# Exponential and Logarithmic Functions

Edited by Paul Ducham

DEFINING EXPONENTIAL FUNCTIONS

Let’s start by noting that the functions f and g given by

f(x) = 2x   and     g(x) = x2

are not the same function. Whether a variable appears as an exponent with a constant base or as a base with a constant exponent makes a big difference. The function g is a quadratic function, which we have already discussed. The function f is an exponential function. The graphs of f and g are shown in Figure 1. As expected, they are very different.
We know how to define the values of 2for many types of inputs. For positive integers, it’s simply repeated multiplication:

2= 2 ∙ 2 = 4;  2= 2 ∙ 2 ∙ 2 = 8;  2= 2 ∙ 2 ∙ 2 ∙ 2 = 16

For negative integers, we use properties of negative exponents:

For rational numbers, a calculator comes in handy:

The only catch is that we don’t know how to define 2for all real numbers. For example, what does

2√2

mean? Your calculator can give you a decimal approximation, but where does it come from? That question is not easy to answer at this point. In fact, a precise definition of 2√2 must wait for more advanced courses. For now, we will simply state that for any positive real number b, the expression bx is defined for all real values of x, and the output is a real number as well. This enables us to draw the continuous graph for f(x) = 2x in Figure 1. In Problems 79 and 80 in Exercises 5-1, we will explore a method for defining bx for irrational x values like 2√2.

DEFINITION 1 Exponential Function

The equation

f(x) = bx        b > 0, b ≠ 1

defines an exponential function for each different constant b, called the base.

The independent variable x can assume any real value. The domain of f is the set of all real numbers, and it can be shown that the range of f is the set of all positive real numbers. We require the base b to be positive to avoid imaginary numbers such as (-2)1/2 Problems 53 and 54 in Exercises 5-1 explore why b = 0 and b = 1 are excluded.

GRAPHS OF EXPONENTIAL FUNCTIONS

The graphs of y = bfor b = 2, 3, and 5 are shown in Figure 2. Note that all three have the same basic shape, and pass through the point (0, 1). Also, the x axis is a horizontal asymptote for each graph, but only as x → — ∞. The main difference between the graphs is their steepness.

Next, let’s look at the graphs of y = bx for b = 1/2, 1/3, and 1/5(Fig. 3). Again, all three have the same basic shape, pass through (0, 1), and have horizontal asymptote y = 0, but we can see that for b < 1, the asymptote is only as x → ∞. In general, for bases less than 1, the graph is a reflection through the y axis of the graphs for bases greater than 1. The graphs in Figures 2 and 3 suggest that the graphs of exponential functions have the properties listed in Theorem 1, which we state without proof.

THEOREM 1 Properties of Graphs of Exponential Functions

Let ∫(x) = bx be an exponential function, b>0, b≠1. Then the graph of f (x):

1. Is continuous for all real numbers
2. Has no sharp corners
3. Passes through the point (0, 1)
4. Lies above the x axis, which is a horizontal asymptote either as x →∞ or x → - ∞, but not both
5. Increases as x increases if b>1; decreases as x increases if 0<b<1
6. Intersects any horizontal line at most once (that is, ∫ is one-to-one)

These properties indicate that the graphs of exponential functions are distinct from the graphs we have already studied. (Actually, property 4 is enough to ensure that graphs of exponential functions are different from graphs of polynomials and rational functions.) Property 6 is important because it guarantees that exponential functions have inverses, called logarithmic functions.

Transformations of exponential functions are very useful in modeling real-world phenomena, like population growth and radioactive decay. It is important to understand how the graphs of those functions are related to the graphs of the exponential functions in this section.

EXAMPLE 1 Transformations of Exponential Functions

Exponential functions whose domains include irrational numbers obey the familiar laws of exponents for rational exponents. We summarize these exponent laws here and add two other useful properties.

Property 2 is another way to express the fact that the exponential function f (x) ax is one-to- one (see property 6 of Theorem 1). Because all exponential functions of the form f (x) ax pass through the point (0, 1) (see property 3 of Theorem 1), property 3 indicates that the graphs of exponential functions with different bases do not intersect at any other points.

