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Calculus Volume 1

1.5 Exponential and Logarithmic Functions

Calculus Volume 11.5 Exponential and Logarithmic Functions
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Learning Objectives

  • 1.5.1. Identify the form of an exponential function.
  • 1.5.2. Explain the difference between the graphs of xbxb and bx.bx.
  • 1.5.3. Recognize the significance of the number e.e.
  • 1.5.4. Identify the form of a logarithmic function.
  • 1.5.5. Explain the relationship between exponential and logarithmic functions.
  • 1.5.6. Describe how to calculate a logarithm to a different base.
  • 1.5.7. Identify the hyperbolic functions, their graphs, and basic identities.

In this section we examine exponential and logarithmic functions. We use the properties of these functions to solve equations involving exponential or logarithmic terms, and we study the meaning and importance of the number e.e. We also define hyperbolic and inverse hyperbolic functions, which involve combinations of exponential and logarithmic functions. (Note that we present alternative definitions of exponential and logarithmic functions in the chapter Applications of Integrations, and prove that the functions have the same properties with either definition.)

Exponential Functions

Exponential functions arise in many applications. One common example is population growth.

For example, if a population starts with P0P0 individuals and then grows at an annual rate of 2%,2%, its population after 1 year is

P(1)=P0+0.02P0=P0(1+0.02)=P0(1.02).P(1)=P0+0.02P0=P0(1+0.02)=P0(1.02).

Its population after 2 years is

P(2)=P(1)+0.02P(1)=P(1)(1.02)=P0(1.02)2.P(2)=P(1)+0.02P(1)=P(1)(1.02)=P0(1.02)2.

In general, its population after tt years is

P(t)=P0(1.02)t,P(t)=P0(1.02)t,

which is an exponential function. More generally, any function of the form f(x)=bx,f(x)=bx, where b>0,b1,b>0,b1, is an exponential function with base bb and exponent x. Exponential functions have constant bases and variable exponents. Note that a function of the form f(x)=xbf(x)=xb for some constant bb is not an exponential function but a power function.

To see the difference between an exponential function and a power function, we compare the functions y=x2y=x2 and y=2x.y=2x. In Table 1.10, we see that both 2x2x and x2x2 approach infinity as x.x. Eventually, however, 2x2x becomes larger than x2x2 and grows more rapidly as x.x. In the opposite direction, as x,x2,x,x2, whereas 2x0.2x0. The line y=0y=0 is a horizontal asymptote for y=2x.y=2x.

xx −3−3 −2−2 −1−1 00 11 22 33 44 55 66
x2x2 99 44 11 00 11 44 99 1616 2525 3636
2x2x 1/81/8 1/41/4 1/21/2 11 22 44 88 1616 3232 6464
Table 1.10 Values of x2x2 and 2x2x

In Figure 1.43, we graph both y=x2y=x2 and y=2xy=2x to show how the graphs differ.

An image of a graph. The x axis runs from -10 to 10 and the y axis runs from 0 to 50. The graph is of two functions. The first function is “y = x squared”, which is a parabola. The function decreases until it hits the origin and then begins increasing. The second function is “y = 2 to the power of x”, which starts slightly above the x axis, and begins increasing very rapidly, more rapidly than the first function.
Figure 1.43 Both 2x2x and x2x2 approach infinity as x,x, but 2x2x grows more rapidly than x2.x2. As x,x2,x,x2, whereas 2x0.2x0.

Evaluating Exponential Functions

Recall the properties of exponents: If xx is a positive integer, then we define bx=b·bbbx=b·bb (with xx factors of b).b). If xx is a negative integer, then x=yx=y for some positive integer y,y, and we define bx=by=1/by.bx=by=1/by. Also, b0b0 is defined to be 1.1. If xx is a rational number, then x=p/q,x=p/q, where pp and qq are integers and bx=bp/q=bpq.bx=bp/q=bpq. For example, 93/2=93=27.93/2=93=27. However, how is bxbx defined if xx is an irrational number? For example, what do we mean by 22?22? This is too complex a question for us to answer fully right now; however, we can make an approximation. In Table 1.11, we list some rational numbers approaching 2,2, and the values of 2x2x for each rational number xx are presented as well. We claim that if we choose rational numbers xx getting closer and closer to 2,2, the values of 2x2x get closer and closer to some number L.L. We define that number LL to be 22.22.

xx 1.41.4 1.411.41 1.4141.414 1.41421.4142 1.414211.41421 1.4142131.414213
2x2x 2.6392.639 2.657372.65737 2.664752.66475 2.6651192.665119 2.6651382.665138 2.6651432.665143
Table 1.11 Values of 2x2x for a List of Rational Numbers Approximating 22

Example 1.33

Bacterial Growth

Suppose a particular population of bacteria is known to double in size every 44 hours. If a culture starts with 10001000 bacteria, the number of bacteria after 44 hours is n(4)=1000·2.n(4)=1000·2. The number of bacteria after 88 hours is n(8)=n(4)·2=1000·22.n(8)=n(4)·2=1000·22. In general, the number of bacteria after 4m4m hours is n(4m)=1000·2m.n(4m)=1000·2m. Letting t=4m,t=4m, we see that the number of bacteria after tt hours is n(t)=1000·2t/4.n(t)=1000·2t/4. Find the number of bacteria after 66 hours, 1010 hours, and 2424 hours.

Solution

The number of bacteria after 6 hours is given by n(6)=1000·26/42828n(6)=1000·26/42828 bacteria. The number of bacteria after 1010 hours is given by n(10)=1000·210/45657n(10)=1000·210/45657 bacteria. The number of bacteria after 2424 hours is given by n(24)=1000·26=64,000n(24)=1000·26=64,000 bacteria.

