Section 3.3: Constructing Exponential Models

In the previous section, we showed how exponential functions of the form f(t) = Cb^t arise in various applications and discussed the applications to banking and finance in great detail.  In this section, we will discuss how to determine the formula for such an exponential function from a graph or table of points.  We will then examine several other applications, and derive some additional formulas, useful for biology, chemistry, geology, and physics.  We will ultimately see how the most general type of functions with an exponential "core" (i.e., of the form f(t) = Cb^t + a) arise in applications and how we can derive such models.

Modeling Exponential Functions

Just as we can determine the precise formula for a linear function from a graph or a table of values, we can do the same thing for exponential functions.  Note: In fact, we will be able to show a direct connection between the two types of functions, once we discuss a new group of functions, called logarithms.  In the previous section, we saw how to view an exponential function, y = f(x) = Cb^x, algebraically, numerically, and geometrically. 

• Algebraically: This consists of a constant factor, C, and a base, b, to the exponent of the input, x.  

We have seen how to interpret the two constants in this formula, as well.

• The base, b: 

  • Numerically: This gives the factor by which the function changes as the input increases by 1 unit.  

  • Geometrically: We have seen that when b > 1, then the graph increases from a horizontal asymptote of  y = 0, as we move to the right; if b < 1, then the graph looks flipped horizontally and decreases to the same horizontal asymptote of  y = 0. 

• The constant factor, C: 

  • Numerically: This is the value of the function, when we plug in x = 0, f(0) = Cb⁰ = C·1 = C.  As with linear functions, we think of this as the "starting" value for the function.

  • Geometrically: As with linear functions, this must then give the value on the vertical axis where the line crosses, that is, its vertical intercept.  

Using Calculus, it is possible to show that:

Any function that increases/decreases continuously by a constant multiple must be given by an exponential function.

Armed with this understanding, we should now be able to work backwards from a table of values or graph of a function to derive its algebraic description.  For example, looking at the table:

we see that it "starts" at 6 and increases by multiples of 3 at each step, so it must correspond to the equation y = 6·3^x - 2.  Similarly, looking at the graph:

we can make a table of points on this graph:

to see that it "starts" at 4 and decreases by a factor of 1/2 as we increase x/move right 1 unit.  This means that it must correspond to the equation y = f(x) = 4(1/2)^x.  

These examples were particularly easy, since we were able to obtain values for points that were 1 unit apart, including the point at x = 0; this was the easy case with linear functions, as well.  As with linear functions, we can use basic algebra to derive a formula, analogous to the slope formula, for the base of an exponential function through any two points, as follows.  Assume that the exponential function, y = f(x) = Cb^x takes the values:

This leads to the two equations:

Then, by dividing one equation by the other:

(using rules of exponents), we discover the change factor, .  We can then solve for the initial amount, C, as . 

Notice that this gives an effective test to see if a table of values of a (continuous) function corresponds to an exponential formula:

If the change factor, , calculated between any two input/output pairs of a function is always the same, the function must be given by an exponential equation.  

As with linear functions, it is sufficient to only check all consecutive pairs in the list.  For example, we can tell, without graphing, that:

can be given by a single, exponential function, of the form g(x) = C2^x, since the change factors between the points are all the same.  Since the vertical intercept is 6, we know that C = 6, and g(x) = 6·2^x.  You can check, by plugging in, that this formula fits all four points.

Even if we cannot read off the vertical intercept, it is easy to solve for, once we have the change factor.  For example, select the function k in the following applet

You can use the "x = " box to read off several points, such as:

Since we obtain a consistent value for the change factor, b = 2/3, we can infer that k(x) = C(2/3)^x.  Plugging in x = -1 gives -72 = k(-1) = C(2/3)^-1 = C(3/2), so that C = -72(2/3) = -48 and k(x) = -48(2/3) ^x.

Make sure that you can distinguish an exponential function from one that is not, and determine its equation from a table of points, by completing the following exercises.

Growth, Decay, and Exponential Functions

We can view many quantities in nature as change continuously by a constant factor, with respect to another independent variable, and so are given by an exponential function, A = Cb^x.  In fact, the same formulas that we use for annually and continuously compounded interest apply equally well.  For example, if we assume that the net, world-wide birth rate to relatively constant over time, we can then model the earth's population by an exponential function.  If we are told that the world population is growing by 1.3% per year, and we know that the population in 1999 was 6 billion people, we can interpret this is two different ways to derive exponential functions to model the earth's population.  

One way to interpret this is that we have an annual growth factor of b = 1 + 0.013.  With this interpretation, if A = the total population (in billions) and t = the number of years since 1999, then we can reason as before with annual compound interest, to determine the relationship A = h(t) = 6(1.013)^t.  However, if you analyze how such populations statistics are actually interpreted (say, at a web-site like GeoHive's Global Statistics), you will see that this is usually viewed as "continuous" growth rate, that is, we should use the continuous compound interest formula, with b = e^0.013 and A = h(t) = 6e^0.013t.