THE EXPONENTIAL FUNCTION WITH BASE E

Surprisingly, among the exponential functions it is not the function g(x)=2x with base 2 or the function h(x)=10with base 10 that is used most frequently in mathematics. Instead, the most commonly used base is a number that you may not be familiar with.
By calculating the value of [1+(1/x)]xfor larger and larger values of x (Table 1), it looks like approaches a number close to 2.7183. In a calculus course, we can show that as x increases without bound, the value of [1+(1/x)]x approaches an irrational number that we call e. Just as irrational numbers such as π and √2 have unending, nonrepeating decimal representations, e also has an unending, nonrepeating decimal representation. To 12 decimal places,

Don’t let the symbol “e” intimidate you! It’s just a number.
Exactly who discovered e is still being debated. It is named after the great Swiss mathematician Leonhard Euler (1707–1783), who computed e to 23 decimal places using [1+(1/x)]x .
The constant e turns out to be an ideal base for an exponential function because in calculus and higher mathematics many operations take on their simplest form using this base. This is why you will see e used extensively in expressions and formulas that model realworld phenomena.

The exponential function with base e is used so frequently that it is often referred to as the exponential function. The graphs of y =ex and y=e

COMPOUND INTEREST

The fee paid to use someone else’s money is called interest. It is usually computed as a percentage, called the interest rate, of the original amount (or principal) over a given period of time. At the end of the payment period, the interest paid is usually added to the principal amount, so the interest in the next period is earned on both the original amount, as well as the interest previously earned. Interest paid on interest previously earned and reinvested in this manner is called compound interest.
Suppose you deposit \$1,000 in a bank that pays 8% interest compounded semiannually. How much will be in your account at the end of 2 years? “Compounded semiannually” means that the interest is paid to your account at the end of each 6-month period, and the interest will in turn earn more interest. To calculate the interest rate per period, we take the annual rate r, 8% (or 0.08), and divide by the number m of compounding periods per year, in this case 2. If represents the amount of money in the account after one compounding period (6 months), then

What do you think the savings and loan will owe you at the end of 6 years (12 compounding periods)? If you guessed A=\$1,000(1+0.04)12 you have observed a pattern that is generalized in the following compound interest formula:

EXAMPLE 4 Compound Interest

If you deposit \$5,000 in an account paying 9% compounded daily,* how much will you have in the account in 5 years? Compute the answer to the nearest cent.

SOLUTION We will use the compound interest formula with P=5,000, r=0.09, (which is 9% written as a decimal), m=365, and n=5(365)=1,825:

EXAMPLE 5 Comparing Investments

If \$1,000 is deposited into an account earning 10% compounded monthly and, at the same time, \$2,000 is deposited into an account earning 4% compounded monthly, will the first account ever be worth more than the second? If so, when?

SOLUTION Let y1 and y2 represent the amounts in the first and second accounts, respectively, then

where x is the number of compounding periods (months). Examining the graphs of y1 and y2 [Fig. 7(a)], we see that the graphs intersect at x=139.438 months. Because compound interest is paid at the end of each compounding period, we compare the amount in the accounts after 139 months and after 140 months [Fig. 7(b)]. The first account is worth more than the second for x ≥ 140 months, or after 11 years and 8 months.

INTEREST COMPOUNDED CONTINUOUSLY

If \$1,000 is deposited in an account that earns compound interest at an annual rate of 8% for 2 years, how will the amount A change if the number of compounding periods is increased? If m is the number of compounding periods per year, then

The amount A is computed for several values of m in Table 2. Notice that the largest gain appears in going from annually to semiannually. Then, the gains slow down as m increases. In fact, it appears that A might be approaching something close to \$1,173.50 as m gets larger and larger.
We now return to the general problem to see if we can determine what happens to A =P[1+(r/m) ]mt as m increases without bound. A little algebraic manipulation of the compound interest formula will lead to an answer and a significant result in the mathematics of finance:

This is known as the continuous compound interest formula, a very important and widely used formula in business, banking, and economics.

CONTINUOUS COMPOUND INTEREST FORMULA

If a principal P is invested at an annual rate r compounded continuously, then the amount A in the account at the end of t years is given by
A=Pert
The annual rate r must be expressed as a decimal.

EXAMPLE 6 Continuous Compound Interest

If \$1,000 is invested at an annual rate of 8% compounded continuously, what amount, to the nearest cent, will be in the account after 2 years?

