Exponential Functions
Definition of Exponential Function
An exponential function is any function for which the rule specifies the independent variable in an exponent. A basic exponential function has the form $ F(x) = a^{x}, a > 0, a $ . The domain of a basic exponential function is considered to be the set of all real numbers, unless otherwise specified.
EXAMPLE 17.1 The following are examples of exponential functions:
\[ \left(\mathrm{a}\right)f(x)=2^{x};\left(\mathrm{b}\right)f(x)=\left(\frac{1}{2}\right)^{x};\left(\mathrm{c}\right)f(x)=4^{-x};\left(\mathrm{d}\right)f(x)=2^{-x^{2}} \]
Properties of Exponents
Properties of exponents can be restated for convenience in terms of variable exponents. Assuming a, b > 0, then for all real x and y:
\[ a^{x}a^{y}=a^{x+y}\quad(ab)^{x}=a^{x}b^{x} \]
\[ \frac{a^{x}}{a^{y}}=a^{x-y}\qquad\left(\frac{a}{b}\right)^{x}=\frac{a^{x}}{b^{x}} \]
\[ (a^{p})^{x}=a^{px} \]
The Number e
The number e is called the natural exponential base. It is defined as $ _{n}(1+)^n $ . e is an irrational number with a value approximately 2.718 281 828 459 045.
Exponential Growth and Decay
Applications generally distinguish between exponential growth and decay. A basic exponential growth function is an increasing exponential function; an exponential decay function is a decreasing exponential function.
Compound Interest
If a principal of P dollars is invested at an annual rate of interest r, and the interest is compounded n times per year, then the amount of money $ A(t) $ generated at time t is given by the formula:
\[ A(t)=P\bigg(1+\frac{r}{n}\bigg)^{nt} \]
Continuous Compound Interest
If a principal of P dollars is invested at an annual rate of interest r, and the interest is compounded continuously, then the amount of money $ A(t) $ available at any later time t is given by the formula:
\[ A(t)=Pe^{rt} \]
Unlimited Population Growth
If a population consisting initially of $ N_{0} $ individuals also is modeled as growing without limit, the population $ N(t) $ at any later time t is given by the formula (k is a constant to be determined):
\[ N(t)=N_{0}e^{k t} \]
Alternatively, a different base can be used.
Logistic Population Growth
If a population consisting initially of $ N_{0} $ individuals is modeled as growing with a limiting population (due to limited resources) of P individuals, the population $ N(t) $ at any later time t is given by the formula (k is a constant to be determined):
\[ N(t)=\frac{N_{0}P}{N_{0}+(P-N_{0})e^{-kt}} \]
Radioactive Decay
If an amount $ Q_{0} $ of a radioactive substance is present at time t = 0, then the amount $ Q(t) $ of the substance present at any later time t is given by the formula (k is a constant to be determined):
\[ Q(t)=Q_{0}e^{-kt} \]
Alternatively, a different base can be used.
SOLVED PROBLEMS
17.1 Explain why the domain of a basic exponential function is considered to be R. What is the range of the function?
Consider, for example, the function $ f(x) = 2^{x} $ . The quantity $ 2^{x} $ is defined for all integer x; for example, $ 2^{3} = 8 $ , $ 2^{-3} = $ , $ 2^{0} = 1 $ , and so on. Moreover, the quantity $ 2^{x} $ is defined for all noninteger rational x, for example, $ 2^{1/2} = $ , $ 2^{5/3} = $ , $ 2^{-3/4} = 1/ $ , and so on.
To define the quantity $ 2^{x} $ for x an irrational number, for example, $ 2^{} $ , use the nonterminating decimal representing $ $ , that is, 1.4142 . . ., and consider the rational powers $ 2^{1} $ , $ 2^{1.4} $ , $ 2^{1.41} $ , $ 2^{1.414} $ , $ 2^{1.4142} $ , and so on. It can be shown in calculus that each successive power gets closer to a real number, which is defined as $ 2^{} $ . This process can be applied to define the quantity $ 2^{x} $ for x any irrational number, hence, $ 2^{x} $ is defined for all real numbers x. The domain of $ f(x) = 2^{x} $ is considered to be R, and similarly for any exponential function $ f(x) = a^{x}, a > 0, a $ .