Checkpoint 1.27

Given the exponential function f(x)=100·3x/2,f(x)=100·3x/2, evaluate f(4)f(4) and f(10).f(10).

Media

Go to World Population Balance for another example of exponential population growth.

Graphing Exponential Functions

For any base b>0,b1,b>0,b1, the exponential function f(x)=bxf(x)=bx is defined for all real numbers xx and bx>0.bx>0. Therefore, the domain of f(x)=bxf(x)=bx is (,)(,) and the range is (0,).(0,). To graph bx,bx, we note that for b>1,bxb>1,bx is increasing on (,)(,) and bxbx as x,x, whereas bx0bx0 as x.x. On the other hand, if 0<b<1,f(x)=bx0<b<1,f(x)=bx is decreasing on (,)(,) and bx0bx0 as xx whereas bxbx as xx (Figure 1.44).

An image of a graph. The x axis runs from -3 to 3 and the y axis runs from 0 to 4. The graph is of four functions. The first function is “f(x) = 2 to the power of x”, an increasing curved function, which starts slightly above the x axis and begins increasing. The second function is “f(x) = 4 to the power of x”, an increasing curved function, which starts slightly above the x axis and begins increasing rapidly, more rapidly than the first function. The third function is “f(x) = (1/2) to the power of x”, a decreasing curved function with decreases until it gets close to the x axis without touching it. The third function is “f(x) = (1/4) to the power of x”, a decreasing curved function with decreases until it gets close to the x axis without touching it. It decreases at a faster rate than the third function.
Figure 1.44 If b>1,b>1, then bxbx is increasing on (,).(,). If 0<b<1,0<b<1, then bxbx is decreasing on (,).(,).

Media

Visit this site for more exploration of the graphs of exponential functions.

Note that exponential functions satisfy the general laws of exponents. To remind you of these laws, we state them as rules.

Rule: Laws of Exponents

For any constants a>0,b>0,a>0,b>0, and for all x and y,

  1. bx·by=bx+ybx·by=bx+y
  2. bxby=bxybxby=bxy
  3. (bx)y=bxy(bx)y=bxy
  4. (ab)x=axbx(ab)x=axbx
  5. axbx=(ab)xaxbx=(ab)x

Example 1.34

Using the Laws of Exponents

Use the laws of exponents to simplify each of the following expressions.

  1. (2x2/3)3(4x−1/3)2(2x2/3)3(4x−1/3)2
  2. (x3y−1)2(xy2)−2(x3y−1)2(xy2)−2

Solution

  1. We can simplify as follows:
    (2x2/3)3(4x−1/3)2=23(x2/3)342(x−1/3)2=8x216x−2/3=x2x2/32=x8/32.(2x2/3)3(4x−1/3)2=23(x2/3)342(x−1/3)2=8x216x−2/3=x2x2/32=x8/32.
  2. We can simplify as follows:
    (x3y−1)2(xy2)−2=(x3)2(y−1)2x−2(y2)−2=x6y−2x−2y−4=x6x2y−2y4=x8y2.(x3y−1)2(xy2)−2=(x3)2(y−1)2x−2(y2)−2=x6y−2x−2y−4=x6x2y−2y4=x8y2.

Checkpoint 1.28

Use the laws of exponents to simplify (6x−3y2)/(12x−4y5).(6x−3y2)/(12x−4y5).

The Number e

A special type of exponential function appears frequently in real-world applications. To describe it, consider the following example of exponential growth, which arises from compounding interest in a savings account. Suppose a person invests PP dollars in a savings account with an annual interest rate r,r, compounded annually. The amount of money after 1 year is

A(1)=P+rP=P(1+r).A(1)=P+rP=P(1+r).

The amount of money after 22 years is

A(2)=A(1)+rA(1)=P(1+r)+rP(1+r)=P(1+r)2.A(2)=A(1)+rA(1)=P(1+r)+rP(1+r)=P(1+r)2.

More generally, the amount after tt years is

A(t)=P(1+r)t.A(t)=P(1+r)t.

If the money is compounded 2 times per year, the amount of money after half a year is

A(12)=P+(r2)P=P(1+(r2)).A(12)=P+(r2)P=P(1+(r2)).

The amount of money after 11 year is

A(1)=A(12)+(r2)A(12)=P(1+r2)+r2(P(1+r2))=P(1+r2)2.A(1)=A(12)+(r2)A(12)=P(1+r2)+r2(P(1+r2))=P(1+r2)2.

After tt years, the amount of money in the account is

A(t)=P(1+r2)2t.A(t)=P(1+r2)2t.

More generally, if the money is compounded nn times per year, the amount of money in the account after tt years is given by the function

A(t)=P(1+rn)nt.A(t)=P(1+rn)nt.

What happens as n?n? To answer this question, we let m=n/rm=n/r and write

(1+rn)nt=(1+1m)mrt,(1+rn)nt=(1+1m)mrt,

and examine the behavior of (1+1/m)m(1+1/m)m as m,m, using a table of values (Table 1.12).

mm 1010 100100 10001000 10,00010,000 100,000100,000 1,000,0001,000,000
(1+1m)m(1+1m)m 2.59372.5937 2.70482.7048 2.716922.71692 2.718152.71815 2.7182682.718268 2.7182802.718280
Table 1.12 Values of (1+1m)m(1+1m)m as mm

Looking at this table, it appears that (1+1/m)m(1+1/m)m is approaching a number between 2.72.7 and 2.82.8 as m.m. In fact, (1+1/m)m(1+1/m)m does approach some number as m.m. We call this number ee. To six decimal places of accuracy,

e2.718282.e2.718282.