Medicine provides another common situation where exponential functions give an appropriate model.  If you take some medicine, the amount of the drug in your system generally decreases over time.  For example, if you originally take 200 mg. of aspirin for a headache, some time later, say 20 min., only 50% of the aspirin will still be in your bloodstream.  We can use the formula from the previous section to determine the correct formula for A = k(t) = the amount of aspirin in your body (in mg.), in terms of t = the number of minutes since you took the original dose.  We know that:

We can then infer that A = k(t) = 200((1/2)^1/20)^t = 200(1/2)^t/20.  Notice that after 40 min., we will only have 50 mg., and after 60 min., we will only have 25 mg.  That is, the amount of aspirin in your bloodstream drops by 1/2 every 20 minutes.

This kind of example, shows up in Physics, Chemistry, Archaeology, and Geology when one works with radioactive isotopes.  For example, we can date archaeological finds, such as pieces of bone or wood, by measuring the fraction of the carbon that is contains which occurs as the isotope, Carbon-14 (written C¹⁴).  It is known that the concentration of C¹⁴ occurs at the same level in all living organisms (about .000001%).  However, when an organism dies, after about 5600 years, the concentration drops to half of that level.  We can then determine the formula for A = f(t) = the fraction of C¹⁴ (in millionths of a percent) in an organic sample, in terms of t = the time since the organism died (in thousands of years).  We know that:

We can then infer that A = k(t) = 1((1/2)^(1/5.6))^t = (1/2)^t/5.6.  Again, this implies that there will only be .25 millionth of a percent of C¹⁴ after 11.2 thousand years, and .125 millionth after 16.8 thousand years, i.e., the amount of C¹⁴ drops by 1/2 every 5600 years.

This situation, of a quantity dropping by 1/2 in equal time intervals, occurs frequently enough that we given it a name.  We say that the "half-life of C¹⁴ is 5600 years".  Likewise, the "half-life of aspirin" in the previous example was 20 min.  More specifically, he former is called "radioactive half-life", while the latter is called "biological half-life".  We can see that in both cases, the change factor was a conveniently written in terms of the "half-life".

If a certain quantity, A, decreases continuously by a constant factor, the length of time, h, that it takes to drop 50% is known as its half-life.  Then A is given by an exponential function with change factor , so that, A = I (1/2)^t/h, where I = the initial amount (i.e., the value of A when t = 0).

A situation like this, where the quantity decreases exponentially over time, is often referred to as exponential decay.

In all of the situations we have examined so far, the functions involved had a horizontal asymptote of   y = 0.  For instance, in examples involving exponential decay, the quantity would continue to get smaller and smaller, as t ® ¥, but would never reach a 0 amount.  In the case of exponential growth, since the graph grows exponentially as we move to the right, we know that the graph much approach 0 as t ® -¥.

There are many situations which involve exponential growth or decay, where the quantity involved approaches a number other than 0.  For example, if you take a soda from the refrigerator and leave it on your kitchen counter, the soda will get warmer and warmer until its temperature is the same as that of the room.  For example, if T = g(t) = the temperature (in ºF) of the soda, in terms of t = the time since it was taken from the refrigerator (in min.), we might see the following behavior in the first three minutes:

At first, this does not seem to be exponential, since the change factors seem to be (46.4/40)^{1/(1 - 0)} = 1.18, (51.52/46.4)^{1/(1 - 0)} = 1.11034, etc.  However, if we know that the kitchen is 72º F, then we expect this function to have a horizontal asymptote at T = 72.  That is, we can draw the graph like this:

This is the graph of an exponential graph that has been shifted up 72 units, so we can infer that its equation is of the form T = g(t) = Cb^t + 72.  Then, by analyzing T - 72 = g(t) - 72 = Cb^t:

we can determine the change factor, b = 0.8 and the value of C = -32.  Notice that, since there has been a vertical shift, C no longer corresponds to the vertical intercept of the graph (i.e., we know that this is g(0) = 40).

You can check that T is given by the function g(t) = -32(0.8)^t + 72.  Right away, we can see that it is an exponential graph with change factor less than 1, so it would decrease from left to right (with a vertical intercept of 1).  However, it has been stretched vertically by 32 and flipped vertically (so that it increases as we move to the right from a vertical intercept of -32).  Finally, it has been shifted up by 72 (moving the horizontal asymptote up to T = 72 and the vertical intercept to -32 + 72 = 40).  You can see this by viewing the Examples in the following applet:

A similar type of behavior occurs when you take a pizza out of the oven.  This time, the pizza would cool down from around 425º F to 72º F.  Sir Isaac Newton was the first to recognize the precise mathematical formulas involved in heating and cooling, so this is called "Newton's Law of Cooling".  Note:  Newton also discovered the Universal Law of Gravitation; while Galileo "discovered" gravity, Newton shocked the world by using Calculus to show that it applies equally well to planets as apples!

Notice that we could not determine the asymptote from the data alone; we had to be able to infer the value of the horizontal asymptote from specific characteristics of the situation (i.e., the ambient temperature of the room).  As we pointed out in the Exercises, by using rules of exponents every function, f, with an exponential "core" may be put in the form f(x) = Cb^x + a, where, a, is its horizontal asymptote.  Thus, we have seen how to determine the formula for any such function, as long as we are given a hint as to the value of the asymptote.

Practice determining appropriate exponential models by completing the following Exercises.  

Go to Logarithmic Functions

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