college algebra fifth edition james stewart lothar redlin saleem watson

College AlgebraFifth EditionJames Stewart Lothar Redlin Saleem Watson

Exponential and Logarithmic Functions5

Modeling with

Exponential and

Logarithmic Functions5.5

Modeling with Exponential Functions

Many processes that occur in nature

can be modeled using exponential

functions.

• Population growth

• Radioactive decay

• Heat diffusion

Modeling with Logarithmic Functions

Logarithmic functions are used in

models for phenomena such as:

• Loudness of sounds

• Intensity of earthquakes

Exponential Models of Population Growth


Biologists have observed that

the population of a species doubles

its size in a fixed period of time.

• For example, under ideal conditions, a certain bacteria population doubles in size every 3 hours.

• If the culture is started with 1000 bacteria, after 3 hours there will be 2000 bacteria, after another 3 hours there will be 4000, and so on.


If we let n = n(t) be the number of bacteria

after t hours, then

2

2 3

3 4

(0) 1000

(3) 1000 2

(6) (1000 2) 2 1000 2

(9) (1000 2 ) 2 1000 2

(12) (1000 2 ) 2 1000 2

n

n

n

n

n


From this pattern, it appears that

the number of bacteria after t hours

is modeled by the function

n(t) = 1000 · 2t/3


In general, suppose the initial size of a

population is n0 and the doubling period is a.

• Then, the size of the population at time t is modeled by:

n(t) = n02ct where c = 1/a.

• If we knew the tripling time b, the formula would be:

n(t) = n03ct

where c = 1/b.


These formulas indicate that the growth of

the bacteria is modeled by an exponential

function.

• However, what base should we use?


The answer is e.

• Then, it can be shown (using calculus) that the population is modeled by:

n(t) = n0ert

where r is the relative rate of growth of population, expressed as a proportion of the population at any time.

• For instance, if r = 0.02, then at any time t, the growth rate is 2% of the population at time t.

Population Growth & Compound Interest

Notice that the formula for population

growth is the same as that for continuously

compounded interest.

• In fact, the same principle is at work in both cases.

Population Growth & Compound Interest

The growth of a population (or an investment)

per time period is proportional to the size

of the population (or the amount of the

investment).

• A population of 1,000,000 will increase more in one year than a population of 1000.

• In exactly the same way, an investment of $1,000,000 will increase more in one year than an investment of $1000.

Exponential Growth Model

A population that experiences exponential

growth increases according to the model

n(t) = n0ert

where:• n(t) = population at time t

• n0 = initial size of the population

• r = relative rate of growth (expressed as a proportion of the population)

• t = time


In the following examples, we

assume that:

• The populations grow exponentially.

E.g. 1—Predicting the Size of a Population

The initial bacterium count in a culture is 500.

A biologist later makes a sample count of

bacteria in the culture and finds that the

relative rate of growth is 40% per hour.

(a) Find a function that models the number of bacteria after t hours.

(b) What is the estimated count after 10 hours?

(c) Sketch the graph of the function n(t).

E.g. 1—Predicting Population Size

We use the exponential growth model

with n0 = 500 and r = 0.4 to get:

n(t) = 500e0.4t

where t is measured in hours

Example (a)


Using the function in part (a), we find that

the bacterium count after 10 hours is:

n(10) = 500e0.4(10)

= 500e4

≈ 27,300

Example (b)


The graph is shown here.

Example (c)

E.g. 2—Comparing Different Rates of Population Growth

In 2000, the population of the world was

6.1 billion and the relative rate of growth

was 1.4% per year.

• It is claimed that a rate of 1.0% per year would make a significant difference in the total population in just a few decades.


Test this claim by estimating the population

of the world in the year 2050 using

a relative rate of growth of:

(a) 1.4% per year

(b) 1.0% per year


Graph the population functions for

the next 100 years for the two relative

growth rates in the same viewing

rectangle.

E.g. 2—Diff. Rates of Popn. Growth

By the exponential growth model,

we have

n(t) = 6.1e0.014t

where:• n(t) is measured in billions.• t is measured in years since 2000.

Example (a)


Since the year 2050 is 50 years after 2000,

we find:

n(50) = 6.1e0.014(50)

= 6.1e0.7

≈ 12.3

• The estimated population in the year 2050 is about 12.3 billion.

Example (a)


We use the function n(t) = 6.1e0.010t.

We find:

n(50) = 6.1e0.010(50)

= 6.1e0.50

≈ 10.1

• The estimated population in the year 2050 is about 10.1 billion.