The letter ee was first used to represent this number by the Swiss mathematician Leonhard Euler during the 1720s. Although Euler did not discover the number, he showed many important connections between ee and logarithmic functions. We still use the notation ee today to honor Euler’s work because it appears in many areas of mathematics and because we can use it in many practical applications.

Returning to our savings account example, we can conclude that if a person puts PP dollars in an account at an annual interest rate r,r, compounded continuously, then A(t)=Pert.A(t)=Pert. This function may be familiar. Since functions involving base ee arise often in applications, we call the function f(x)=exf(x)=ex the natural exponential function. Not only is this function interesting because of the definition of the number e,e, but also, as discussed next, its graph has an important property.

Since e>1,e>1, we know exex is increasing on (,).(,). In Figure 1.45, we show a graph of f(x)=exf(x)=ex along with a tangent line to the graph of at x=0.x=0. We give a precise definition of tangent line in the next chapter; but, informally, we say a tangent line to a graph of ff at x=ax=a is a line that passes through the point (a,f(a))(a,f(a)) and has the same “slope” as ff at that point .. The function f(x)=exf(x)=ex is the only exponential function bxbx with tangent line at x=0x=0 that has a slope of 1. As we see later in the text, having this property makes the natural exponential function the most simple exponential function to use in many instances.

An image of a graph. The x axis runs from -3 to 3 and the y axis runs from 0 to 4. The graph is of the function “f(x) = e to power of x”, an increasing curved function that starts slightly above the x axis. The y intercept is at the point (0, 1). At this point, a line is drawn tangent to the function. This line has the label “slope = 1”.
Figure 1.45 The graph of f(x)=exf(x)=ex has a tangent line with slope 11 at x=0.x=0.

Example 1.35

Compounding Interest

Suppose $500$500 is invested in an account at an annual interest rate of r=5.5%,r=5.5%, compounded continuously.

  1. Let tt denote the number of years after the initial investment and A(t)A(t) denote the amount of money in the account at time t.t. Find a formula for A(t).A(t).
  2. Find the amount of money in the account after 1010 years and after 2020 years.

Solution

  1. If PP dollars are invested in an account at an annual interest rate r,r, compounded continuously, then A(t)=Pert.A(t)=Pert. Here P=$500P=$500 and r=0.055.r=0.055. Therefore, A(t)=500e0.055t.A(t)=500e0.055t.
  2. After 1010 years, the amount of money in the account is
    A(10)=500e0.055·10=500e0.55$866.63.A(10)=500e0.055·10=500e0.55$866.63.

    After 2020 years, the amount of money in the account is
    A(20)=500e0.055·20=500e1.1$1,502.08.A(20)=500e0.055·20=500e1.1$1,502.08.
Checkpoint 1.29

If $750$750 is invested in an account at an annual interest rate of 4%,4%, compounded continuously, find a formula for the amount of money in the account after tt years. Find the amount of money after 3030 years.

Logarithmic Functions

Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. These come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.

The exponential function f(x)=bxf(x)=bx is one-to-one, with domain (,)(,) and range (0,).(0,). Therefore, it has an inverse function, called the logarithmic function with base b.b. For any b>0,b1,b>0,b1, the logarithmic function with base b, denoted logb,logb, has domain (0,)(0,) and range (,),(,), and satisfies

logb(x)=yif and only ifby=x.logb(x)=yif and only ifby=x.

For example,

log2(8)=3since23=8,log10(1100)=−2since10−2=1102=1100,logb(1)=0sinceb0=1for any baseb>0.log2(8)=3since23=8,log10(1100)=−2since10−2=1102=1100,logb(1)=0sinceb0=1for any baseb>0.

Furthermore, since y=logb(x)y=logb(x) and y=bxy=bx are inverse functions,

logb(bx)=xandblogb(x)=x.logb(bx)=xandblogb(x)=x.

The most commonly used logarithmic function is the function loge.loge. Since this function uses natural ee as its base, it is called the natural logarithm. Here we use the notation ln(x)ln(x) or lnxlnx to mean loge(x).loge(x). For example,

ln(e)=loge(e)=1,ln(e3)=loge(e3)=3,ln(1)=loge(1)=0.ln(e)=loge(e)=1,ln(e3)=loge(e3)=3,ln(1)=loge(1)=0.

Since the functions f(x)=exf(x)=ex and g(x)=ln(x)g(x)=ln(x) are inverses of each other,

ln(ex)=xandelnx=x,ln(ex)=xandelnx=x,

and their graphs are symmetric about the line y=xy=x (Figure 1.46).

An image of a graph. The x axis runs from -3 to 3 and the y axis runs from -3 to 4. The graph is of two functions. The first function is “f(x) = e to power of x”, an increasing curved function that starts slightly above the x axis. The y intercept is at the point (0, 1) and there is no x intercept. The second function is “f(x) = ln(x)”, an increasing curved function. The x intercept is at the point (1, 0) and there is no y intercept. A dotted line with label “y = x” is also plotted on the graph, to show that the functions are mirror images over this line.
Figure 1.46 The functions y=exy=ex and y=ln(x)y=ln(x) are inverses of each other, so their graphs are symmetric about the line y=x.y=x.

Media

At this site you can see an example of a base-10 logarithmic scale.

In general, for any base b>0,b1,b>0,b1, the function g(x)=logb(x)g(x)=logb(x) is symmetric about the line y=xy=x with the function f(x)=bx.f(x)=bx. Using this fact and the graphs of the exponential functions, we graph functions logblogb for several values of b>1b>1 (Figure 1.47).