SOLUTION Use the continuous compound interest formula to find A when P=\$1,000, r=0.08, and t=2:

Notice that the values calculated in Table 2 get closer to this answer as m gets larger.

MATHEMATICAL MODELING

Populations tend to grow exponentially and at different rates. A convenient and easily understood measure of growth rate is the doubling time—that is, the time it takes for a population to double. Over short periods the doubling time growth model is often used to model population growth:

and the population is double the original, as it should be. We will use this model to solve a population growth problem in Example 1.

EXAMPLE 1 Population Growth

According to a 2008 estimate, the population of Nicaragua was about 5.7 million, and that population is growing due to a high birth rate and relatively low mortality rate. If the population continues to grow at the current rate, it will double in 37 years. If the growth remains steady, what will the population be in
(A) 15 years?           (B) 40 years?
Calculate answers to three significant digits.

SOLUTIONS We can use the doubling time growth model, A=A0(2)t/d with A0=5.7 and d=37: A =5.7(2)t/37 See Figure 1.

The doubling time model is not the only one used to model populations. An alternative model based on the continuous compound interest formula will be used in Example 2. In this case, the formula is written as

The relative growth rate is written as a percentage in decimal form. For example, if a population is growing so that at any time the population is increasing at 3% of the current population per year, the relative growth rate k would be 0.03.

EXAMPLE 2 Medicine—Bacteria Growth
Cholera, an intestinal disease, is caused by a cholera bacterium that multiplies exponentially by cell division as modeled by A=A0e1.386t where A is the number of bacteria present after t hours and A0 is the number of bacteria present at t=0. If we start with 1 bacterium, how many bacteria will be present in (A) 5 hours? (B) 12 hours?

Calculate the answers to three significant digits.

Exponential functions can also be used to model radioactive decay, which is sometimes referred to as negative growth. Radioactive materials are used extensively in medical diagnosis and therapy, as power sources in satellites, and as power sources in many countries. If we start with an amount A0 of a particular radioactive substance, the amount declines exponentially over time. The rate of decay varies depending on the particular radioactive substance. A convenient and easily understood measure of the rate of decay is the half-life of the material—that is, the time it takes for half of a particular material to decay. We can use the following half-life decay model:

The radioactive isotope gallium 67 (67Ga), used in the diagnosis of malignant tumors, has a biological half-life of 46.5 hours. If we start with 100 milligrams of the isotope, how many milligrams will be left after
(A) 24 hours?        (B) 1 week?
Calculate answers to three significant digits.

In Example 2, we saw that a base e exponential function can be used as an alternative to the doubling time model. Not surprisingly, the same can be said for the half-life model. In this case, the formula will be

Our atmosphere is constantly being bombarded with cosmic rays. These rays produce neutrons, which in turn react with nitrogen to produce radioactive carbon-14. Radioactive carbon- 14 enters all living tissues through carbon dioxide, which is first absorbed by plants. As long as a plant or animal is alive, carbon-14 is maintained in the living organism at a constant level. Once the organism dies, however, carbon-14 decays according to the equation

where A is the amount of carbon-14 present after t years and A0 is the amount present at time t=0. This can be used to calculate the approximate age of fossils.

DATA ANALYSIS AND REGRESSION

Many graphing calculators have options for exponential and logistic regression. We can use exponential regression to fit a function of the form y=abx to a set of data points, and logistic regression to fit a function of the form

to a set of data points. The techniques are similar to those for linear and quadratic functions.

EXAMPLE 7 Infectious Diseases
The U.S. Department of Health and Human Services published the data in Table 1.

An exponential model for the data on mumps is given by A=81,082(0.844)t
where A is the number of reported cases of mumps and t is time in years with t=0 representing 1970.

(A) Use the model to predict the number of reported cases of mumps in 2010.
(B) Compare the actual number of cases of mumps reported in 1980 to the number given by the model.

SOLUTIONS (A) The year 2010 is represented by t=40. Evaluating A=81,082(0.844)t at t=40 gives a prediction of 92 cases of mumps in 2010.
(B) The year 1980 is represented by t 10. Evaluating A 81,082(0.844)t at t=10 gives 14,871 cases in 1980. The actual number of cases reported in 1980 was 8,576, nearly 6,300 less than the number given by the model.