Since $ 2^{x} $ is positive for all real x, the range of the function is the positive numbers, $ (0, ) $ .
17.2. Analyze and sketch the graph of a basic exponential function of form $ f(x) = a^{x}, a > 1 $ .
The graph has no obvious symmetry. Since $ a^{0}=1 $ , the graph passes through the point $ (0,1) $ . Since $ a^{1}=a $ , the graph passes through the point $ (1,a) $ .
It can be shown that, if $ x_{1} < x_{2} $ , then $ a^{x_{1}} < a^{x_{2}} $ , that is, the function is increasing on R; hence the term exponential growth function. Therefore, the basic exponential function is a one-to-one function.
It can further be shown that as $ x $ , $ a^{x} $ , and as $ x -$ , $ a^{x} $ . Thus, the negative x-axis is a horizontal asymptote for the graph.
Since \(a^{x}\) is positive for all real \(x\) (see the previous problem), the range of the function is \((0,\infty)\).
The graph is shown in Fig. 17-1.
17.3. Analyze and sketch the graph of a basic exponential function of form $ f(x) = a^{x}, a < 1 $ .
The graph has no obvious symmetry. Since \(a^{0}=1\), the graph passes through the point \((0,1)\). Since \(a^{1}=a\), the graph passes through the point \((1,a)\).
It can be shown that, if $ x_{1} < x_{2} $ , then $ a^{x_{1}} > a^{x_{2}} $ , that is, the function is decreasing on R; hence the term exponential decay function.
It can further be shown, that as $ x $ , $ a^{x} $ , and as $ x -$ , $ a^{x} $ . Thus, the positive x-axis is a horizontal asymptote for the graph.
Since \(a^{x}\) is positive for all real \(x\), the range of the function is \((0,\infty)\).
The graph is shown in Fig. 17-2.
17.4. Show that the graph of $ f(x) = a^{-x}, a > 1 $ , is an exponential decay curve.
Let \(b = 1/a\). Then, since \(a > 1\), it follows that \(b < 1\). Moreover, \(a^{-x} = (1/b)^{-x} = b^{x}\). Since the graph of \(f(x) = b^{x}\), \(b < 1\), is an exponential decay curve, so is the graph of \(f(x) = a^{-x}\), \(a > 1\).
17.5. Sketch a graph of (a) $ f(x) = 2^{x} $ (b) $ f(x) = 2^{-x} $
- Form a table of values.
| x | y |
| -2 | $ \(</td></tr><tr><td>-1</td><td>\) $ |
| 0 | 1 |
| 1 | 2 |
| 2 | 4 |
| 3 | 8 |
- Form a table of values.
| x | y |
| -3 | 8 |
| -2 | 4 |
| -1 | 2 |
| 0 | 1 |
| 1 | $ \(</td></tr><tr><td>2</td><td>\) $ |
Domain: R, Range: \((0,\infty)\)
Asymptote: negative x axis.
The graph is shown in Fig. 17-3.
Domain: R, Range: $ (0,) $
Asymptote: positive x axis.
The graph is shown in Fig. 17-4.
17.6. Explain the definition of the natural exponential base e.
Consider the following table of values for the quantity $ (1 + )^{n} $
| n | 1 | 10 | 100 | 1000 | 10,000 | 100,000 | 1,000,000 |
| $ (1 + )^n $ | 2 | 2.59374246 | 2.70481383 | 2.71692393 | 2.71814593 | 2.71826824 | 2.71828047 |
As $ n $ , the quantity $ (1 + )^{n} $ does not increase beyond all bounds, but seems to approach a value.