An image of a graph. The x axis runs from -3 to 3 and the y axis runs from 0 to 4. The graph is of three functions. All three functions a log functions that are increasing curved functions that start slightly to the right of the y axis and have an x intercept at (1, 0). The first function is “y = log base 10 (x)”, the second function is “f(x) = ln(x)”, and the third function is “y = log base 2 (x)”. The third function increases the most rapidly, the second function increases next most rapidly, and the third function increases the slowest.
Figure 1.47 Graphs of y=logb(x)y=logb(x) are depicted for b=2,e,10.b=2,e,10.

Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms.

Rule: Properties of Logarithms

If a,b,c>0,b1,a,b,c>0,b1, and rr is any real number, then

1.logb(ac)=logb(a)+logb(c)(Product property)2.logb(ac)=logb(a)logb(c)(Quotient property)3.logb(ar)=rlogb(a)(Power property)1.logb(ac)=logb(a)+logb(c)(Product property)2.logb(ac)=logb(a)logb(c)(Quotient property)3.logb(ar)=rlogb(a)(Power property)

Example 1.36

Solving Equations Involving Exponential Functions

Solve each of the following equations for x.x.

  1. 5x=25x=2
  2. ex+6ex=5ex+6ex=5

Solution

  1. Applying the natural logarithm function to both sides of the equation, we have
    ln5x=ln2.ln5x=ln2.

    Using the power property of logarithms,
    xln5=ln2.xln5=ln2.

    Therefore, x=ln2/ln5.x=ln2/ln5.
  2. Multiplying both sides of the equation by ex,ex, we arrive at the equation
    e2x+6=5ex.e2x+6=5ex.

    Rewriting this equation as
    e2x5ex+6=0,e2x5ex+6=0,

    we can then rewrite it as a quadratic equation in ex:ex:
    (ex)25(ex)+6=0.(ex)25(ex)+6=0.

    Now we can solve the quadratic equation. Factoring this equation, we obtain
    (ex3)(ex2)=0.(ex3)(ex2)=0.

    Therefore, the solutions satisfy ex=3ex=3 and ex=2.ex=2. Taking the natural logarithm of both sides gives us the solutions x=ln3,ln2.x=ln3,ln2.
Checkpoint 1.30

Solve e2x/(3+e2x)=1/2.e2x/(3+e2x)=1/2.

Example 1.37

Solving Equations Involving Logarithmic Functions

Solve each of the following equations for x.x.

  1. ln(1x)=4ln(1x)=4
  2. log10x+log10x=2log10x+log10x=2
  3. ln(2x)3ln(x2)=0ln(2x)3ln(x2)=0

Solution

  1. By the definition of the natural logarithm function,
    ln(1x)=4if and only ife4=1x.ln(1x)=4if and only ife4=1x.

    Therefore, the solution is x=1/e4.x=1/e4.
  2. Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as
    log10x+log10x=log10xx=log10x3/2=32log10x.log10x+log10x=log10xx=log10x3/2=32log10x.

    Therefore, the equation can be rewritten as
    32log10x=2orlog10x=43.32log10x=2orlog10x=43.

    The solution is x=104/3=10103.x=104/3=10103.
  3. Using the power property of logarithmic functions, we can rewrite the equation as ln(2x)ln(x6)=0.ln(2x)ln(x6)=0.
    Using the quotient property, this becomes
    ln(2x5)=0.ln(2x5)=0.

    Therefore, 2/x5=1,2/x5=1, which implies x=25.x=25. We should then check for any extraneous solutions.
Checkpoint 1.31

Solve ln(x3)4ln(x)=1.ln(x3)4ln(x)=1.

When evaluating a logarithmic function with a calculator, you may have noticed that the only options are log10log10 or log, called the common logarithm, or ln, which is the natural logarithm. However, exponential functions and logarithm functions can be expressed in terms of any desired base b.b. If you need to use a calculator to evaluate an expression with a different base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential or logarithmic function in terms of the natural exponential and natural logarithmic functions.

Rule: Change-of-Base Formulas

Let a>0,b>0,a>0,b>0, and a1,b1.a1,b1.

  1. ax=bxlogbaax=bxlogba for any real number x.x.
    If b=e,b=e, this equation reduces to ax=exlogea=exlna.ax=exlogea=exlna.
  2. logax=logbxlogbalogax=logbxlogba for any real number x>0.x>0.
    If b=e,b=e, this equation reduces to logax=lnxlna.logax=lnxlna.

Proof

For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know that for any base b>0,b1,logb(ax)=xlogba.b>0,b1,logb(ax)=xlogba. Therefore,

blogb(ax)=bxlogba.blogb(ax)=bxlogba.

In addition, we know that bxbx and logb(x)logb(x) are inverse functions. Therefore,

blogb(ax)=ax.blogb(ax)=ax.

Combining these last two equalities, we conclude that ax=bxlogba.ax=bxlogba.

To prove the second property, we show that

(logba)·(logax)=logbx.(logba)·(logax)=logbx.

Let u=logba,v=logax,u=logba,v=logax, and w=logbx.w=logbx. We will show that u·v=w.u·v=w. By the definition of logarithmic functions, we know that bu=a,av=x,bu=a,av=x, and bw=x.bw=x. From the previous equations, we see that

buv=(bu)v=av=x=bw.buv=(bu)v=av=x=bw.

Therefore, buv=bw.buv=bw. Since exponential functions are one-to-one, we can conclude that u·v=w.u·v=w.

Example 1.38

Changing Bases

Use a calculating utility to evaluate log37log37 with the change-of-base formula presented earlier.

