### 12.3 - Exponential Functions

Click here to review the definition of a function. Click here to see how exponential functions compare with other types of functions in the gallery of functions.

Exponential functions are closely related to geometric sequences. A geometric sequence is a list of numbers in which each number is obtained by multiplying the previous number by a fixed factor m. An example is the sequence {1, 3, 9, 27, 81, …}. If we label the numbers in the sequence as {y0, y1, y2, …} then their values are given by the formula
yn = y0 · m n.

A geometric sequence is completely described by giving its starting value y0 and the multiplication factor m. For the above example y0 = 1 and m = 3. Another example of a geometric sequence is the sequence {40, 20, 10, 5, 2.5, …}. For this sequence y0 = 40 and m = 0.5.

An exponential function is obtained from a geometric sequence by replacing the counting integer n by the real variable x. The graph below shows the exponential functions corresponding to these two geometric sequences.

Thus we define an exponential function to be any function of the form

y = y0 · m x.
It gets its name from the fact that the variable x is in the exponent. The “starting valuey0 may be any real constant but the base m must be a positive real constant to avoid taking roots of negative numbers.

The exponential function y = y0 · m x has these two properties:
• When x = 0 then y = y0.

• When x is increased by 1 then y is multiplied by a factor of m. This is true for any real value of x, not just integer values of x. To prove this suppose that y has some value ya when x has some value xa. That is, Now increase x from xa to xa + 1. We get We see that y is now m times its previous value of ya. If the multiplication factor m > 1 then we say that y grows exponentially, and if m < 1 then we say that y decays exponentially.

### The Graph of the Exponential Function

We have seen graphs of exponential functions before:
• In the section on real exponents we saw a saw a graph of y = 10 x.
• In the gallery of basic function types we saw five different exponential functions, some growing, some decaying.
The graph below shows two more examples. The first example is the exponential growth function y = 1 (3)x. For it y = 1 when x = 0 and y grows by a factor of 3 when x increases by 1. The second example is the exponential decay function y = 40 (0.5)x. For it y = 40 when x = 0 and y is multiplied by a factor of 0.5 when x is increased by 1. (Equivalently we could say that y decays by a factor of 2 when x is increased by 1.)

Notice that the two curves have the same general shape but are reversed left to right and that neither ever touches the x axis. Notice also that if we interchange the x and y axes then the graph of an exponential function turns into the graph of a logarithmic function. The reason for this is that if we take the exponential function y = b x, then interchange x and y to get x = b y, and then solve for y we get y = log b(x). ### The Special Propery of y = e x

We have seen graphs of various exponential functions y = b x with various bases b. Notice that for all of them, as we go higher and higher up the curves, they get steeper and steeper. Of all the possible bases there is one particular base, namely e, that causes the curve to have the property that the slope exactly equals the height at every point along the curve. This property defines the number e and makes the curve y = e x an important standard in calculus where we study the slopes of different functions. Let’s prove this important property.

### Constructing the Function y = e x

In this section we will construct a function whose slope equals its own height everywhere and then show that it is indeed the function y = e x. We begin in the figure to the right with a single straight-line segment whose left endpoint is at (x = 0, y = 1). Since the height there is 1 we demand that the slope equal 1. The problem of course is that this line segment rises to the right but the slope is not increasing as it should if we wish the slope to equal the height everywhere along the curve. So we improve on this by dividing the region 0 < x < 1 into two equal intervals as shown here. At x = 0.5 we change the slope to 1.5 to reflect the fact that the height there is 1.5. Now the slope equals the height at two points, namely the left endpoints of both line segments. Now we improve even more by dividing the region 0 < x < 1 into n equal intervals as shown here. We can find the height of the right endpoint of each line segment by the following method:

Start with the formula for the slope of any line segment. In this formula yR is the height at the right endpoint and yL is the height at the left endpoint of the line segment. Now we demand that the slope of each line segment equals the height of its left endpoint. Substitute this in. Now solve for yR. Knowing yL we can use this formula to find the yR. We can iterate this formula across all n line segments, working from left to right. The heights of the right endpoints of all n line segments work out to be: These heights are shown in the third figure. Now we let n → ∞ (i.e. we let the number of intervals become infinite). The curve will become smooth and the slope will equal the height everywhere. We claim that the resulting curve is the function y = e x. To prove this let’s first find the value of y at x = 1. The third figure says that this value is (1 + 1 / n) n. In the following table we have used a calculator to find the value of the expression (1 + 1 / n) n for various values of n:

 n (1 + 1 / n) n 1 2 2 2.25 10 2.59374 1000 2.71692 1,000,000 2.71828

The values are approaching a definite limit as n becomes large. We define the number e to be the value of the expression (1 + 1 / n) n in the limit as n → ∞. (The number is named e in honor of Leonard Euler who first discovered it.) We express this using this notation: Now let’s find the value of y at arbitrary x. We must begin at x = 0 and go right a total of n · x intervals or steps. From the figure above it is clear that the value of y after n x intervals is y = (1 + 1 / n) n x. By the rules of exponents this can be written in the limit as n → ∞. Thus the function whose slope equals its own height everywhere and which goes through the point (x = 0 , y = 1) is y = e x as claimed.