EXAMPLE 8 AIDS Cases and Deaths

The U.S. Department of Health and Human Services published the data in Table 2

where A is the number of AIDS cases diagnosed by year t with t=0 representing 1985.
(A) Use the model to predict the number of AIDS cases diagnosed by 2010.
(B) Compare the actual number of AIDS cases diagnosed by 2005 to the number given by the model.

SOLUTIONS (A) The year 2010 is represented by t 25. Evaluating

at t=25 gives a prediction of approximately 940,000 cases of AIDS diagnosed by 2010.
(B) The year 2005 is represented by t=20. Evaluating

at t=20 gives 916,690 cases in 2005. The actual number of cases diagnosed by 2005 was 944,306, nearly 28,000 greater than the number given by the model.

A COMPARISON OF EXPONENTIAL GROWTH PHENOMENA

The equations and graphs given in Table 3 compare several widely used growth models. These are divided basically into two groups: unlimited growth and limited growth. Following each equation and graph is a short, incomplete list of areas in which the models are used. We have only touched on a subject that has been extensively developed and that you are likely to study in greater depth in the future.

DEFINING LOGARITHMIC FUNCTIONS

The exponential function f(x)=bx for b>0, b≠1 is a one-to-one function, and therefore has an inverse. Its inverse, denoted f-1(x)=logbx (read “log to the base b of x”) is called the logarithmic function with base b. Just like exponentials, there are different logarithmic functions for each positive base other than 1. A point (x, y) is on the graph of f-1=logbx if and only if the point (y, x) is on the graph of f=bx. In other words,

We can use this fact to learn some things about the logarithmic functions from our knowledge of exponential functions. For example, the graph of f-1=logbx is the graph of f=bx reflected through the line y=x. Also, the domain of f-1=logbx is the range of and vice versa.
In Example 1, we will use information about f(x)=2xto graph its inverse, f-1=log2x.

EXAMPLE 1 Graphing a Logarithmic Function

Make a table of values for f(x)=2x and reverse the ordered pairs to obtain a table of values for f-1(x)=log2x. Then use both tables to graph f(x) and f-1(x) on the same set of axes.

SOLUTION We chose to evaluate f for integer values from -3 to 3. The tables are shown here, along with the graph (Fig. 1). Note the important comments about domain and range below the graph.

It is very important to remember that the equations y=logbx and x=by define the same function, and as such can be used interchangeably. Because the domain of an exponential function includes all real numbers and its range is the set of positive real numbers, the domain of a logarithmic function is the set of all positive real numbers and its range is the set of all real numbers. For example, log103 is defined, but log100 and log10-5 are not defined.
In short, the function y=logbx for any b is only defined for positive x values. Typical logarithmic curves are shown in Figure 2. Notice that in each case, the y axis is a vertical asymptote for the graph.
The graphs in Example 1 and Figure 2 suggest that logarithmic graphs share some common properties. Several of these properties are listed in Theorem 1. It might be helpful in understanding them to review Theorem 1 in Section 5-1. Each of these properties is a consequence of a corresponding property of exponential graphs.

THEOREM 1 Properties of Graphs of Logarithmic Functions

Let f(x)=logbx be a logarithmic function, b>0, b≠1. Then the graph of f (x):
1. Is continuous on its domain (0, ∞)
2. Has no sharp corners
3. Passes through the point (1, 0)
4. Lies to the right of the y axis, which is a vertical asymptote
5. Is increasing as x increases if b>1; is decreasing as x increases if 0<b<1
6. Intersects any horizontal line exactly once, so is one-to-one

CONVERTING BETWEEN LOGARITHMIC FORM AND EXPONENTIAL FORM

We now look into the matter of converting logarithmic forms to equivalent exponential forms, and vice versa. It will be useful to sometimes convert a logarithmic expression into the equivalent exponential form. At other times, it will be useful to do the reverse.

To gain a little deeper understanding of logarithmic functions and their relationship to the exponential functions, we will consider a few problems where we want to find x, b, or y in y=logbx given the other two values. All values were chosen so that the problems can be solved without a calculator. In each case, converting to the equivalent exponential form is useful.

PROPERTIES OF LOGARITHMIC FUNCTIONS

Some of the properties of exponential functions that we studied in Section 5-1 can be used to develop corresponding properties of logarithmic functions. Several of these important properties of logarithmic functions are listed in Theorem 2. We will justify them individually.