Example (b)


These graphs show that:• A small change in the relative rate of growth will,

over time, make a large difference in population size.

E.g. 3—World Population Projections

The population of the world in 2000

was 6.1 billion, and the estimated relative

growth rate was 1.4% per year.

• If the population continues to grow at this rate, when will it reach 122 billion?


We use the population growth function

with:

n0 = 6.1 billion

r = 0.014

n(t) = 122 billion

• This leads to an exponential equation, which we solve for t.


0.014

0.014

0.014

6.1 122

20

ln ln20

0.014 ln20

ln20

0.014213.98

t

t

t

e

e

e

t

t

t• The population will reach 122 billion in approximately

214 years—in the year 2000 + 214 = 2214.

E.g. 4—Finding the Initial Population

A certain breed of rabbit was introduced onto

a small island about 8 years ago.

The current rabbit population on the island is

estimated to be 4100, with a relative growth

rate of 55% per year.

(a) What was the initial size of the population?

(b) Estimate the population 12 years from now.

E.g. 4—Finding Initial Population

From the exponential growth model,

we have:

n(t) = n0e0.55t

Also, we know that the population at

time t = 8 is:

n(8) = 4100

Example (a)


We substitute what we know into the

equation and solve for n0:

• Thus, we estimate that 50 rabbits were introduced onto the island.

0.55(8)0

0 0.55(8)

4100

4100 410050

81.45

n e

ne

Example (a)


Now that we know n0, we can write

a formula for population growth:

n(t) = 50e0.55t

• Twelve years from now, t = 20 and

n(20) = 50e0.55(20) ≈ 2,993,707

• We estimate that the rabbit population on the island 12 years from now will be about 3 million.

Example (b)


Can the rabbit population in Example 4(b)

actually reach such a high number?

• In reality, as the island becomes overpopulated with rabbits, the rabbit population growth will be slowed due to food shortage and other factors.

• One model that takes into account such factors is the logistic growth model.

E.g. 5—The Number of Bacteria in a Culture

A culture starts with 10,000 bacteria,

and the number doubles every 40 min.

(a) Find a function that models the number of bacteria at time t.

(b) Find the number of bacteria after one hour.

(c) After how many minutes will there be 50,000 bacteria?

(d) Sketch a graph of the number of bacteria at time t.

E.g. 5—Bacteria in a Culture

To find the function that models this

population growth, we need to find

the rate r.

• Thus, we use the formula for population growth with:

n0 = 10,000t = 40 n(t) = 20,000

• Then, we solve for r.

Example (a)


(40)

40

40

10,000 20,000

2

ln ln2

40 ln2

ln2

400.01733

r

r

r

e

e

e

r

r

r

Example (a)

• We can now write the function for the population growth: n(t) = 10,000e0.01733t


Using the function we found in part (a)

with t = 60 min (one hour), we get:

n(60) = 10,000e0.01733(60)

≈ 28,287

• The number of bacteria after one hour is approximately 28,000.

Example (b)


We use the function we found in part (a)

with n(t) = 50,000 and solve the resulting

exponential equation for t.

Example (c)


• The bacterium count will reach 50,000 in approximately 93 min.

0.01733

0.01733

0.01733

10,000 50,000

5

ln ln5

0.01733 ln2

ln5

0.0173392.9

t

t

t

e

e

e

t

t

t

Example (c)


The graph of the function

n(t) = 10,000e0.01733t is shown.

Example (d)

Radioactive Decay

Radioactive Decay

Radioactive substances decay by

spontaneously emitting radiation.

• The rate of decay is directly proportional to the mass of the substance.

• This is analogous to population growth, except that the mass of radioactive material decreases.

Radioactive Decay

It can be shown that the mass m(t) remaining

at time t is modeled by the function

m(t) = m0e–rt

where:• r is the rate of decay expressed as

a proportion of the mass.

• m0 is the initial mass.

Half-Life

Physicists express the rate of decay

in terms of half-life—the time required

for half the mass to decay.

• We can obtain the rate r from this as follows.

Radioactive Decay

If h is the half-life, then a mass of 1 unit

becomes ½ unit when t = h. • Substituting this into the model, we get:

• The last equation allows us to find the rate r from the half-life h.

12

12

1

1

ln

1ln 2

ln2

rhe

rh

rh

rh

Radioactive Decay Model

If m0 is the initial mass of a radioactive

substance with half-life h, the mass remaining

at time t is modeled by the function

m(t) = m0e–rt

whereln2

rh

E.g. 6—Radioactive Decay

Polonium-210 (210Po) has a half-life of 140

days.