Solution

Use the second equation with a=3a=3 and e=3:e=3:

log37=ln7ln31.77124.log37=ln7ln31.77124.

Checkpoint 1.32

Use the change-of-base formula and a calculating utility to evaluate log46.log46.

Example 1.39

Chapter Opener: The Richter Scale for Earthquakes

A photograph of an earthquake fault.
Figure 1.48 (credit: modification of work by Robb Hannawacker, NPS)

In 1935, Charles Richter developed a scale (now known as the Richter scale) to measure the magnitude of an earthquake. The scale is a base-10 logarithmic scale, and it can be described as follows: Consider one earthquake with magnitude R1R1 on the Richter scale and a second earthquake with magnitude R2R2 on the Richter scale. Suppose R1>R2,R1>R2, which means the earthquake of magnitude R1R1 is stronger, but how much stronger is it than the other earthquake? A way of measuring the intensity of an earthquake is by using a seismograph to measure the amplitude of the earthquake waves. If A1A1 is the amplitude measured for the first earthquake and A2A2 is the amplitude measured for the second earthquake, then the amplitudes and magnitudes of the two earthquakes satisfy the following equation:

R1R2=log10(A1A2).R1R2=log10(A1A2).

Consider an earthquake that measures 8 on the Richter scale and an earthquake that measures 7 on the Richter scale. Then,

87=log10(A1A2).87=log10(A1A2).

Therefore,

log10(A1A2)=1,log10(A1A2)=1,

which implies A1/A2=10A1/A2=10 or A1=10A2.A1=10A2. Since A1A1 is 10 times the size of A2,A2, we say that the first earthquake is 10 times as intense as the second earthquake. On the other hand, if one earthquake measures 8 on the Richter scale and another measures 6, then the relative intensity of the two earthquakes satisfies the equation

log10(A1A2)=86=2.log10(A1A2)=86=2.

Therefore, A1=100A2.A1=100A2. That is, the first earthquake is 100 times more intense than the second earthquake.

How can we use logarithmic functions to compare the relative severity of the magnitude 9 earthquake in Japan in 2011 with the magnitude 7.3 earthquake in Haiti in 2010?

Solution

To compare the Japan and Haiti earthquakes, we can use an equation presented earlier:

97.3=log10(A1A2).97.3=log10(A1A2).

Therefore, A1/A2=101.7,A1/A2=101.7, and we conclude that the earthquake in Japan was approximately 5050 times more intense than the earthquake in Haiti.

Checkpoint 1.33

Compare the relative severity of a magnitude 8.48.4 earthquake with a magnitude 7.47.4 earthquake.

Hyperbolic Functions

The hyperbolic functions are defined in terms of certain combinations of exex and ex.ex. These functions arise naturally in various engineering and physics applications, including the study of water waves and vibrations of elastic membranes. Another common use for a hyperbolic function is the representation of a hanging chain or cable, also known as a catenary (Figure 1.49). If we introduce a coordinate system so that the low point of the chain lies along the yy-axis, we can describe the height of the chain in terms of a hyperbolic function. First, we define the hyperbolic functions.

A photograph of a spider web collecting dew drops.
Figure 1.49 The shape of a strand of silk in a spider’s web can be described in terms of a hyperbolic function. The same shape applies to a chain or cable hanging from two supports with only its own weight. (credit: “Mtpaley”, Wikimedia Commons)

Definition

Hyperbolic cosine

coshx=ex+ex2coshx=ex+ex2

Hyperbolic sine

sinhx=exex2sinhx=exex2

Hyperbolic tangent

tanhx=sinhxcoshx=exexex+extanhx=sinhxcoshx=exexex+ex

Hyperbolic cosecant

cschx=1sinhx=2exexcschx=1sinhx=2exex

Hyperbolic secant

sechx=1coshx=2ex+exsechx=1coshx=2ex+ex

Hyperbolic cotangent

cothx=coshxsinhx=ex+exexexcothx=coshxsinhx=ex+exexex

The name cosh rhymes with “gosh,” whereas the name sinh is pronounced “cinch.” Tanh, sech, csch, and coth are pronounced “tanch,” “seech,” “coseech,” and “cotanch,” respectively.

Using the definition of cosh(x)cosh(x) and principles of physics, it can be shown that the height of a hanging chain, such as the one in Figure 1.49, can be described by the function h(x)=acosh(x/a)+ch(x)=acosh(x/a)+c for certain constants aa and c.c.

But why are these functions called hyperbolic functions? To answer this question, consider the quantity cosh2tsinh2t.cosh2tsinh2t. Using the definition of coshcosh and sinh,sinh, we see that

cosh2tsinh2t=e2t+2+e−2t4e2t2+e−2t4=1.cosh2tsinh2t=e2t+2+e−2t4e2t2+e−2t4=1.

This identity is the analog of the trigonometric identity cos2t+sin2t=1.cos2t+sin2t=1. Here, given a value t,t, the point (x,y)=(cosht,sinht)(x,y)=(cosht,sinht) lies on the unit hyperbola x2y2=1x2y2=1 (Figure 1.50).

An image of a graph. The x axis runs from -1 to 3 and the y axis runs from -3 to 3. The graph is of the relation “(x squared) - (y squared) -1”. The left most point of the relation is at the x intercept, which is at the point (1, 0). From this point the relation both increases and decreases in curves as x increases. This relation is known as a hyperbola and it resembles a sideways “U” shape. There is a point plotted on the graph of the relation labeled “(cosh(1), sinh(1))”, which is at the approximate point (1.5, 1.2).
Figure 1.50 The unit hyperbola cosh2tsinh2t=1.cosh2tsinh2t=1.