3 and 4. These are simply another way to state that f(x)=bx and f-1(x)=logbx are inverse functions. Property 3 can be written as f-1(f(x))=x for all x in the domain of f. Property 4 can be written as f(f-1(x))=x for all x in the domain of This matches our characterization of inverse functions in Theorem 5, Section 3-6. Together, these properties say that if you apply an exponential function and a logarithmic function with the same base consecutively (in either order) you end up with the same value you started with. This follows from the fact that logarithmic functions are one-to-one. Properties 6, 7, and 8 are used often in manipulating logarithmic expressions.

EXAMPLE 5

COMMON AND NATURAL LOGARITHMS

To work with logarithms effectively, we will need to be able to calculate (or at least approximate) the logarithms of any positive number to a variety of bases. Historically, tables were used for this purpose, but now calculators are used because they are faster and can find far more values than any table can possibly include.
Of all possible bases, there are two that are used most often. Common logarithms are logarithms with base 10. Natural logarithms are logarithms with base e. Most calculators have a function key labeled “log” and a function key labeled “ln.” The former represents the common logarithmic function and the latter the natural logarithmic function. In fact, “log” and “ln” are both used in most math books, and whenever you see either used in this book without a base indicated, they should be interpreted as follows:

EXAMPLE 6 Calculator Evaluation of Logarithms
Use a calculator to evaluate each to six decimal places.
A) log 3,184 (B) ln 0.000 349 (C) log (-3.24)

SOLUTIONS (A) log 3,184=3.502 973        (B) ln 0.000 349=-7.960 439
(C) log (-3.24)=Error

Why is an error indicated in part C? Because -3.24 is not in the domain of the log function. [Note: Calculators display error messages in various ways. Some calculators use a more advanced definition of logarithmic functions that involves complex numbers. They will display an ordered pair, representing a complex number, as the value log (-3.24) of rather than an error message. You should interpret such a display as indicating that the number entered is not in the domain of the logarithmic function as we have defined it.]
When working with common and natural logarithms, we will follow the common practice of using the equal sign “=” where it might be technically correct to use the approximately equal sign “” No harm is done as long as we keep in mind that in a statement such as log 3.184=0.503, the number on the right is only assumed accurate to three decimal places and is not exact.

We now turn to the opposite problem: Given the logarithm of a number, find the number. To solve this problem, we make direct use of the logarithmic–exponential relationships, and change logarithmic expressions into exponential form.

THE CHANGE-OF-BASE FORMULA

How would you find the logarithm of a positive number to a base other than 10 or e? For example, how would you find log3 5.2? In Example 9 we evaluate this logarithm using several properties of logarithms. Then we develop a change-of-base formula to find such logarithms more easily. EXAMPLE 9 Evaluating a Base 3 Logarithm Evaluate log35.2 to four decimal places.

SOLUTION Let y=log35.2 and proceed as follows:

If we repeat the process we used in Example 9 on a generic logarithm, something interesting happens. The goal is to evaluate logbN where b is any acceptable base, and N is any positive real number. As in Example 9, let y=logbN.

LOGARITHMIC SCALES

SOUND INTENSITY: The human ear is able to hear sound over a very wide range of intensities. The loudest sound a healthy person can hear without damage to the eardrum has an intensity 1 trillion (1,000,000,000,000) times that of the softest sound a person can hear. If we were to use these intensities as a scale for measuring volume, we would be stuck using numbers from zero all the way to the trillions, which seems cumbersome, if not downright silly. In the last section, we saw that logarithmic functions increase very slowly. We can take advantage of this to create a scale for sound intensity that is much more condensed, and therefore more manageable.
The decibel scale for sound intensity is an example of such a scale. The decibel, named after the inventor of the telephone, Alexander Graham Bell (1847–1922), is defined as follows:

where D is the decibel level of the sound, I is the intensity of the sound measured in watts per square meter (W/m2) and is the intensity of the least audible sound that an average healthy young person can hear. The latter is standardized to be I0=10-12 watts per square meter. Table 1 lists some typical sound intensities from familiar sources. In Example 1 and Problems 5 and 6 in Exercises 5-4, we will calculate the decibel levels for these sounds.