 Summary: The function y = e x is the function whose slope equals its own height everywhere and which goes through the point (x = 0 , y = 1).

 Generalization: The function y = y0 · e bx is the function whose slope is b times its own height everywhere and which goes through the point (x = 0 , y = y 0 ). (b may be positive or negative.)

This generalization can be understood as follows:
• The function y = y0 · e x is just the function y = e x but stretched vertically by the factor y0. Thus both its slope and height are everywhere multiplied by the factor y0. Its slope still equals its own height everywhere but it goes through the point (x = 0 , y = y0) .

• The function y = y0 · e bx is just the function y = y0 · e x but squeezed horizontally by the factor b. Thus its slope is everywhere multiplied by factor b but its height is not changed.

### Alternative Forms for Exponential Growth and Decay

Let us now replace the independent variable x in the exponential function by the variable t and let it represent time. Now the exponential function looks like this:
y = y0 · m t.
We already know that y has the value y0 when t = 0 and that y is multiplied by a factor m when time t is increased by 1. If m > 1 then y grows with time and if m < 1 then y decays with time. We call the growth or decay exponential because t is in the exponent. We have seen how we can change the base by using the change of base formula. In this section we want to show how using various bases or putting the exponent in various forms can reveal various features of the exponential growth or decay.

Form 1: Base Greater than 1.     First, we can always choose the base m to be greater than 1. The reason for doing this is that then
y = y0 · mt,
with positive exponent makes it obvious that we have exponential growth, and

y = y0 · m− t,
with negative exponent makes it obvious that we have exponential decay. The growth or decay is by the factor m each time t increases by 1. For example the function y = (¼) t describes exponential decay but the base is smaller than 1. But using the rules of exponents we can rewrite it as y = (4−1) t, or as
y = 4 − t.
Now the base is bigger than 1 and the exponent is negative. Notice that these two forms show that the statements “y is multiplied by a factor of 1/4 ” and “y decays by a factor of 4 ” are equivalent.

Form 2: Growth or Decay by Given Factor in Given Time.     Next we want to give t units, say seconds. To do this we can write an exponential growth in the form
y = y0 · b t / T,
where b again is a base > 1, and T is a positive constant with the same units of time as t. This form is useful because it makes it plain that y grows by a factor b in a time of T seconds as we can see from this table of values:
 t y0 · b t / T 0 y0 · b 0   =   y0 T y0 · b T / T   =   y0 · b 2 T y0 · b 2 T / T   =   y0 · b 2

Note: We can always change b but we must change T accordingly. For example exponential growth by a factor of 10 every 1 second is equivalent to growth by a factor of 100 every 2 seconds.

Similarly, any exponential decay can be written as
y = y0 · b − t / T,
making it plain that y decays by a factor b in a time of T seconds.

Form 3: The Time Constant Form.     This is a special case of Form 2. If b = e then the constant T is called the time constant and is denoted by the greek letter τ (tau). The exponential growth formula now reads
y = y0 · et / τ,
making it plain that y grows by a factor of e or approximately 2.7 every τ seconds. The decay formula reads
y = y0 · e − t / τ,
making it plain that y decays by a factor of e or approximately 2.7 every τ seconds (or decays to e − 1 ≈ 37% of its former value every τ seconds). This can be seen from the table of values:
 t y0 · e − t / τ 0 y0 · e 0   =   y0 τ y0 · e − τ / τ   =   y0 · e − 1   ≈   37% of  y0 5 τ y0 · e − 5 τ / τ   =   y0 · e − 5   ≈   0.7% of  y0

Electrical engineers prefer this form because τ is easy to measure and to calculate.

 Summary: The time constant τ is the period of time that it takes for an exponentially decaying quantity to decay to a fraction e − 1 of its initial value, or to about 37% of its initial value. After a period of 5 time constants it has decayed to less than 1% of its initial value and for many engineering purposes equals zero.

Form 4: The Rate Form.     Any exponential growth can be written in the form

y = y0 · er t.
Comparing this with the time constant form we see that r = 1 / τ. Assuming τ has units of seconds, then r has units of 1/seconds. r is called the instantaneous growth rate. Bankers and people interested in rates of growth prefer this form. The equation y = y0 · er t represents a quantity y whose initial value is y0 and whose rate of growth at any instant equals r times its value at that instant.