Graphs of Hyperbolic Functions

To graph coshxcoshx and sinhx,sinhx, we make use of the fact that both functions approach (1/2)ex(1/2)ex as x,x, since ex0ex0 as x.x. As x,coshxx,coshx approaches 1/2ex,1/2ex, whereas sinhxsinhx approaches −1/2ex.−1/2ex. Therefore, using the graphs of 1/2ex,1/2ex,1/2ex,1/2ex, and 1/2ex1/2ex as guides, we graph coshxcoshx and sinhx.sinhx. To graph tanhx,tanhx, we use the fact that tanh(0)=0,−1<tanh(x)<1tanh(0)=0,−1<tanh(x)<1 for all x,tanhx1x,tanhx1 as x,x, and tanhx1tanhx1 as x.x. The graphs of the other three hyperbolic functions can be sketched using the graphs of coshx,sinhx,coshx,sinhx, and tanhxtanhx (Figure 1.51).

An image of six graphs. Each graph has an x axis that runs from -3 to 3 and a y axis that runs from -4 to 4. The first graph is of the function “y = cosh(x)”, which is a hyperbola. The function decreases until it hits the point (0, 1), where it begins to increase. There are also two functions that serve as a boundary for this function. The first of these functions is “y = (1/2)(e to power of -x)”, a decreasing curved function and the second of these functions is “y = (1/2)(e to power of x)”, an increasing curved function. The function “y = cosh(x)” is always above these two functions without ever touching them. The second graph is of the function “y = sinh(x)”, which is an increasing curved function. There are also two functions that serve as a boundary for this function. The first of these functions is “y = (1/2)(e to power of x)”, an increasing curved function and the second of these functions is “y = -(1/2)(e to power of -x)”, an increasing curved function that approaches the x axis without touching it. The function “y = sinh(x)” is always between these two functions without ever touching them. The third graph is of the function “y = sech(x)”, which increases until the point (0, 1), where it begins to decrease. The graph of the function has a hump. The fourth graph is of the function “y = csch(x)”. On the left side of the y axis, the function starts slightly below the x axis and decreases until it approaches the y axis, which it never touches. On the right side of the y axis, the function starts slightly to the right of the y axis and decreases until it approaches the x axis, which it never touches. The fifth graph is of the function “y = tanh(x)”, an increasing curved function. There are also two functions that serve as a boundary for this function. The first of these functions is “y = 1”, a horizontal line function and the second of these functions is “y = -1”, another horizontal line function. The function “y = tanh(x)” is always between these two functions without ever touching them. The sixth graph is of the function “y = coth(x)”. On the left side of the y axis, the function starts slightly below the boundary line “y = 1” and decreases until it approaches the y axis, which it never touches. On the right side of the y axis, the function starts slightly to the right of the y axis and decreases until it approaches the boundary line “y = -1”, which it never touches.
Figure 1.51 The hyperbolic functions involve combinations of exex and ex.ex.

Identities Involving Hyperbolic Functions

The identity cosh2tsinh2t,cosh2tsinh2t, shown in Figure 1.50, is one of several identities involving the hyperbolic functions, some of which are listed next. The first four properties follow easily from the definitions of hyperbolic sine and hyperbolic cosine. Except for some differences in signs, most of these properties are analogous to identities for trigonometric functions.

Rule: Identities Involving Hyperbolic Functions

  1. cosh(x)=coshxcosh(x)=coshx
  2. sinh(x)=sinhxsinh(x)=sinhx
  3. coshx+sinhx=excoshx+sinhx=ex
  4. coshxsinhx=excoshxsinhx=ex
  5. cosh2xsinh2x=1cosh2xsinh2x=1
  6. 1tanh2x=sech2x1tanh2x=sech2x
  7. coth2x1=csch2xcoth2x1=csch2x
  8. sinh(x±y)=sinhxcoshy±coshxsinhysinh(x±y)=sinhxcoshy±coshxsinhy
  9. cosh(x±y)=coshxcoshy±sinhxsinhycosh(x±y)=coshxcoshy±sinhxsinhy

Example 1.40

Evaluating Hyperbolic Functions

  1. Simplify sinh(5lnx).sinh(5lnx).
  2. If sinhx=3/4,sinhx=3/4, find the values of the remaining five hyperbolic functions.

Solution

  1. Using the definition of the sinhsinh function, we write
    sinh(5lnx)=e5lnxe−5lnx2=eln(x5)eln(x−5)2=x5x−52.sinh(5lnx)=e5lnxe−5lnx2=eln(x5)eln(x−5)2=x5x−52.
  2. Using the identity cosh2xsinh2x=1,cosh2xsinh2x=1, we see that
    cosh2x=1+(34)2=2516.cosh2x=1+(34)2=2516.

    Since coshx1coshx1 for all x,x, we must have coshx=5/4.coshx=5/4. Then, using the definitions for the other hyperbolic functions, we conclude that tanhx=3/5,cschx=4/3,sechx=4/5,tanhx=3/5,cschx=4/3,sechx=4/5, and cothx=5/3.cothx=5/3.

Checkpoint 1.34

Simplify cosh(2lnx).cosh(2lnx).

Inverse Hyperbolic Functions

From the graphs of the hyperbolic functions, we see that all of them are one-to-one except coshxcoshx and sechx.sechx. If we restrict the domains of these two functions to the interval [0,),[0,), then all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions. Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.