In calculus it is shown that this value is an irrational number, called e, with a decimal approximation of 2.718 281 828 459 045. . . In calculus this number and the exponential functions $ f(x) = e^{x} $ , $ f(x) = e^{-x} $ , and so on, are shown to have special properties.
17.7. Derive the formula $ A(t) = P(1 + r/n)^{nt} $ for the amount of money resulting from investing a principal P for a time t at an annual rate r, compounded n times per year.
First assume that the amount P is invested for one year at the simple interest rate of r. Then the interest after one year is $ I = = (1) = $ . The amount of money present after one year is then
\[ A=P+I=P+Pr=P(1+r). \]
If this amount is then invested for a second year at the simple interest rate of r, then the interest after the second year is $ P(1 + r)r(1) = P(1 + r)r $ . The amount of money present after two years is then
\[ A=P(1+r)+P(1+r)r=P(1+r)(1+r)=P(1+r)^{2} \]
Thus the amount present at the end of each year is multiplied by a factor of $ 1 + r $ during the next year. Generalizing, the amount present at time t, assuming compounding once per year, is
\[ A(t)=P(1+r)^{t} \]
Now assume that interest is compounded n times per year. The interest after one compounding period is then I = Pr/n. The amount of money present after one compounding period is
\[ A=P+I=P+Pr/n=P(1+r/n) \]
Thus, the amount present at the end of each compounding period is multiplied by a factor of $ 1 + r/n $ during the next period. Hence the amount present after one year, n compounding periods, is
\[ A=P(1+r/n)^{n} \]
and the amount present at time t is given by
\[ A(t)=P((1+r/n)^{n})^{t}=P(1+r/n)^{nt} \]
17.8. Derive the formula $ A(t) = Pe^{rt} $ for the amount of money resulting from investing a principal P for a time t at an annual rate r, compounded continuously.
Continuous compounding is understood as the limiting case of compounding n times per year, as $ n $ . From the previous problem, if interest is compounded n times per year, the amount present at time t is given by $ A(t) = P(1 + r/n)^{nt} $ . If n is allowed to increase beyond all bounds, then
\[ \begin{aligned}A(t)&=\lim_{n\to\infty}P(1+r/n)^{nt}\\&=\lim_{n\to\infty}P(1+r/n)^{(n/r)rt}\\&=\lim_{n\to\infty}P[(1+r/n)^{n/r}]^{rt}\\&=\lim_{n/r\to\infty}P[(1+r/n)^{n/r}]^{rt}\\&=P[\lim_{n/r\to\infty}(1+r/n)^{n/r}]^{rt}\\&=Pe^{rt}\\ \end{aligned} \]
17.9. Calculate the amount of money present if $1000 is invested at 5% interest for seven years, compounded
yearly; (b) quarterly; (c) monthly; (d) daily; (e) continuously.
Use $ A(t) = P(1 + r/n)^{nt} $ with P = 1000, r = 0.05, t = 7, and n = 1.
\[ A(7)=1000(1+0.05/1)^{1\cdot7}=\1407.10 \]
- Use $ A(t) = P(1 + r/n)^{nt} $ with P = 1000, r = 0.05, t = 7, and n = 4.
\[ A(7)=1000(1+0.05/4)^{4\cdot7}=\1415.99 \]
- Use $ A(t) = P(1 + r/n)^{nt} $ with P = 1000, r = 0.05, t = 7, and n = 12.
\[ A(7)=1000(1+0.05/12)^{12\cdot7}=\1418.04 \]
- Use $ A(t) = P(1 + r/n)^{nt} $ with P = 1000, r = 0.05, t = 7, and n = 365.
\[ A(7)=1000(1+0.05/365)^{365\cdot7}=\1419.03 \]
- Use $ A(t) = Pe^{rt} $ with P = 1000, r = 0.05, and t = 7.
\[ A(7)=1000\cdot e^{0.05\cdot7}=\1419.07 \]
Note that the difference in interest that results from increasing the frequency of compounding from daily to continuously is quite small.