Similarly any exponential decay can be written in the form
y = y0 · e − r t,
This equation represents a quantity y whose initial value is y0 and whose rate of decay at any time equals r times its value at that time.

Example: Describe the function y = \$50 · e 0.20 t in words and sketch its graph. Assume that t is measured in years.

Solution: This is an exponential growth function expressed in rate form. Its value is \$50 at time 0 and it grows at a rate of 20% per year. In the sketch below we have shown that the slope of the curve is 0.20 times the height at three different points along the curve. Example: Describe the function y = 100 · e − t / 1.5 and sketch its graph.

Solution: This is an exponential decay function expressed in time constant form. Its value at time 0 is 100 and it decays to 37% of its former value in any 1.5 second interval. In the sketch we have shown this for two different intervals. Example: Take the exponential function y = 12 · (1/3) t and put it into the decay by a given factor in a given time” form which will show at a glance how long it takes
a) to decay by a factor of 9, and
b) to decay by a factor of 100.
Solution:

a) We wish to change the base to 9 so the formula reads Then the value of T will be the time required to decay by the factor of 9. Notice that 1/3 = 9 −1/2. Substituting this into the original exponential function gives This form clearly shows that y decays by a factor of 9 every 2 seconds.

b) We wish to change the base to 100 so the formula reads No simple short-cut with exponents is possible in this case. Instead we can equate the given original form for y with the desired form. This gives We must solve this equation for T. To do this divide by 12 and take log 10 on both sides Use property 3 of logarithms to bring down to exponent: Now divide both sides by t and solve for T. We get
T = 4.19 seconds.
Substituting this into the desired form gives This form clearly shows that y decays by a factor of 100 every 4.19 seconds.

Example: Consider the function y = 1000 · (1/4) t which is graphed below: Using the exponential form of the change of base formula and some simple algebra it is possible to rewrite this function in the following equivalent forms: Discuss the merits of each of the forms.

Solution:

All of the forms have a base greater than 1 so the negative exponent indicates exponential decay (as opposed to growth). All of the forms have y with an initial value of 1000.
• Form (a) shows that y decays by a factor of 4 each time t increases by 1 second.

• Form (b) shows that y decays by a factor of 2 each time t increases by 0.5 sec.

• Form (c) shows that y decays at an instantaneous rate of 138.6% per second. (The dashed triangle in the figure shows that at the initial value of 1000 this implies an initial slope of −1386.)

• Form (d) shows that y decays by a factor of e (or decays to 1/e ≈ 37% of of its former value) each time t increases by 0.721 sec. In other words y decays with a time constant of 0.721 sec.
Note that these statements hold anywhere along the exponential curve. It is also interesting to note that if the decay were linear (followed a straight line) then y would hit zero at t = τ.

Example: Suppose that we put \$1500 into a bank account which receives interest at a rate of 8% per year and which is compounded continuously. (Compounded continuously is just another way of saying grows exponentially.) Let y denote the amount of money in the account at any time t. Then y can be expressed in the rate form where y0 = \$1500 is called the principal and r = 0.08/yr is the interest rate. The units of y will also be dollars and t must be given in years so that the units of r and t cancel. Questions:
(a) What amount will be in the account at the end of 15 months?

(b) After how many years will there be \$ 4000 in the account?
Solution:

(a) Substituting t = 15 months = 1.25 yrs into the equation and evaluating gives (b) Substituting y = \$4000 into the equation and solving for t gives  t = 12.26 yrs.
Answers:
(a) After 15 months the account holds \$ 1657.75.

(b) The account holds \$ 4000 after 12.26 yrs.

Example: Make a sketch of the exponential decay function y = 45 · e − t / 20 Solution: Follow these steps:

(a) Draw a smooth decay curve from upper-left to lower-right. Label the axes but don’t put any numbers on the axes yet. (b) Draw a bracket (or just imagine it) to indicate where y has 100% (or all) of its initial value. Then draw brackets ½ and ¼ as high where y has 50% and 25% of its initial value. About half-way between 25% and 50% lies 37%. Remember that e−1 is about 37%. (c) Go across at 37% until you hit the curve then go down. This value on the t axis is the time constant. Our function has a time constant of 20 so put that on the t axis. (d) Accuracy check: a straight line with the initial slope should hit the same spot on the t axis. (e) Finish up the graph by putting more tick marks and values on both axes.

 Algebra Coach Exercises

### Analyzing Exponential Growth and Decay

Just as 2 points determine a straight line, so 2 points determine an exponential function. To derive its equation follow these steps:
1. choose the desired form of the equation,
2. substitute in the 2 points, and
3. solve the resulting 2 equations for the 2 unknowns, one being the initial value and the other being the growth or decay rate or time constant. Example: The electric current, i, flowing in a certain electric circuit decays exponentially with time, t, as shown. Two points on the curve are given. Find an exponential equation of the time constant form
i = i0 · e − t / τ
to describe the current.