Definition

Inverse Hyperbolic Functions

sinh−1x=arcsinhx=ln(x+x2+1)cosh−1x=arccoshx=ln(x+x21)tanh−1x=arctanhx=12ln(1+x1x)coth−1x=arccotx=12ln(x+1x1)sech−1x=arcsechx=ln(1+1x2x)csch−1x=arccschx=ln(1x+1+x2|x|)sinh−1x=arcsinhx=ln(x+x2+1)cosh−1x=arccoshx=ln(x+x21)tanh−1x=arctanhx=12ln(1+x1x)coth−1x=arccotx=12ln(x+1x1)sech−1x=arcsechx=ln(1+1x2x)csch−1x=arccschx=ln(1x+1+x2|x|)

Let’s look at how to derive the first equation. The others follow similarly. Suppose y=sinh−1x.y=sinh−1x. Then, x=sinhyx=sinhy and, by the definition of the hyperbolic sine function, x=eyey2.x=eyey2. Therefore,

ey2xey=0.ey2xey=0.

Multiplying this equation by ey,ey, we obtain

e2y2xey1=0.e2y2xey1=0.

This can be solved like a quadratic equation, with the solution

ey=2x±4x2+42=x±x2+1.ey=2x±4x2+42=x±x2+1.

Since ey>0,ey>0, the only solution is the one with the positive sign. Applying the natural logarithm to both sides of the equation, we conclude that

y=ln(x+x2+1).y=ln(x+x2+1).

Example 1.41

Evaluating Inverse Hyperbolic Functions

Evaluate each of the following expressions.

sinh−1(2)sinh−1(2)
tanh−1(1/4)tanh−1(1/4)

Solution

sinh−1(2)=ln(2+22+1)=ln(2+5)1.4436sinh−1(2)=ln(2+22+1)=ln(2+5)1.4436

tanh−1(1/4)=12ln(1+1/411/4)=12ln(5/43/4)=12ln(53)0.2554tanh−1(1/4)=12ln(1+1/411/4)=12ln(5/43/4)=12ln(53)0.2554

Checkpoint 1.35

Evaluate tanh−1(1/2).tanh−1(1/2).

Section 1.5 Exercises

For the following exercises, evaluate the given exponential functions as indicated, accurate to two significant digits after the decimal.

229.

f(x)=5xf(x)=5x a. x=3x=3 b. x=12x=12 c. x=2x=2

230.

f(x)=(0.3)xf(x)=(0.3)x a. x=−1x=−1 b. x=4x=4 c. x=−1.5x=−1.5

231.

f(x)=10xf(x)=10x a. x=−2x=−2 b. x=4x=4 c. x=53x=53

232.

f(x)=exf(x)=ex a. x=2x=2 b. x=−3.2x=−3.2 c. x=πx=π

For the following exercises, match the exponential equation to the correct graph.

  1. y=4xy=4x
  2. y=3x1y=3x1
  3. y=2x+1y=2x+1
  4. y=(12)x+2y=(12)x+2
  5. y=3xy=3x
  6. y=15xy=15x
233.
An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -2 to 8. The graph is of a decreasing curved function. The function decreases until it approaches the line “y = 2”, but never touches this line. The y intercept is at the point (0, 3) and there is no x intercept.
234.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -9 to 2. The graph is of a function that starts slightly below the line “y = 1” and begins decreasing rapidly in a curve. The x intercept and y intercept are both at the origin.
235.
An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that starts slightly above the x axis and begins increasing rapidly. There is no x intercept and the y intercept is at the point (0, (1/3)). Another point of the graph is at (1, 1).
236.
An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved decreasing function that decreases until it comes close the x axis without touching it. There is no x intercept and the y intercept is at the point (0, 1). Another point of the graph is at (-1, 4).
237.
An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that increases until it comes close the x axis without touching it. There is no x intercept and the y intercept is at the point (0, -1). Another point of the graph is at (-1, -3).
238.
An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that starts slightly above the x axis and begins increasing rapidly. There is no x intercept and the y intercept is at the point (0, 2). Another point of the graph is at (-1, 1).

For the following exercises, sketch the graph of the exponential function. Determine the domain, range, and horizontal asymptote.

239.

f(x)=ex+2f(x)=ex+2

240.

f(x)=2xf(x)=2x

241.

f(x)=3x+1f(x)=3x+1

242.

f(x)=4x1f(x)=4x1

243.

f(x)=12xf(x)=12x

244.

f(x)=5x+1+2f(x)=5x+1+2

245.

f(x)=ex1f(x)=ex1

For the following exercises, write the equation in equivalent exponential form.

246.

log381=4log381=4

247.

log82=13log82=13

248.

log51=0log51=0

249.

log525=2log525=2

250.

log0.1=−1log0.1=−1

251.

ln(1e3)=−3ln(1e3)=−3

252.

log93=0.5log93=0.5

253.

ln1=0ln1=0

For the following exercises, write the equation in equivalent logarithmic form.

254.

23=823=8

255.

4−2=1164−2=116

256.

102=100102=100

257.

90=190=1

258.

(13)3=127(13)3=127

259.

643=4643=4

260.

ex=yex=y

261.

9y=1509y=150

262.

b3=45b3=45

263.

4−3/2=0.1254−3/2=0.125

For the following exercises, sketch the graph of the logarithmic function. Determine the domain, range, and vertical asymptote.

264.

f(x)=3+lnxf(x)=3+lnx

265.

f(x)=ln(x1)f(x)=ln(x1)

266.

f(x)=ln(x)f(x)=ln(x)

267.

f(x)=1lnxf(x)=1lnx

268.

f(x)=logx1f(x)=logx1

269.

f(x)=ln(x+1)f(x)=ln(x+1)

For the following exercises, use properties of logarithms to write the expressions as a sum, difference, and/or product of logarithms.

