WALCH EDUCATIONmathplayer.weebly.com/uploads/3/8/0/0/38006537/ccgps... · 2020-02-07 · Then you...

1 2 3 4 5 6 7 8 9 10

ISBN 978-0-8251-7375-2

Copyright © 2014

J. Weston Walch, Publisher

Portland, ME 04103

www.walch.com

Printed in the United States of America

EDUCATIONWALCH

These materials may not be reproduced for any purpose.The reproduction of any part for an entire school or school system is strictly prohibited.

No part of this publication may be transmitted, stored, or recorded in any formwithout written permission from the publisher.

iiiTable of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

Unit 4: Exponential and Logarithmic FunctionsLesson 1: Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . U4-1Lesson 2: Introducing Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . U4-30Lesson 3: Solving Exponential Equations Using Logarithms . . . . . . . . . . . . . . . . . U4-61

Answer Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . AK-1

Table of Contents

vIntroduction

Welcome to the CCGPS Advanced Algebra Student Resource Book. This book will help you learn how to use algebra, geometry, data analysis, and probability to solve problems. Each lesson builds on what you have already learned. As you participate in classroom activities and use this book, you will master important concepts that will help to prepare you for the EOCT and for other mathematics assessments and courses.

This book is your resource as you work your way through the Advanced Algebra course. It includes explanations of the concepts you will learn in class; math vocabulary and definitions; formulas and rules; and exercises so you can practice the math you are learning. Most of your assignments will come from your teacher, but this book will allow you to review what was covered in class, including terms, formulas, and procedures.

• In Unit 1: Inferences and Conclusions from Data, you will learn about summarizing and interpreting data and using the normal curve. You will explore populations, random samples, and sampling methods, as well as surveys, experiments, and observational studies. Finally, you will compare treatments and read reports.

• In Unit 2: Polynomial Functions, you will begin by exploring polynomial structures and operations with polynomials. Then you will go on to prove identities, graph polynomial functions, solve systems of equations with polynomials, and work with geometric series.

• In Unit 3: Rational and Radical Relationships, you will be introduced to operating with rational expressions. Then you will learn about solving rational and radical equations and graphing rational functions. You will solve and graph radical functions. Finally, you will compare properties of functions.

• In Unit 4: Exponential and Logarithmic Functions, you will start working with exponential functions and begin exploring logarithmic functions. Then you will solve exponential equations using logarithms.

• In Unit 5: Trigonometric Functions, you will begin by exploring radians and the unit circle. You will graph trigonometric functions, including sine and cosine functions, and use them to model periodic phenomena. Finally, you will learn about the Pythagorean Identity.

Introduction

Introductionvi

• In Unit 6: Mathematical Modeling, you will use mathematics to model equations and piecewise, step, and absolute value functions. Then, you will explore constraint equations and inequalities. You will go on to model transformations of graphs and compare properties within and between functions. You will model operating on functions and the inverses of functions. Finally, you will learn about geometric modeling.

Each lesson is made up of short sections that explain important concepts, including some completed examples. Each of these sections is followed by a few problems to help you practice what you have learned. The “Words to Know” section at the beginning of each lesson includes important terms introduced in that lesson.

As you move through your Advanced Algebra course, you will become a more confident and skilled mathematician. We hope this book will serve as a useful resource as you learn.

Lesson 1: Exponential FunctionsUNIT 4 • EXPONENTIAL AND LOGARITHMIC FUNCTIONS

U4-1Lesson 1: Exponential Functions

4.1

Common Core Georgia Performance Standards

MCC9–12.A.SSE.3c

MCC9–12.F.IF.7e

MCC9–12.F.IF.8b

Essential Questions

1. How can you rewrite an exponential function using the properties of exponents?

2. What information can an exponential function provide when it is rewritten?

3. How can you determine whether the growth rate of an exponential function is positive or negative?

4. How can you graph an exponential function?

5. What are some key features of an exponential function that can be determined from its graph?

6. How can problems involving interest rates be solved using exponential functions?

WORDS TO KNOW

asymptote an equation that represents sets of points that are not allowed by the conditions in a parent function or model; a line that a function gets closer and closer to, but never crosses or touches