17.10. Simplify the expressions:
\[ (a)\left(\frac{e^{x}+e^{-x}}{2}\right)^{2}-\left(\frac{e^{x}-e^{-x}}{2}\right)^{2}\quad(b)\frac{(e^{x}+e^{-x})(e^{x}+e^{-x})-(e^{x}-e^{-x})(e^{x}-e^{-x})}{(e^{x}+e^{-x})^{2}} \]
\[ \begin{aligned}\left(\frac{e^{x}+e^{-x}}{2}\right)^{2}-\left(\frac{e^{x}-e^{-x}}{2}\right)^{2}&=\frac{(e^{x})^{2}+2e^{x}e^{-x}+(e^{-x})^{2}}{4}-\frac{(e^{x})^{2}-2e^{x}e^{-x}+(e^{-x})^{2}}{4}\\&=\frac{e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x}}{4}\\&=\frac{4}{4}=1\end{aligned} \]
\[ \begin{aligned}(b)\frac{(e^{x}+e^{-x})(e^{x}+e^{-x})-(e^{x}-e^{-x})(e^{x}-e^{-x})}{(e^{x}+e^{-x})^{2}}&=\frac{e^{2x}+2e^{x}e^{-x}+e^{-2x}-e^{2x}+2e^{x}e^{-x}-e^{-2x}}{(e^{x}+e^{-x})^{2}}\\&=\frac{4}{(e^{x}+e^{-x})^{2}}\end{aligned} \]
For an alternate form, regard the last expression as a complex fraction and multiply numerator and denominator by $ e^{2x} $ to obtain
\[ \frac{4}{(e^{x}+e^{-x})^{2}}=\frac{4e^{2x}}{e^{2x}(e^{x}+e^{-x})^{2}}=\frac{4e^{2x}}{(e^{2x}+1)^{2}} \]
17.11. Find the zeros of the function $ f(x) = xe^{-x} - e^{-x} $
Solve $ xe^{-x} - e^{-x} = 0 $ by factoring to obtain
\[ \begin{aligned}e^{-x}(x-1)&=0\\e^{-x}=0or x-1&=0\\x&=1\end{aligned} \]
Since $ e^{-x} $ is never 0, the only zero of the function is 1.
17.12. The number of bacteria in a culture is counted as 400 at the start of an experiment. If the number of bacteria doubles every 3 hours, the number of individuals can be represented by the formula $ N(t) = 400(2)^{t/3} $ . Find the number of bacteria present in the culture after 24 hours.
\[ N(24)=400(2)^{24/3}=400\cdot2^{8}=102,400\ individuals \]
17.13. Human populations can be modeled over short periods by unlimited exponential growth functions. If a country has a population of 22 million in 2000 and maintains a population growth rate of 1% per year, then its population in millions at a later time, taking t = 0 in 2000, can be modeled as $ N(t) = 22 e^{0.01 t} $ . Estimate the population in the year 2010.
In the year 2010, t = 10. Hence $ N(10) = 22e^{0.01(10)} = 24.3 $ . Hence the population is estimated to be 24.3 million.
17.14. A herd of deer is introduced onto an island. The initial population is 500 individuals, and it is estimated that the long-term sustainable population is 2000 individuals. If the size of the population is given by the logistic growth function
\[ N(t)=\frac{2000}{1+3e^{-0.05t}} \]
estimate to two significant digits the number of deer present after (a) 1 year; (b) 20 years; (c) 50 years. Use the given formula with the given values of t.
\[ (a)t=1:N(1)=\frac{2000}{1+3e^{-0.05(1)}}\approx520individuals \]
\[ (b)t=20:N(20)=\frac{2000}{1+3e^{-0.05(20)}}\approx950individuals \]
\[ (c)t=50:N(50)=\frac{2000}{1+3e^{-0.05(50)}}\approx1600individuals \]
17.15. Draw a graph of the function $ N(t) $ from the previous problem.
Use the calculated values. Note also that $ N(0) $ is given as 500, and that as $ t $ , since $ e^{-0.05t} $ , the value of the function approaches 2000 asymptotically. The graph is shown in Fig. 17-5.