Solution: We will explain two methods of solving this problem.

Method 1: Substitute the values of i and t at the 2 given points into the equation. This yields a system of 2 equations in the 2 unknowns. The unknowns are i0 and τ: We can eliminate i0 by dividing these equations Take the natural logarithm of both sides and solve for τ
ln(3.222) = 2.85 / τ

τ = 2.436
Now back-substitute this value of τ into, say, the first of the two equations to get i0:
8.7 = i0 · e − 1.25 / 2.436

i0 = 14.5
Thus the equation is i = 14.5 · e − t / 2.44.

Method 2: This method uses the fact that an exponential function decays by a given factor in a given time anywhere along the curve. Thus for the purposes of finding the time constant τ we may use the value 8.7 as i0 and the time difference 4.1 − 1.25 = 2.85 sec. as the time taken for the function to decay to the value 2.7. Substituting these numbers into the equation yields
2.7 = 8.7 · e − 2.85 / τ
This equation can be solved for τ
τ = 2.436
Now get the actual i0 by substituting τ and one of the points on the curve, say (t = 1.25, i = 8.7), into the equation i = i0 · e − t / τ to get:

8.7 = i0 · e − 1.25 / 2.436

i0 = 14.5
Thus again we find the equation to be i = 14.5 · e − t / 2.44.

 Algebra Coach Exercises

### Exponential Decay Toward a Limiting Value The figure to the right shows four functions whose differences from the limiting value y = 5 decay exponentially at various rates. The functions are:
(a)     y = 5 − 5 e − t / 2
(b)     y = 5 + 3 e − t / 2
(c)     y = 5 + 3 e − t / 6
(d)     y = 5 − 7 e − t / 6

These functions could describe the temperature of hot drinks cooling and cold ones warming toward room temperature. (c) and (d) are in better insulated containers so they take longer to heat or cool. These functions are all of the form
y = y + a e − t / τ,
There are 3 parameters: y , a and τ. If we write transpose y to the left-hand-side and write the equation as
yy = a e − t / τ,
then we see that the right side is the familiar exponential decay in time constant form, and that the difference of y from y equals a when t = 0 and that this difference decays with time constant τ. When t = ∞ then y = y. Because there are 3 parameters, we must be given the value of y at 3 different times to fix them.

We look at two examples. In this first example the limiting value is given so we only need 2 more points. Example: The voltage in a certain electric circuit decays exponentially toward the limiting value v = 8.2. Two points on the curve are given. The curve may be described by an equation of the form
v = v + a e − t / τ,
Calculate the values of a and τ.

Solution: Similar to the previous example, there are two methods of solving this problem.

Method 1: Substitute v = 8.2 and the values of v and t at the 2 given points into the equation. This yields a system of 2 equations in the 2 unknowns a and τ: In the previous example we divided one equation by the other in order to eliminate one of the variables. Here we must first move the 8.2 to the left-hand-side: Now we can divide the equations. We get
0.3226 = e −1.8 / τ
Solving gives τ = 1.591 and back-substituting gives a = −13.18.

Method 2: In this method use the fact that the difference of v from v decays exponentially to find the time constant τ first. At the first point this difference is 6.2, and 1.8 seconds later the difference is 2.0. Substituting these values into the time constant form of the exponential decay formula, y = y0 e − t / τ, gives
2.0 = 6.2 e − 1.8 / τ
Solving for τ gives
τ = 1.591
Now find the value of a by substituting τ, v and the value of t and v at either one of the given points into the equation v = v + a e − t / τ. This again gives a = −13.18.

In this second example we are given 3 points separated by equal time intervals. If the time intervals were unequal the resulting system of equations could only be solved by computer. Example: The curve to the right is described by the equation
v = v + a e − t / τ,
Calculate the values of v , a and τ accurate to 3 significant figures.

Solution: Move v to the left-hand-side as we did in the previous example. Then substitute in the values of v and t at the 3 given points. This yields the following system of 3 non-linear equations in the 3 unknowns v , a and τ: We can eliminate a by dividing Eq.(3) by (2), and (2) by (1), to get: Now we can eliminate τ by equating Eqs.(4) and (5). (Note that they are equal only because the 3 points were separated by equal time intervals.) We get After cross-multiplying we can solve to get v =116.45 volts. Substituting this back into Eq.(4) or (5) we get τ = 8.151 secs and substituting this back into Eq.(1), (2) or (3) we get a = 152.9 volts, so the final equation to 3 sig. figs. is
v = 116 − 153 e − t / 8.15.

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