EXAMPLE 1 Sound Intensity
(A) Find the number of decibels from a whisper with sound intensity 5.2×10-10 watts per square meter, then from heavy traffic at 8.5×10-4 watts per square meter. Round your answers to two decimal places. (B) How many times larger is the sound intensity of heavy traffic compared to a whisper?

EARTHQUAKE INTENSITY: The energy released by the largest earthquake recorded, measured in joules, is about 100 billion (100,000,000,000) times the energy released by a small earthquake that is barely felt. In 1935 the California seismologist Charles Richter devised a logarithmic scale that bears his name and is still widely used in the United States. The magnitude of an earthquake M on the Richter scale* is given as follows:

EXAMPLE 2 Earthquake Intensity

The 1906 San Francisco earthquake released approximately 5.96×1016of energy. Another quake struck the Bay Area just before game 3 of the 1989 World Series, releasing 1.12×1015 joules of energy.

(A) Find the magnitude of each earthquake on the Richter scale. Round your answers to two decimal places.
(B) How many times more energy did the 1906 earthquake release than the one in 1989?

EXAMPLE 3 Earthquake Intensity

If the energy release of one earthquake is 1,000 times that of another, how much larger is the Richter scale reading of the larger than the smaller?

ROCKET FLIGHT: The theory of rocket flight uses advanced mathematics and physics to show that the velocity v of a rocket at burnout (depletion of fuel supply) is given by

where c is the exhaust velocity of the rocket engine, Wt is the takeoff weight (fuel, structure, and payload), and Wb is the burnout weight (structure and payload).
Because of the Earth’s atmospheric resistance, a launch vehicle velocity of at least 9.0 kilometers per second is required to achieve the minimum altitude needed for a stable orbit. Formula (3) indicates that to increase velocity v, either the weight ratio Wt/Wb must be increased or the exhaust velocity c must be increased. The weight ratio can be increased by the use of solid fuels, and the exhaust velocity can be increased by improving the fuels, solid or liquid.

EXAMPLE 4 Rocket Flight Theory

A typical single-stage, solid-fuel rocket may have a weight ratio Wt/Wb =18.7and an exhaust velocity c=2.38 kilometers per second. Would this rocket reach a launch velocity of 9.0 kilometers per second?

The velocity of the launch vehicle is far short of the 9.0 kilometers per second required to achieve orbit. This is why multiple-stage launchers are used—the deadweight from a preceding stage can be jettisoned into the ocean when the next stage takes over.

DATA ANALYSIS AND REGRESSION

Based on the logarithmic graphs we studied in the last section, when a quantity increases relatively rapidly at first, but then levels off and increases very slowly, it might be a good candidate to be modeled by a logarithmic function. Most graphing calculators with regression commands can fit functions of the form y=a+b to a set of data points using the same techniques we used earlier for other types of regression.

EXAMPLE 5 Home Ownership Rates

The U.S. Census Bureau published the data in Table 3 on home ownership rates. A logarithmic model for the data is given by R=-36.7+23.0 ln t where R is the home ownership rate and t is time in years with t=0 representing 1900.

(A) Use the model to predict the home ownership rate in 2015.
(B) Compare the actual home ownership rate in 1950 to the rate given by the model.

SOLUTIONS (A) The year 2015 is represented by t =115. Evaluating
R =-36.7+23.0 ln t
at t =115 predicts a home ownership rate of 72.4% in 2015.
(B) The year 1950 is represented by t 50. Evaluating
R =-36.7+23.0 ln t
at t =50 gives a home ownership rate of 53.3% in 1950. The actual home ownership rate in 1950 was 55%, approximately 1.7% greater than the number given by the model.

SOLVING EXPONENTIAL EQUATIONS

The distinguishing feature of exponential equations is that the variable appears in an exponent. Before defining logarithms, we didn’t have a reliable method for removing variables from an exponent: Now we do. We’ll illustrate how these properties are helpful in Examples 1-4.
EXAMPLE 1
Solving an Exponential Equation
Find all solutions to 23x-2 = 5 to four decimal places.
SOLUTION: In order to have any chance of solving for x, we will first need to get x out of the exponent. This is where logs come in very handy.
23x-2 = 5                   Take the common or natural log of both sides.
log 3x-2 = log 5             Use logb Np = p log b N to get 3x - 2 out of the exponent position.
(3x - 2) log 2 = log 5             Divide both sides by log 2.

