the following Exercises.  

Banking, Interest, and Exponential Functions

As we pointed out earlier, one of the most common examples of a quantity growing at a constant factor occurs in banking and finance via interest.  This can either work against you, such as interest on credit card balances or loans, or for you, when you put your money in a savings account or Certificate of Deposit (effectively making a loan to the bank).  In either case, interest is calculated in the same way:

At the end of each agreed upon time period (usually expressed as a fraction of a year) some fraction of the current balance, based on the interest rate, is added to the balance.  This is known as compounding and the time period is called the compounding period.

For example, if you are carrying a balance of $1000 on your credit card, you may need to pay up to 0.05% interest per day.  That means that each day, you balance grows by a factor of 1.0005.  At first this may not seem like a lot of money; you will only owe an additional $1000(.0005) = $0.50 after the first day.  However, over the course of the year you could naively estimate that you would have to pay a total finance charge $0.50(365) = $182.50.  However, it is even worse, because this estimate does not take into account the effect of compounding.  

Specifically, as your balance grows (because of the interest you owe), you will also be expected to pay interest on the interest you already owe.  For example, on the first day you will owe $1000(1.0005), but on the second day, you will owe $1000(1.0005)(1.0005) = $1000(1.0005)², and on the third day $1000(1.0005)³, etc.  By the end of the year you will owe $1000(1.0005)³⁶⁵ » $1200 (the "»" sign stands for "approximately equal to").  That is, your finance charge for the year is close to $200 or 20% of your original balance of $1000.

In this example, we can see that your balance, A, (assuming that you make no payments), is given by the formula:

  A = 1000(1.0005)^d = 1000(1 + .0005)^d, 

where d = the number of days.  In this form, we can see clearly how the daily interest rate of 0.05% (which is .0005, as a decimal) shows up in the formula.  However, interest formulas are usually expressed as a function of years.  Thus, this would become A = 1000(1 + .0005)^365t, where t = the number of years, since d = 365t.  The bank will often describe the interest rate by saying that they charge an "annual (i.e., yearly) rate of 365(0.05)% = 18.25%", but that it is "compounded daily" (i.e., 365 times per year).  This means that they would rewrite this formula as A = 1000(1 + .0005)^365t = 1000(1 + .1825/365)^365t.  In this form, we can see both the annual rate (e.g., 18.25%) and the number of compounding periods (e.g., 365; notice how it appears in two different places in the formula).  Note: If they are really honest, they will admit that you will really end up paying 20% (due to compounding); they generally phrase this by saying that you will pay an "effective annual rate of 20%". 

To summarize what we have learned to this point, we have seen that if some sort of banking account has an associated (annual) interest rate of r% (expressed as a decimal) that is to be compounded k times per year (e.g., "yearly" = 1, "monthly" = 12, "daily" = 365, etc.) and it has an initial balance of I (often referred to as your "principal", when talking about a mortgage loan) then the balance in the account, , is given by the formula:  

Notice that this is of the form A = Cb^t, with C = I (the initial amount) and the growth factor, b = (1 + r/k)^k, which shows the direct influence of the number of periods, k, and the periodic interest rate, r, on the growth rate of the account balance.

While there is only one standard way of calculating interest, for two special choices of the compounding period, this formula can look a bit different.  When the compounding period is "yearly", so that k = 1, this formula simplifies to the one with which we began:

The other case is a bit tricky, but it is also one of the most commonly used methods for computing interest.  When a bank says that it compounds "continuously" (that is, "all the time" or "moment by moment"), they are essentially using a value of k = ¥.  You may well wonder how we can compute a function like A = I(1 + r/¥)^¥t, but it is an amazing fact (that you can prove using Calculus) that, no matter the value of r, (1 + r/¥)^¥/r = 2.71828.....  Because of this, the number 2.71828... is very special (like 3.14159...) and was given a special name of "e", by the great Swiss mathematician, Leonhard Euler (1707-1783) in 1727 (however, its value was known as early as 1614 to the Scottish mathematician, John Napier).  Like p = 3.14159..., we can never write down all the digits for e = 2.71828..., and this special number shows up throughout mathematics.  

Using this notation, the growth factor b = (1 + r/¥)^¥ = ( (1 + r/¥)^¥/r)^r = e^r, and the formula for continuously compounded interest "simplifies" to:

While this looks very different than the original formula, we can see that it really is the "same", in the sense that it gives (essentially) the same values as if we used a really large value for k.  For example, compare the results of continuous compounding on $1000 at 5% interest with the same principal and interest, but compounding (only) 10,000 times per year: 

You can see that these two functions are very close; they agree to about 7 decimal places for small values of t.

As we saw in our original example, the amount of interest that you pay is generally a bit higher than you might suspect.  The yearly growth rate, b - 1 (as opposed to the growth factor) is called the "effective" interest rate; by law, this must be disclosed somewhere in any discussion of loans (but it is often hidden away in the fine print).

The fraction (as a percentage) that the balance changes in one year is known as the effective interest rate, since that is what is actually (i.e., "effectively") paid on the loan.  We will denote it by r_effective.

For example, the growth factor after one year in the compound interest formula is , so in that case .  So, for example, on an 8% auto loan that is compounded monthly, you will pay an effective yearly rate of (1 + 0.08/12)¹² - 1 » .083 or 8.3%.  The higher the number of compounding periods, the worse it gets.  In the extreme case of continuous compounding, the effective yearly rate is .  Thus, if your auto loan were compounded continuously, you will pay an effective yearly rate of e^.08 - 1 » .08329 or 8.329%.  

Practice working with these "interest"ing formulas by completing the following Exercises.  

Go to Constructing Exponential Models

Table of Contents

Send questions or comments to jwicks@northpark.edu

Glossary