Suppose a sample has a mass of 300 mg.

(a) Find a function that models the amount remaining at time t.

(b) Find the mass remaining after one year.

(c) How long will it take for the sample to decay to a mass of 200 mg?

(d) Draw a graph of the sample mass as a function of time.


Using the model for radioactive decay

with

m0 = 300 and r = (ln 2/140) ≈ 0.00495

we have:

m(t) = 300e-0.00495t

Example (a)



with t = 365 (one year).

m(365) = 300e-0.00495(365)

≈ 49.256

• Thus, approximately 49 mg of 210Po remains after one year.

Example (b)



with m(t) = 200 and solve the resulting

exponential equation for t.

0.00495

0.00495 23

0.00495 23

300 200

ln ln

t

t

t

e

e

e

Example (c)


• The time required for the sample to decay to 200 mg is about 82 days.

23

23

0.00495 ln

ln

0.00495

81.9

t

t

t

Example (c)


A graph of the function

m(t) = 300e-0.00495t

is shown.

Example (d)

Newton's Law of Cooling

Newton’s Law of Cooling

Newton’s Law of Cooling states that:

The rate of cooling of an object is proportional

to the temperature difference between the

object and its surroundings—provided the

temperature difference is not too large.

• Using calculus, the following model can be deduced from this law.

Newton’s Law of Cooling

If D0 is the initial temperature difference

between an object and its surroundings, and

if its surroundings have temperature Ts , then

the temperature of the object at time t is

modeled by the function

T(t) = Ts + D0e–kt

where k is a positive constant that depends

on the type of object.

E.g. 7—Newton’s Law of Cooling

A cup of coffee has a temperature of 200°F

and is placed in a room that has a

temperature of 70°F.

After 10 min, the temperature of the coffee

is 150°F.

(a) Find a function that models the temperature of the coffee at time t.

(b) Find the temperature of the coffee after 15 min.


(c) When will the coffee have cooled to 100°F?

(d) Illustrate by drawing a graph of the temperature function.


The temperature of the room is:

Ts = 70°F

The initial temperature difference is:

D0 = 200 – 70

= 130°F• So, by Newton’s Law of Cooling,

the temperature after t minutes is modeled by the function

T(t) = 70 + 130e–kt

Example (a)


We need to find the constant k

associated with this cup of coffee.

• To do this, we use the fact that, when t = 10, the temperature is T(10) = 150.

Example (a)


So, we have:

10

10

10 813

813

8110 13

70 130 150

130 80

10 ln

ln

0.04855

k

k

k

e

e

e

k

k

k

Example (a)


Substituting this value of k into

the expression for T(t), we get:

T(t) = 70 + 130e-0.04855t

Example (a)



with t = 15.

T(15) = 70 + 130e-0.04855(15)

≈ 133 °F

Example (b)


We use the function in (a) with T(t) = 100

and solve the resulting exponential

equation for t.

0.04855

0.04855

0.04855 313

70 130 100

130 30

t

t

t

e

e

e

Example (c)


• The coffee will have cooled to 100°F after about half an hour.

Example (c)

313

313

0.04855 ln

ln

0.04855

30.2

t

t

t


Here’s the graph of the temperature

function.

• Notice that the line t = 70 is a horizontal asymptote.

• Why?

Example (d)

Logarithmic Scales

Logarithmic Scales

When a physical quantity varies over

a very large range, it is often convenient

to take its logarithm in order to have a more

manageable set of numbers.

Logarithmic Scales

We discuss three such situations:

• The pH scale—which measures acidity

• The Richter scale—which measures the intensity of earthquakes

• The decibel scale—which measures the loudness of sounds

Logarithmic Scales

Other quantities that are measured

on logarithmic scales include:

• Light intensity

• Information capacity

• Radiation

The pH Scale

Chemists measured the acidity of a solution

by giving its hydrogen ion concentration

until Sorensen, in 1909, proposed a more

convenient measure.

• He defined: pH = –log[H+]

where [H+] is the concentration of hydrogen ions measured in moles per liter (M).

The pH Scale

He did this to avoid very small numbers

and negative exponents.

• For instance, if

[H+] = 10–4 M then

pH = –log10(10–4) = –(–4) = 4

pH Classifications

Solutions with a pH of 7 are defined as

neutral.

Those with pH < 7 are acidic.

Those with pH > 7 are basic.