17.16. A certain radioactive isotope decays according to the formula $ Q(t) = Q_{0}e^{-0.034t} $ , where t is the time in years and $ Q_{0} $ is the number of grams present initially. If 20 grams are present initially, approximate to the nearest tenth of a gram the amount present after 10 years.
Use the given formula with $ Q_{0}=20 $ and t=10: $ Q(10)=20e^{-0.034(10)}=14.2 $ grams.
17.17. If a radioactive isotope decays according to the formula $ Q(t) = Q_{0} ^{-t/T} $ , where t is the time in years and $ Q_{0} $ is the number of grams present initially, show that the amount present at time t = T is $ Q_{0}/2 $ . (T is called the half-life of the isotope.)
Use the given formula with t = T. Then $ Q(T) = Q_{0} ^{-T/T} = Q_{0} ^{-1} = Q_{0}/2 $ .
SUPPLEMENTARY PROBLEMS
17.18. Sketch a graph of the functions (a) $ f(x) = 1 - e^{-x} $ and (b) $ f(x) = 2{-x{2}/2} $
Ans. (a) Fig. 17-6; (b) Fig. 17-7.
17.19. Simplify the expression $ (){2}+(\frac{e{x}-e{-x}}{2}){2} $
Ans. $ $ or $ $
17.20. Prove that the difference quotient (see Chapter 9) for $ f(x) = e^{x} $ can be written as
\[ \frac{e^{h}-1}{h}e^{x} \]
17.21. Find the zeros of the function $ f(x) = -x{2}e{-x} + 2xe^{-x} $ .
Ans. 0, 2
17.22. $8000 is invested in an account yielding 5.5% interest. Find the amount of money in the account after one year if interest is compounded (a) quarterly; (b) daily; (c) continuously.
Ans. (a) $8449.16; (b) $8452.29; (c) $8452.32
17.23. In the previous problem, find the annualized percentage rate for the account (this is the equivalent rate without compounding that would yield the same amount of interest).
Ans. (a) 5.61%; (b) 5.65%; (c) 5.65%
17.24. How much would have to be invested at 5.5% compounded continuously to obtain 5000 after 10 years?
Ans. 2884.75
17.25. A family has just had a new child. How much would have to be invested at 6%, compounded daily, in order to have 60,000 for her college education in 17 years?
Ans. 21,637.50
17.26. If the number of bacteria in a culture is given by the formula $ Q(t) = 250 ^{t/4} $ , where t is measured in days, estimate (a) the initial population; (b) the population after 4 days; (c) the population after 14 days.
Ans. (a) 250; (b) 750; (c) 11,700
17.27. If the population of trout in a lake is given by the formula $ N(t) = $ , where t is measured in years, estimate (a) the initial population; (b) the population after 10 years; (c) the long-term limiting value of the population.
Ans. (a) 1600; (b) 1960; (c) 4000
17.28. If a radioactive isotope decays according to the formula $ Q(t) = Q_{0} ^{-t/12} $ , where t is measured in years, find the portion of an initial amount remaining after (a) 1 year; (b) 12 years; (c) 100 years.
Ans. (a) $ 0.94Q_{0} $ ; (b) $ 0.5Q_{0} $ ; (c) $ 0.003Q_{0} $
17.29. The half-life (see Problem 17.17) of Carbon-14 is 5730 years.
If 100 grams of Carbon-14 were present initially, how much would remain after 3000 years?
If a sample contains 38 grams of Carbon-14, how much was present 4500 years ago?
Ans. (a) 69.6 grams; (b) 65.5 grams