base the quantity that is being raised to a power in an exponential expression; in ax, a is the base

compound interest interest earned on both the initial amount and on previously earned interest

continuously compounded interest

compound interest that is being added to the balance at every instant

delta (∆) a Greek letter commonly used to represent the change in a value

U4-2Unit 4: Exponential and Logarithmic Functions 4.1

domain the set of all input values (x-values) that satisfy the given function without restriction

e an irrational number with an approximate value of 2.71828

end behavior the behavior of the graph as x approaches positive or negative infinity

exponent the quantity that shows the number of times the base is being multiplied by itself in an exponential expression; also known as the power. In ax, x is the power/exponent.

exponential decay an exponential equation with a base, b, that is between 0 and 1 (0 < b < 1); can be represented by the formula f(t) = a(1 – r)t, where a is the initial value, (1 – r) is the decay factor, t is time, and f(t) is the final value

exponential expression an expression that contains a base raised to a power/exponent

exponential function a function in the form f(x) = a(bx) + c, where a, b, and c are constants and b is greater than 0 but not equal to 1

exponential growth an exponential function with a base, b, greater than 1 (b > 1); can be represented by the formula f(t) = a(1 + r)t, where a is the initial value, (1 + r) is the growth factor, t is time, and f(t) is the final value

natural exponential function

an exponential function with a base of e

percent rate of change the percentage by which a function increases or decreases within a certain interval

power the quantity that shows the number of times the base is being multiplied by itself in an exponential expression; also known as the exponent. In ax, x is the power/exponent.

range the set of all outputs of a function; the set of y-values that are valid for the function


4.1

rate of change a ratio that describes how much one quantity changes with respect to the change in another quantity; also known as the slope of a line

y-intercept the point at which the graph crosses the y-axis; written as (0, y)

Recommended Resources

• Discovery Education. “Compound Interest Calculator.”

http://www.walch.com/rr/00252

This site includes a tool to calculate the amount of compound interest earned upon entering the principal amount, the interest rate, the compounding frequency, and the time in years. The calculator breaks down the interest earned during each compounding period and gives the final amount.

• Hotmath.com. “Exponential Functions: y = a(bx) + c.”


This site graphs the function y = a(bx) + c for different values of a, b, and c. Use the sliders to change the values of the parameters and observe how the graph changes accordingly.

• MathBitsNotebook.com. “Exponents – Integers – Algebraic.”


This practice page tests knowledge of the laws of exponents with true/false statements and multiple-choice questions, and provides immediate feedback.

U4-4Unit 4: Exponential and Logarithmic Functions 4.1.1

Lesson 4.1.1: Rewriting Exponential Functions

Introduction

An exponential function is a function in the form f(x) = a(bx) + c, where a, b, and c are constants and b is greater than 0 but not equal to 1. An exponential expression is an expression that contains a base raised to a power/exponent. The expressions in exponential functions can often be rewritten using the properties of exponents in order to interpret information in real-world functions. For example, exponential functions can be used to model situations involving compound interest, or interest earned on both the initial amount and on previously earned interest. By rewriting these types of exponential functions, interest rates for different time periods can be determined.

Key Concepts

• In the exponential function form f(x) = a(bx) + c, bx is an exponential expression; b represents the base, or the quantity that is being raised to a power in an exponential expression, and x represents the exponent or power, the quantity that shows the number of times the base is being multiplied by itself.

• If the base of an exponential function is greater than 1 (b > 1), the function models exponential growth, or an increase in an amount at a constant rate. Exponential growth can be represented by the formula f(t) = a(1 + r)t, where a is the initial value, (1 + r) is the growth factor, t is time, and f(t) is the final value.

• If the base of an exponential function is greater than 0 but less than 1 (0 < b < 1), the function models exponential decay, or a decrease in an amount at a constant rate. Exponential decay can be represented by the formula f(t) = a(1 – r)t, where a is the initial value, (1 – r) is the decay factor, t is time, and f(t) is the final value.

• An exponential function that has a base of e is called a natural exponential function. The value of e, an irrational number, is approximately 2.71828.