270.

logx4ylogx4y

271.

log39a3blog39a3b

272.

lnab3lnab3

273.

log5125xy3log5125xy3

274.

log4xy364log4xy364

275.

ln(6e3)ln(6e3)

For the following exercises, solve the exponential equation exactly.

276.

5x=1255x=125

277.

e3x15=0e3x15=0

278.

8x=48x=4

279.

4x+132=04x+132=0

280.

3x/14=1103x/14=110

281.

10x=7.2110x=7.21

282.

4·23x20=04·23x20=0

283.

73x2=1173x2=11

For the following exercises, solve the logarithmic equation exactly, if possible.

284.

log3x=0log3x=0

285.

log5x=−2log5x=−2

286.

log4(x+5)=0log4(x+5)=0

287.

log(2x7)=0log(2x7)=0

288.

lnx+3=2lnx+3=2

289.

log6(x+9)+log6x=2log6(x+9)+log6x=2

290.

log4(x+2)log4(x1)=0log4(x+2)log4(x1)=0

291.

lnx+ln(x2)=ln4lnx+ln(x2)=ln4

For the following exercises, use the change-of-base formula and either base 10 or base e to evaluate the given expressions. Answer in exact form and in approximate form, rounding to four decimal places.

292.

log547log547

293.

log782log782

294.

log6103log6103

295.

log0.5211log0.5211

296.

log2πlog2π

297.

log0.20.452log0.20.452

298.

Rewrite the following expressions in terms of exponentials and simplify.

a. 2cosh(lnx)2cosh(lnx) b. cosh4x+sinh4xcosh4x+sinh4x c. cosh2xsinh2xcosh2xsinh2x d. ln(coshx+sinhx)+ln(coshxsinhx)ln(coshx+sinhx)+ln(coshxsinhx)

299.

[T] The number of bacteria N in a culture after t days can be modeled by the function N(t)=1300·(2)t/4.N(t)=1300·(2)t/4. Find the number of bacteria present after 15 days.

300.

[T] The demand D (in millions of barrels) for oil in an oil-rich country is given by the function D(p)=150·(2.7)−0.25p,D(p)=150·(2.7)−0.25p, where p is the price (in dollars) of a barrel of oil. Find the amount of oil demanded (to the nearest million barrels) when the price is between $15 and $20.

301.

[T] The amount A of a $100,000 investment paying continuously and compounded for t years is given by A(t)=100,000·e0.055t.A(t)=100,000·e0.055t. Find the amount A accumulated in 5 years.

302.

[T] An investment is compounded monthly, quarterly, or yearly and is given by the function A=P(1+jn)nt,A=P(1+jn)nt, where AA is the value of the investment at time t,Pt,P is the initial principle that was invested, jj is the annual interest rate, and nn is the number of time the interest is compounded per year. Given a yearly interest rate of 3.5% and an initial principle of $100,000, find the amount AA accumulated in 5 years for interest that is compounded a. daily, b., monthly, c. quarterly, and d. yearly.

303.

[T] The concentration of hydrogen ions in a substance is denoted by [H+],[H+], measured in moles per liter. The pH of a substance is defined by the logarithmic function pH=log[H+].pH=log[H+]. This function is used to measure the acidity of a substance. The pH of water is 7. A substance with a pH less than 7 is an acid, whereas one that has a pH of more than 7 is a base.

  1. Find the pH of the following substances. Round answers to one digit.
  2. Determine whether the substance is an acid or a base.
    1. Eggs: [H+]=1.6×10−8[H+]=1.6×10−8 mol/L
    2. Beer: [H+]=3.16×10−3[H+]=3.16×10−3 mol/L
    3. Tomato Juice: [H+]=7.94×10−5[H+]=7.94×10−5 mol/L
304.

[T] Iodine-131 is a radioactive substance that decays according to the function Q(t)=Q0·e−0.08664t,Q(t)=Q0·e−0.08664t, where Q0Q0 is the initial quantity of a sample of the substance and t is in days. Determine how long it takes (to the nearest day) for 95% of a quantity to decay.

305.

[T] According to the World Bank, at the end of 2013 (t=0t=0 ) the U.S. population was 316 million and was increasing according to the following model:

P(t)=316e0.0074t,P(t)=316e0.0074t, where P is measured in millions of people and t is measured in years after 2013.

  1. Based on this model, what will be the population of the United States in 2020?
  2. Determine when the U.S. population will be twice what it is in 2013.
306.

[T] The amount A accumulated after 1000 dollars is invested for t years at an interest rate of 4% is modeled by the function A(t)=1000(1.04)t.A(t)=1000(1.04)t.

  1. Find the amount accumulated after 5 years and 10 years.
  2. Determine how long it takes for the original investment to triple.
307.

[T] A bacterial colony grown in a lab is known to double in number in 12 hours. Suppose, initially, there are 1000 bacteria present.

  1. Use the exponential function Q=Q0ektQ=Q0ekt to determine the value k,k, which is the growth rate of the bacteria. Round to four decimal places.
  2. Determine approximately how long it takes for 200,000 bacteria to grow.
308.

[T] The rabbit population on a game reserve doubles every 6 months. Suppose there were 120 rabbits initially.

  1. Use the exponential function P=P0atP=P0at to determine the growth rate constant a.a. Round to four decimal places.
  2. Use the function in part a. to determine approximately how long it takes for the rabbit population to reach 3500.
309.

[T] The 1906 earthquake in San Francisco had a magnitude of 8.3 on the Richter scale. At the same time, in Japan, an earthquake with magnitude 4.9 caused only minor damage. Approximately how much more energy was released by the San Francisco earthquake than by the Japanese earthquake?

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