• Notice that, when the pH increases by one unit, [H+] decreases by a factor of 10.

E.g. 8—pH Scale and Hydrogen Ion Concentration

(a) The hydrogen ion concentration

of a sample of human blood was

measured to be:

[H+] = 3.16 x 10^ –8

• Find the pH and classify the blood as acidic or basic.

E.g. 8—pH Scale and Hydrogen Ion Concentration

(b) The most acidic rainfall ever measured

occurred in Scotland in 1974.

Its pH was 2.4.

• Find the hydrogen ion concentration.

E.g. 8—pH Scale

A calculator gives:

pH = –log[H+]

= –log(3.16 x 10–8)

≈ 7.5

• Since this is greater than 7, the blood is basic.

Example (a)

E.g. 8—Hydrogen Ion Concentration

To find the hydrogen ion concentration,

we need to solve for [H+] in the logarithmic

equation

log[H+] = –pH

• So, we write it in exponential form: [H+] = 10–pH

• In this case, pH = 2.4; so,

[H+] = 10–2.4 ≈ 4.0 x 10–3 M

Example (b)

The Richter Scale

In 1935, American geologist Charles Richter

(1900–1984) defined the magnitude M

of an earthquake to be

where: • I is the intensity of the earthquake (measured by

the amplitude of a seismograph reading taken 100 km from the epicenter of the earthquake)

• S is the intensity of a “standard” earthquake (whose amplitude is 1 micron = 10-4 cm).

logI

MS

The Richter Scale

The magnitude of a standard

earthquake is:

log log1 0S

MS

The Richter Scale

Richter studied many earthquakes that

occurred between 1900 and 1950.

• The largest had magnitude 8.9 on the Richter scale.

• The smallest had magnitude 0.

The Richter Scale

This corresponds to a ratio of intensities

of 800,000,000.

Thus, the scale provides more manageable

numbers to work with.

• For instance, an earthquake of magnitude 6 is ten times stronger than an earthquake of magnitude 5.

E.g. 9—Magnitude of Earthquakes

The 1906 earthquake in San Francisco had

an estimated magnitude of 8.3 on

the Richter scale.

• In the same year, a powerful earthquake occurred on the Colombia- Ecuador border and was four times as intense.

• What was the magnitude of the Colombia-Ecuador earthquake on the Richter scale?

E.g. 9—Magnitude of Earthquakes

If I is the intensity of the San Francisco

earthquake, from the definition of

magnitude, we have:

• The intensity of the Colombia-Ecuador earthquake was 4I.

• So, its magnitude was:

log 8.3MS

I

4log log4 log log4 8.3

8.9

MS S

I I

E.g. 10—Intensity of Earthquakes

The 1989 Loma Prieta earthquake that

shook San Francisco had a magnitude

of 7.1 on the Richter scale.

• How many times more intense was the 1906 earthquake than the 1989 event?


If I1 and I2 are the intensities of the 1906 and

1989 earthquakes, we need to find I1/I2.

• To relate this to the definition of magnitude, we divide numerator and denominator by S.

1 1 1 2

2 2

/log log log log

/

8.3 7.1

1.2

S

S S S

I I I I

I I


Therefore,

• The 1906 earthquake was about 16 times as intense as the 1989 earthquake.

I II

I

1 2log( / )1

2

1.2

10

10

16

The Decibel Scale

The ear is sensitive to an extremely wide

range of sound intensities.

• We take as a reference intensity

I0 = 10–12 W/m2 (watts per square meter)

at a frequency of 1000 hertz.

• This measures a sound that is just barely audible (the threshold of hearing).

The Decibel Scale

The psychological sensation of loudness

varies with the logarithm of the intensity

(the Weber-Fechner Law).

Hence, the intensity level B, measured in

decibels (dB), is defined as:

0

10 logB I

I

The Decibel Scale

The intensity level of the barely audible

reference sound is:

0

0

10 log

10 log1

0 dB

B

I

I

E.g. 11—Sound Intensity of a Jet Takeoff

Find the decibel intensity level of a jet

engine during takeoff if the intensity was

measured at 100 W/m2.

• From the definition of intensity level, we see that:

• The intensity level is 140 dB.

214

120

1010 log 10 log 10 log10

10

140dB

B

I

I

Intensity Levels of Sound

The table lists decibel intensity levels

for some common sounds—ranging from

the threshold of human hearing to the jet

takeoff of Example 11.

• The threshold of pain is about 120 dB.

college algebra fifth edition james stewart lothar redlin saleem watson

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