• Natural exponential functions are commonly used in the fields of physics, chemistry, and business. Professionals in these fields often deal with situations involving continual growth or decay, such as analyzing the return on an investment or the progress of a chemical reaction.


4.1.1

• In banking, natural exponential functions are used to calculate compound interest that is being added to a balance at every instant. This is called continuously compounded interest.

• A single function has a change of rate that is dependent on the change that occurs between two points on the graph of that function. This is the same as the slope when the function is a line.

• The rate of change of a function is a ratio that describes how

much one quantity changes with respect to the change in another

quantity; mathematically, the rate of change can be represented by ( ) ( ) ( ) ( ) ( )1 2

1 2

2 1

2 1

rf x f x

x x

f x f x

x x

f x

x

�

�=

−−

=−−

= , in which f1(x) is the function value at

the domain value x1 and f2(x) is the function value at the domain value x2. The

uppercase Greek letter delta (∆) is commonly used to represent the “change”

in a value; for example, ∆f(x) can be read as “change in f of x.” Therefore,

∆f(x) and ∆x are more concise ways of representing the numerator and

denominator, respectively, in the rate of change formula.

• Note that the order in which the function values are compared must be the same as the order in which the domain values are compared. Mixing up the order of the values in the numerator with the order of the values in the denominator will result in an incorrect rate calculation.

• The percent rate of change of a function is the percentage by which the function increases or decreases within a certain interval. If this percent is positive, then the function shows exponential growth; if the percent is negative, then the function shows exponential decay.


• The following table of properties can be helpful when rewriting exponential functions.

Properties of Exponents

Property General rule

Zero Exponent Property a0 = 1

Negative Exponent Property bb

m

nm

n

1=

−

Product of Powers Property a a am n m n• = +

Quotient of Powers Propertya

aa

m

nm n= −

Power of a Power Property b bm n mn( ) =

Power of a Product Property bc b cn n n( ) =

Power of a Quotient Propertya

b

a

b

m m

m

=

Rational Exponents Property a an n

1

= , where n ≠ 0


4.1.1

Example 1

Rewrite the function f(x) = 363x so that the exponential term has a base of 6.

1. Rewrite 36 as 6 raised to a power.

f(x) = 363x Original function

f(x) = (62)3x Rewrite 36 as 6 raised to the power 2.

2. Rewrite the exponential term so that 6 is raised to a power that is the product of two terms.

According to the Power of a Power Property, b bm n mn( ) = . Use this relationship to rewrite the function so that 6 is raised to a power that is the product of 2 terms.

f(x) = (62)3x Rewritten function from the previous step

f(x) = 62 • 3x Apply the Power of a Power Property.

3. Simplify the exponents being multiplied.

f(x) = 62 • 3x Rewritten function from the previous step

f(x) = 66x Simplify.

When the exponential term 363x in the function f(x) = 363x is rewritten with a base of 6, the result is f(x) = 66x.

Guided Practice 4.1.1


Example 2

For the function g xx

( ) 1.212= , by what percentage does g(x) change as x increases from 0 to 1? Does the function model exponential growth or exponential decay?

1. Rewrite the exponential term of the function.

The exponential term, x

1.212 , has a fractional exponent with a

variable in the numerator. To make this exponent easier to work with,

rewrite the exponential term as a base raised to the 1

2 power.

g xx

( ) 1.212= Original function

g xx

( ) 1.211

2=

Rewrite so that the base is raised to the 1

2 power.

2. Simplify the term being raised to a fractional power.

Recall that raising a quantity to the 1

2 power is the same as taking the

square root of that quantity. This is part of the Rational Exponents

Property, which states that a an n

1

= , where n ≠ 0. Use a calculator to

determine the square root.

g xx

( ) 1.211

2=

Rewritten function from the previous step

g xx

( ) 1.21= Apply the Rational Exponents Property.

g(x) = 1.1x Calculate the square root.


4.1.1

3. Determine the function’s value for the given values of x.

Since we must determine by what percentage g(x) changes as x increases from 0 to 1, solve the rewritten function for the values of g(x) when x = 0 and when x = 1. Then, compare the results.