EXAMPLE 2
Compound Interest Recall that when an amount of money P (principal) is invested at an annual rate r compounded annually, the amount of money A in the account after n years, assuming no withdrawals, is given by

How many years to the nearest year will it take the money to double if it is invested at 6% compounded annually?
SOLUTON: The interest rate is r = 0.06, and we want the amount A to be twice the principal, or 2P. So we substitute r = 0.06 and A = 2P, and solve for n.
2P = P(1.06)n             Divide both sides by P to isolate (1.06)n.
2 = 1.06n                Take the common or natural log of both sides.
log 2 = log 1.06n          Note how log properties are used to get n out of the exponent position.
log 2 = n log 1.06         Divide both sides by log 1.06 (which is just a number!).

= 12 years

EXAMPLE 3
Atmospheric Pressure
The atmospheric pressure P, in pounds per square inch, at x miles above sea level is given approximately by
P = 14.7e-0.21x
At what height will the atmospheric pressure be half the sea-level pressure? Compute the answer to two significant digits.
SOLUTION : Since x is miles above sea level, sea-level pressure is the pressure at x 0, which is 14.7e0, or 14.7.
One-half of sea level pressure is 14.7/2 = 7.35. Now our problem is to find x so that P = 7.35; that is, we solve 7.35 = 14.7e-0.21x for x:
7.35 = 14.7e-0.21x    Divide both sides by 14.7 to isolate the exponential.
0.5 = e-0.21x              Because the base is e, take the natural log of both sides.
ln 0.5 = ln e-0.21x         In ea = a, so ln e-0.21x = -0.21x
ln 0.5 = -0.21x           Divide both sides by 0.21.

= 3.3 miles

The graph of

is a curve called a catenary (Fig. 1). A uniform cable suspended between two fixed points is a physical example of such a curve, which resembles a parabola, but isn’t.

EXAMPLE 4
Solving an Exponential Equation
In equation (1), find x when y = 2.5. Compute the answer to four decimal places.
SOLUTION:

Note that the method produces exact solutions, an important consideration in certain calculus applications (see Problems 57–60 of Exercises 5-5).

SOLVING LOGARITHMIC EQUATIONS

We will now illustrate the solution of several types of logarithmic equations.
EXAMPLE 5
Solving a Logarithmic Equation
Solve log (x + 3) + log x = 1, and check.
SOLUTION: First use properties of logarithms to express the left side as a single logarithm, then convert to exponential form and solve for x.
log (x + 3) + log x = 1     Combine left side using log M + log N = log MN.
log [x(x + 3)] = 1    Change to equivalent exponential form (the base is 10).
x(x + 3) = 101   Write in ax2 + bx + c = 0 form and solve.
x2 + 3x - 10 = 0       Factor.
(x + 5)(x -2) = 0        If ab = 0, then a = 0 or b = 0.
x = -5, 2
CHECK:               x = -5: log (-5 + 3) + log (-5) is not defined
because the domain of the log function is (0, ∞).
x = 2: log (2 + 3) + log 2 = log 5 + log 2
= log (5 ∙ 2) = log 10  1
The only solution to the original equation is x = 2. Extraneous solutions are common in log equations, so answers should always be checked in the original equation to see whether any should be discarded.

EXAMPLE 6
Solving a Logarithmic Equation
Solve (ln x)2 = ln x2.
SOLUTION: There are no logarithmic properties for simplifying (ln x)2. However, we can simplify ln x2, obtaining an equation involving ln x and (ln x)2.
(ln x)2 = ln x2                     ln Mp = p ln M, so ln x2 = 2 ln x.
(ln x)2 = 2 ln x               This is a quadratic equation in ln x. Move all nonzero terms to the left.
(ln x)2 - 2 ln x = 0              Factor out ln x.
(ln x)(ln x - 2) = 0             If ab = 0, then a = 0 or b = 0.
ln x = 0 or ln x - 2 = 0      If ln x = a, x = ea.
x = e0     ln x = 2
= 1 x           = e2
Checking that both x = 1 and x = e2 are solutions to the original equation is left to you.
EXAMPLE 7
Earthquake Intensity
Recall from Section 5-4 that the magnitude of an earthquake on the Richter scale is given by

Solve for E in terms of the other symbols.
SOLUTION: