8.7 Exponential Growth and Decay
Exponential Growth Models
Recalling the investigations in Section 8.3, we started by developing a formula for discrete compound interest. This led to another formula for continuous compound interest,
P (t) = P0ert, (1)
where P0 is the initial amount (principal) and r is the annual interest rate in decimal form. If money in a bank account grows at an annual rate r (via payment of interest), and if the growth is continually added in to the account (i.e., interest is continuously compounded), then the balance in the account at time t years is P (t), as given by formula (1).
But we can use the exact same analysis for quantities other than money. If P (t) represents the amount of some quantity at time t years, and if P (t) grows at an annual rate r with the growth continually added in, then we can conclude in the same manner that P (t) must have the form
P (t) = P0ert, (2)
where P0 is the initial amount at time t = 0, namely P (0).
A classic example is uninhibited population growth. If a population P (t) of a certain species is placed in a good environment, with plenty of nutrients and space to grow, then it will grow according to formula (2). For example, the size of a bacterial culture in a petri dish will follow this formula very closely if it is provided with optimal living conditions. Many other species of animals and plants will also exhibit this behavior if placed in an environment in which they have no predators. For example, when the British imported rabbits into Australia in the late 18th century for hunting, the rabbit population exploded because conditions were good for living and reproducing, and there were no natural predators of the rabbits.
Exponential Growth
If a function P (t) grows continually at a rate r > 0, then P (t) has the form
P (t) = P0ert, (3)
where P0 is the initial amount P (0). In this case, the quantity P (t) is said to exhibit exponential growth, and r is the growth rate.
Remarks 4.
1. If a physical quantity (such as population) grows according to formula (3), we say that the quantity is modeled by the exponential growth function P (t).
2. Some may argue that population growth of rabbits, or even bacteria, is not really continuous. After all, rabbits are born one at a time, so the population actually grows in discrete chunks. This is certainly true, but if the population is large, then the growth will appear to be continuous. For example, consider the world population of humans. There are so many people in the world that there are many new births and deaths each second. Thus, the time difference between each 1 unit change in the population is just a tiny fraction of a second, and consequently the discrete growth will act virtually the same as continuous growth. (This is analogous to the almost identical results for continuous compounding and discrete daily compounding that we found in Section 8.3; compounding each second or millisecond would be even closer.)
3. Likewise, using the continuous exponential growth formula (3) to model discrete quantities will sometimes result in fractional answers. In this case, the results will need to be rounded off in order to make sense. For example, an answer of 224.57 rabbits is not actually possible, so the answer should be rounded to 225.
4. In formula (3), if time is measured in years (as we have done so far in this chapter), then r is the annual growth rate. However, time can instead be measured in any convenient units. The same formula applies, except that the growth rate r is given in terms of the particular time units used. For example, if time t is measured in hours, then r is the hourly growth rate.
In Section 8.2, we showed that a function of the form btwith b > 1 is an exponential growth function. Likewise, if A > 0, then the more general exponential function Abt also exhibits exponential growth, since the graph of Abt is just a vertical scaling of the graph of bt. However, the exponential growth function in formula (3) appears to be different. We will show below that the function P0ert can in fact be written in the form Abt with b > 1.
Let’s first look at a specific example. Suppose P (t) = 4e0.8t. Using the Laws of Exponents, we can rewrite P (t) as
P (t) = 4e0.8t= 4(e0.8)t. (5)
Since e0.8 ≈ 2.22554, it follows that
P (t) ≈ 4(2.22554)t.
Because the base ≈ 2.22554 is larger than 1, this shows that P (t) is an exponential growth function, as seen inFigure 1(a)).
Now suppose that P (t) is any function of the form P0ert with r > 0. As in (5) above, we can use the Laws of Exponents to rewrite P (t) as
P (t) = P0ert= P0(er)t= P0bt with b = er.
To prove that b > 1, consider the graph of y = ex shown inFigure 1(b). Recall that e ≈ 2.718, so e > 1, and therefore y = ex is itself an exponential growth curve. Also, the y-intercept is (0, 1) since e0 = 1. It follows that b = er > 1 since r > 0 (see Figure 1(b)).
Therefore, functions of the form P (t) = P0ert with r > 0 are exponential growth functions.
2t y
20 P
2x y
5 y = ex
er
r 1
(a) P (t) = 4e0.8t (b) b = er> 1 since r > 0 Figure 1.
Applications of Exponential Growth
We will now examine the role of exponential growth functions in some real-world ap- plications. In the following examples, assume that the population is modeled by an exponential growth function as in formula (3).
I Example 6. Suppose that the population of a certain country grows at an annual rate of 2%. If the current population is 3 million, what will the population be in 10 years?
This is a future value problem. If we measure population in millions and time in years, then P (t) = P0ert with P0 = 3 and r = 0.02. Inserting these particular values into formula (3), we obtain
P (t) = 3e0.02t.
The population in 10 years is P (10) = 3e(0.02)(10) ≈ 3.664208 million.
I Example 7. In the same country as in Example 6, how long will it take the population to reach 5 million?
As before,
P (t) = 3e0.02t.
Now we want to know when the future value P (t) of the population at some time t will equal 5 million. Therefore, we need to solve the equation P (t) = 5 for time t, which leads to the exponential equation
5 = 3e0.02t.
Using the procedure for solving exponential equations that was presented in Section 8.6,
5 = 3e0.02t
=⇒ 5
3 = e0.02t isolate the exponential
=⇒ ln 5 3
= ln(e0.02t) apply the natural log function
=⇒ ln 5 3
= 0.02t since ln(ex) = x
=⇒ ln 53
0.02 = t division
=⇒ t ≈ 25.54128.
Thus, it would take about 25.54 years for the population to reach 5 million.
The population of bacteria is typically measured by weight, as in the next two examples.
I Example 8. Suppose that a size of a bacterial culture is given by the function P (t) = 100e0.15t,
where the size P (t) is measured in grams and time t is measured in hours. How long will it take for the culture to double in size?
The initial size is P0 = 100 grams, so we want to know when the future value P (t) at some time t will equal 200. Therefore, we need to solve the equation P (t) = 200 for time t, which leads to the exponential equation
200 = 100e0.15t. Using the same procedure as in the last example,
200 = 100e0.15t
=⇒ 2 = e0.15t isolate the exponential
=⇒ ln(2) = ln(e0.15t) apply the natural log function
=⇒ ln(2) = 0.15t since ln(ex) = x
=⇒ ln(2)
0.15 = t division
=⇒ t ≈ 4.620981.
Thus, it would take about 4.62 hours for the size to double.
The last example deserves an additional comment. Suppose that we had started with 1000 grams instead of 100. Then to double in size would require a future value of 2000 grams. Therefore, in this case, we would have to solve the equation
2000 = 1000e0.15t.
But the first step is to isolate the exponential by dividing both sides by 1000 to get 2 = e0.15t,
and this is the same as the second line of the solution in the last example, so the answer will be the same. Likewise, repeating this argument for any initial amount will lead to the same second line, and therefore the same answer. Thus, the doubling time depends only on r, not on the initial amount P0.
Exponential Decay Models
We’ve observed that if a quantity increases continually at a rate r, then it is modeled by a function of the form P (t) = P0ert. But what if a quantity decreases instead?
Although we won’t present the details here, the analysis can be carried out in the same way as the derivation of the continuous compounding formula in Section 8.3. The only difference is that the growth rate r in the formulas must be replaced by −r since the quantity is decreasing. The conclusion is that the quantity is modeled by a function of the form P (t) = P0e−rt instead of P0ert.
Exponential Decay
If a function P (t) decreases continually at a rate r > 0, then P (t) has the form
P (t) = P0e−rt, (9)
where P0 is the initial amount P (0). In this case, the quantity P (t) is said to exhibit exponential decay, and r is the decay rate.
In Section 8.2, we showed that a function of the form btwith b < 1 is an exponential decay function. Likewise, if A > 0, then the more general exponential function Abtalso exhibits exponential decay, since the graph of Abtis just a vertical scaling of the graph of bt. However, the exponential decay function in formula (9) appears to be different.
We will show below that the function P0e−rt can in fact be written in the form Abt with b < 1.
Let’s first look at a specific example. Suppose P (t) = 4e−0.8t. Using the Laws of Exponents, we can rewrite P (t) as
P (t) = 4e−0.8t= 4(e−0.8)t. (10)
Since e−0.8 ≈ 0.44933, it follows that
P (t) ≈ 4(0.44933)t.
Because the base ≈ 0.44933 is less than 1, this shows that P (t) is an exponential decay function, as seen inFigure 2(a)).
Now suppose that P (t) is any function of the form P0e−rt with r > 0. As in (10) above, we can use the Laws of Exponents to rewrite P (t) as
P (t) = P0e−rt= P0(e−r)t= P0bt with b = e−r.
To prove that b < 1, consider the graph of y = e−x shown inFigure 2(b). Now e−x= (e−1)x= 1
e
x
and 1/e ≈ 0.36788 < 1, so y = e−x is itself an exponential decay curve. (Alternatively, you can observe that the graph of y = e−x is the reflection of the graph of y = ex across the y-axis.) Also, the y-intercept is (0, 1) since e−0 = 1. It follows that b = e−r < 1 since r > 0 (seeFigure 2(b)).
Therefore, functions of the form P (t) = P0e−rt with r > 0 are exponential decay functions.
2t y
P 20
2x y
y = e−x 5
e−r r 1
(a) P (t) = 4e−0.8t (b) b = e−r < 1 since r > 0 Figure 2.
Applications of Exponential Decay
The main example of exponential decay is radioactive decay. Radioactive elements and isotopes spontaneously emit subatomic particles, and this process gradually changes the substance into a different isotope. For example, the radioactive isotope Uranium- 238 eventually decays into the stable isotope Lead-206. This is a random process for individual atoms, but overall the mass of the substance decreases according to the exponential decay formula (9).
I Example 11. Suppose that a certain radioactive element has an annual decay rate of 10%. Starting with a 200 gram sample of the element, how many grams will be left in 3 years?
This is a future value problem. If we measuring size in grams and time in years, then P (t) = P0e−rt with P0 = 200 and r = 0.10. Inserting these particular values into formula (9), we obtain
P (t) = 200e−0.10t.
The amount in 3 years is P (3) = 200e−(0.10)(3)≈ 148.1636 grams.
I Example 12. Using the same element as in Example 11, if a particular sample of the element decays to 50 grams after 5 years, how big was the original sample?
This is a present value problem, where the unknown is the initial amount P0. As before, r = 0.10, so
P (t) = P0e−0.10t. Since P (5) = 50, we have the equation
50 = P (5) = P0e−(0.10)(5). This equation can be solved by division:
50
e−(0.10)(5) = P0
Finish by calculating the value of the left side to get P0 ≈ 82.43606 grams.
I Example 13. Suppose that a certain radioactive isotope has an annual decay rate of 5%. How many years will it take for a 100 gram sample to decay to 40 grams?
Use P (t) = P0e−rtwith P0 = 100 and r = 0.05, so P (t) = 100e−0.05t.
Now we want to know when the future value P (t) of the size of the sample at some time t will equal 40. Therefore, we need to solve the equation P (t) = 40 for time t, which leads to the exponential equation
40 = 100e−0.05t.
Using the procedure for solving exponential equations that was presented in Section 8.6,
40 = 100e−0.05t
=⇒ 0.4 = e−0.05t isolate the exponential
=⇒ ln(0.4) = ln(e−0.05t) apply the natural log function
=⇒ ln(0.4) = −0.05t since ln(ex) = x
=⇒ ln(0.4)
−0.05 = t division
=⇒ t ≈ 18.32581.
Thus, it would take approximately 18.33 years for the sample to decay to 40 grams.
We saw earlier that exponential growth processes have a fixed doubling time. Sim- ilarly, exponential decay processes have a fixed half-life, the time in which one-half the original amount decays.
I Example 14. Using the same element as inExample 13, what is the half-life of the element?
As before, r = 0.05, so
P (t) = P0e−0.05t.
The initial size is P0 grams, so we want to know when the future value P (t) at some time t will equal one-half the initial amount, P0/2. Therefore, we need to solve the equation P (t) = P0/2 for time t, which leads to the exponential equation
P0
2 = P0e−0.05t. Using the same procedure as in the last example,
P0
2 = P0e−0.05t
=⇒ 1
2 = e−0.05t isolate the exponential
=⇒ ln 1 2
= ln(e−0.05t) apply the natural log function
=⇒ ln 1 2
= −0.05t since ln(ex) = x
=⇒ ln 12
−0.05 = t division
=⇒ t ≈ 13.86294.
Thus, the half-life is approximately 13.86 years.
The process of radioactive decay also forms the basis of the carbon-14 dating tech- nique. The Earth’s atmosphere contains a tiny amount of the radioactive isotope carbon-14, and therefore plants and animals also contain some carbon-14 due to their interaction with the atmosphere. However, this interaction ends when a plant or ani- mal dies, so the carbon-14 begins to decay (the decay rate is 0.012%). By comparing the amount of carbon-14 in a bone, for example, with the normal amount in a living animal, scientists can compute the age of the bone.
I Example 15. Suppose that only 1.5% of the normal amount of carbon-14 remains in a fragment of bone. How old is the bone?
Use P (t) = P0e−rtwith r = 0.00012, so
P (t) = P0e−0.00012t.
The initial size is P0 grams, so we want to know when the future value P (t) at some time t will equal 1.5% of the initial amount, 0.015P0. Therefore, we need to solve the equation P (t) = 0.015P0 for time t, which leads to the exponential equation
0.015P0 = P0e−0.00012t. Using the same procedure as in Example 14,
0.015P0 = P0e−0.00012t
=⇒ 0.015 = e−0.00012t isolate the exponential
=⇒ ln (0.015) = ln(e−0.00012t) apply the natural log function
=⇒ ln (0.015) = −0.00012t since ln(ex) = x
=⇒ ln (0.015)
−0.00012 = t division
=⇒ t ≈ 34998.
Thus, the bone is approximately 34998 years old.
While the carbon-14 technique only works on plants and animals, there are other similar dating techniques, using other radioactive isotopes, that are used to date rocks and other inorganic matter.