Recall that the Zero Exponent Property states that any quantity raised to a power of 0 is equal to 1 (that is, a0 = 1).

g(x) = 1.1x Rewritten function from the previous step

g(0) = 1.1(0) Substitute 0 for x.

g(0) = 1 Apply the Zero Exponent Property.

When x = 0, g(x) = 1.

g(x) = 1.1x Rewritten function from the previous step

g(1) = 1.1(1) Substitute 1 for x.

g(1) = 1.1 Simplify.

When x = 1, g(x) = 1.1.


4. Find the percent rate of change and determine whether the original function shows exponential growth or exponential decay.

In order to determine whether or not the function models exponential growth or exponential decay, we must find the percent rate of change. If this percent is positive, then the function shows exponential growth; if the percent is negative, then the function shows exponential decay.

Substitute the known values into the formula for the rate of change, and then convert the result to a percentage.

Let the first set of values, (0, 1), be (x1, y1). Let the second set of values, (1, 1.1), be (x2, y2).

rate of change 2 1

2 1

y

x

y y

x x

�

�= =

−−

Rate of change formula

rate of change1.1 1

1 0

( ) ( )( ) ( )=

−−

Substitute 0 for x1, 1 for y1, 1 for x2, and 1.1 for y2.

rate of change0.1

1= Simplify.

rate of change = 0.1 Divide.

Now, multiply by 100 to convert the rate of change to a percentage.

0.1 • 100 = 10%

The percent rate of change is positive; therefore, the function

g xx

( ) 1.212= models exponential growth.


4.1.1

Example 3

Meg has a $5,000 student loan. For the next x years she is paying only the interest,

which accumulates at a rate of 8% per year after monthly compounding. The function

( ) 5000 1.081

=

f x n

nx

can be used to find the approximate monthly interest rate of

Meg’s loan, where n is the number of payments made each year (12). What is the

approximate monthly interest rate?

1. Evaluate the function for n = 12.

This will yield a function that accounts for monthly interest.

( ) 5000 1.081

=

f x n

nx

Original function

=

( )( )

f x

x

( ) 5000 1.081

12

12

Substitute 12 for n.

f(x) ≈ 5000(1.0064)12x Use a calculator to simplify 1.081

12 .

2. Subtract 1 from the number inside the parentheses and convert the difference to a percent.

The number in parentheses is the growth factor of the monthly interest on the original loan amount. To determine the growth rate, subtract 1 from this amount.

1.0064 – 1 = 0.0064

Multiply the result by 100 to convert the interest rate to a percentage.

0.0064 • 100 = 0.64%

The approximate monthly interest rate of the loan is 0.64%.


Example 4

Eton originally purchased his car for $16,000, but since then the car has been

depreciating, or losing value, at a rate of 12% per year. The exponential function

( ) 16,000 0.881

v x n

nx

=

can be used to determine the value of Eton’s car at the end

of each month, where n is the number of times per year that the car’s value is being

determined. In this case, n is 12. What is the car’s approximate monthly depreciation

rate?

1. Evaluate the function for n = 12.

This will yield a function that describes the car’s monthly depreciation.

( ) 16,000 0.881

v x n

nx

=

Original function

v x

x

=

( )( )

( ) 16,000 0.881

12

12

Substitute 12 for n.

v(x) ≈ 16,000(0.9894)12 Use a calculator to simplify 0.881

12 .

2. Subtract the number inside the parentheses from 1 and convert the difference to a percent.

The number in parentheses is the decay factor of the car’s value each month. To determine the decay rate, subtract this amount from 1. Notice that we are subtracting from 1 here since this function represents decay.

1 – 0.9894 = 0.0106


0.0106 • 100 = 1.06%

The car’s approximate monthly depreciation rate is 1.06%.


4.1.1

Example 5

Naoki opened a savings account at her bank with an initial deposit of $500. She has not made any deposits or withdrawals since. Her annual interest rate is 3.62%, and her account earns continuously compounded interest. The amount of money in Naoki’s account can be modeled by the natural exponential function g(x) = 500e0.0362x, where x is the number of years since she made her initial deposit. This amount of money includes the $500 originally deposited plus interest earned. What is the approximate annual interest rate that Naoki earns once compounding is taken into consideration?

1. Rewrite the exponential term.

g(x) = 500e0.0362x Original function

g(x) = 500(e0.0362)x Apply the Power of a Power Property.

2. Raise the base to the power inside the parentheses.

Recall that the value of e is approximately 2.71828.

g(x) = 500(e0.0362)x Rewritten function from the previous step

g(x) ≈ 500(1.0369)xUse a calculator to raise the base to the 0.0362 power.

3. Subtract 1 from the number inside the parentheses and convert the difference to a percent.

The number in parentheses is the growth factor of the initial deposit for each year, x. To determine the growth rate, subtract 1 from this amount.

1.0369 – 1 = 0.0369


0.0369 • 100 = 3.69%

Naoki’s saving account earns interest at a rate of approximately 3.69% annually once compounding is taken into consideration.

U4-14

PRACTICE

UNIT 4 • EXPONENTIAL AND LOGARITHMIC FUNCTIONSLesson 1: Exponential Functions

Unit 4: Exponential and Logarithmic Functions 4.1.1

For problems 1 and 2, use each given function to answer the questions.

1. How can the function f xx

( ) 25 4= be rewritten so that the exponential term has a base of 5?

2. For the function g xx

( ) 3.612= , by what percentage does g(x) change as x increases from 0 to 1? Does this function model exponential growth or exponential decay?

Use the given information to complete problems 3–10. Round your answers to the nearest hundredth of a percent.

3. The number of subscribers for a particular video-on-demand service has been increasing exponentially at a rate of 40% per year. The annual number of subscribers can be modeled by the function v(x) = 150,000(1.4x), where x is the number of years since the service was first offered. By approximately what percentage is the number of subscribers increasing per month?

4. Jason ran out of vitamin tablets and forgot to buy a new bottle for a while. After Jason took his last 250-milligram vitamin tablet, the number of milligrams of the vitamin remaining in his bloodstream decreased by 92% per day, and can be modeled by the exponential function m(x) = 250(0.08x), where x is the number of days since he took the tablet. By approximately what percentage does the amount of the vitamin in Jason’s bloodstream decrease per hour?

5. The number of cells in a Petri dish has been increasing exponentially at a rate of 12% per hour since the beginning of an experiment and can be modeled by the function c(x) = 4000(1.12x), where x is the number of hours into the experiment. By approximately what percentage is the number of cells increasing per minute?

Practice 4.1.1: Rewriting Exponential Functions

continued

U4-15

PRACTICE


Lesson 1: Exponential Functions 4.1.1

6. Since the year 2000, the number of people using film cameras has been decreasing by 35% per year. The exponential function f(x) = 620,000(0.65x) can be used to model the number of film camera users, with x being the number of years since 2000. By approximately what percentage is the number of film camera users decreasing per quarter?

7. Nadima opened a savings account at a bank, initially depositing $445. She has not made any deposits or withdrawals since. Her annual interest rate is 2.95%, and her account earns continuously compounded interest at this rate. The amount of money in Nadima’s account can be modeled by the natural exponential function g(x) = 445e0.0295x, for continuous compounding, where x is the number of years since she made her initial deposit. What is the approximate annual interest rate that Nadima earns once compounding is taken into consideration?

8. A certain state’s population has been growing exponentially at a rate of 4% per year since 1980. The state’s population can be modeled by the function p(x) = 2,000,000(1.04x), where x is the number of years since 1980. By approximately what percentage is the state’s population increasing per decade?

9. Charlotte’s leather sofa, which she originally purchased for $2,325, is losing value at a rate of 18% per year. The exponential function z(x) = 2325(0.82x) can be used to model how much the sofa is worth x years after purchase. Approximately what percentage of its value is Charlotte’s sofa losing per month?

10. The amount of carbon-14 in micrograms that remains in a wooly mammoth fossil since the mammoth died has been decreasing exponentially at a rate of 1.2% every century and can be modeled by the function m(x) = 20(0.988x), where x is the number of centuries since the mammoth’s death. By approximately what percentage has the amount of carbon-14 in micrograms been decreasing every 10,000 years?


Lesson 4.1.2: Properties of Exponential FunctionsIntroduction

An exponential function has various key features, or characteristics. Two of these are the domain, which is the set of all input values (x-values) that satisfy the given function without restriction, and the range, the set of all outputs (y-values) that are valid for the function. Other key features include the y-intercept(s), asymptote(s), and end behavior. The y-intercept is the point at which the graph of a function crosses the y-axis, written (0, y). The asymptote is an equation that represents sets of points that are not allowed by the conditions in a parent function or model; shown on a graph, the asymptote is a line that a function gets closer and closer to, but never touches. The end behavior is the behavior of the graph as x approaches positive or negative infinity, and can be described as whether the function’s graph increases or decreases within its domain. These key features can be determined by analyzing the values of a, b, and c in the general form of the equation of the function, or by viewing a graph of the function.

Key Concepts

• Recall that the general form of an exponential function is f(x) = a(bx) + c, where a, b, and c are constants and b is greater than 0 but not equal to 1.

• If a = 1 and c = 0, f(x) = a(bx) + c can be rewritten as f(x) = bx.

• The exponential function f(x) = bx has a domain of all real numbers, a range of all real numbers greater than 0, a y-intercept of 1, and an asymptote that is the x-axis.

• In general, if a is positive, a function of the form f(x) = a(bx) + c has a domain of all real numbers, a range of all real numbers greater than c, a y-intercept of a + c, and an asymptote of y = c.

• If b > 1, the function increases within its domain, and if 0 < b < 1, the function decreases within its domain. In other words, the function shows exponential growth if b > 1 and exponential decay if 0 < b < 1.

• The domain and range of an exponential function may be restricted if the function is used to model a real-world scenario.

• The graph of an exponential function can be obtained by using a graphing calculator.


4.1.2

On a TI-83/84:

Step 1: Press [MODE].

Step 2: Make sure Func is highlighted on the fourth line. If not, arrow down to it and press [ENTER].

Step 3: Press [Y=]. Press [CLEAR] to delete any equations.

Step 4: Enter the equation of the exponential function at Y1. Use [X, T, θ, n] for variables and [^] for exponents.

Step 5: If needed, press [WINDOW] to adjust the viewing window. Change the settings as appropriate.

Step 6: Press [GRAPH].

On a TI-Nspire:

Step 1: Press [home]. Arrow over to the graphing icon and press [enter].

Step 2: Press [menu]. Use the arrow key to select 3: Graph Type and then 1: Function. Press [enter].

Step 3: Use your keypad to enter the equation of the exponential function after f1(x).

Step 4: If needed, adjust the viewing window. Press [menu], then select 4: Window/Zoom, and then 1: Window Settings. Change the settings as appropriate.

Step 5: Press [enter].


Example 1

What are the domain, range, y-intercept, asymptote, and end behavior of the exponential function f(x) = 7x? Confirm your results by graphing.

1. Determine the values of a, b, and c in the function.

The function is in the form f(x) = a(bx) + c, so a = 1, b = 7, and c = 0.

2. Use the values of a and c to find the domain, range, y-intercept, and asymptote of the function.

From the previous step, it is known that a = 1 and c = 0.

Recall that if a is positive, a function of the form f(x) = a(bx) + c has a domain of all real numbers, a range of all real numbers greater than c, a y-intercept of a + c, and an asymptote of y = c.

Since a is positive, the domain is all real numbers.

The range is all real numbers greater than c, which is 0.

The y-intercept is a + c, or 1 + 0, which simplifies to 1.

The asymptote is y = c, or y = 0.

Therefore, the domain of f(x) = 7x is all real numbers, the range is all real numbers greater than 0, the y-intercept is 1, and the asymptote is the x-axis (y = 0).

3. Use the value of b to determine the function’s end behavior.

The value of b in the function f(x) = 7x is 7, and 7 > 1. When any number greater than 1 is raised to a larger and larger power, the resulting value also becomes larger and larger. Therefore, the function increases within its domain.

Guided Practice 4.1.2


4.1.2

4. Graph the function and use it to confirm your findings.

Graph the function f(x) = 7x using the directions appropriate to your calculator model.

On a TI-83/84:Step 1: Press [MODE].Step 2: Make sure Func is highlighted on the fourth line. If not,

arrow down to it and press [ENTER].Step 3: Press [Y=]. Press [CLEAR] to delete any equations.Step 4: Enter the equation of the exponential function at Y1. Use

[X, T, θ, n] for variables and [^] for exponents.Step 5: If needed, press [WINDOW] to adjust the viewing window.

Change the settings as appropriate. Step 6: Press [GRAPH].

On a TI-Nspire:Step 1: Press [home]. Arrow over to the graphing icon and press [enter].Step 2: Press [menu]. Use the arrow key to select 3: Graph Type

and then 1: Function. Press [enter].Step 3: Use your keypad to enter the equation of the exponential

function after f1(x). Step 4: If needed, adjust the viewing window. Press [menu], then

select 4: Window/Zoom, and then 1: Window Settings. Change the settings as appropriate.

Step 5: Press [enter].

(continued)


– 10 – 8 – 6 – 4 – 2 2 4

10

9

8

7

6

5

4

3

2

0

1

– 1

– 2

y

x

Compare the key features of the graph (domain, range, y-intercept, asymptote, and end behavior) with your findings.

The graph extends infinitely in both directions, so the domain is all real numbers.

The graph includes only y-values greater than 0, which means the range is all real numbers greater than 0.

The graph intersects the y-axis at (0, 1), so the y-intercept is 1.

The graph approaches but never touches the x-axis; therefore, the asymptote is the x-axis (y = 0).

This graph shows the function values rising from left to right, so the function is increasing within its domain.

Thus, the graph confirms the findings from the previous steps.


4.1.2

Example 2

What are the domain, range, y-intercept, asymptote, and end behavior of the

exponential function g xx

( ) 61

85=

+ ? Confirm your results by graphing.

1. Determine the values of a, b, and c in the function.

The function is in the form f(x) = a(bx) + c, so a = 6, b = 1

8, and c = 5.

2. Use the values of a and c to find the domain, range, y-intercept, and asymptote of the function.

From the previous step, it is known that a = 6 and c = 5.

Recall that if a is positive, a function of the form f(x) = a(bx) + c has a domain of all real numbers, a range of all real numbers greater than c, a y-intercept of a + c, and an asymptote of y = c.

Since a is positive, the function follows this rule. Therefore, the domain is all real numbers, the range is all real numbers greater than 5, the y-intercept is 6 + 5 = 11, and the asymptote is y = 5.

3. Use the value of b to determine the function’s end behavior.

The value of b is 1

8, and 0

1

81< < . When any number greater than

0 and less than 1 is raised to a larger and larger power, the resulting

values become smaller and smaller. Therefore, the function decreases

within its domain.


4. Graph the function and use it to confirm your findings.

Graph the function g xx

( ) 61

85=

+ on a graphing calculator.

–24 –16 –12–20 –8 –4 0 84 16 2412 20

–24

–18

–20

–12

–8

–4

8

4

16

20

12

24

=

+g x( ) 6

1

85

x

y

x

The graph confirms that the domain is all real numbers, the range is all real numbers greater than 5, the y-intercept is 11, and the asymptote is y = 5.


4.1.2

Example 3

Examine the following graph of an exponential function to determine the function’s domain, range, y-intercept, asymptote, and end behavior.

–4 –3 –2 –1 0 1 2 3 4

1

2

3

4

5

6

7

8

9

10

1. Analyze the values of x and y in the graph to determine the domain and range of the function.

The graph shows that the values of x extend infinitely in both the positive and negative directions, so the domain is all real numbers. The values of y extend infinitely in the positive direction, but they never reach 0, so the range is all real numbers greater than 0.

2. Determine where the graph crosses the y-axis to find the y-intercept of the function.

The graph crosses the y-axis at the point (0, 1), so the y-intercept is 1.


3. Determine the line that the graph approaches to find the function’s asymptote.

The graph approaches the line y = 0, or the x-axis; this is the asymptote.

4. Determine the function’s end behavior.

Since we do not know the exact value of b in the function, look at the graph to see whether the curve of the function is rising or falling from left to right. If the curve is rising, then the function increases within its domain. If the curve is falling, then the function decreases within its domain.

This graph shows the curve of the function falling from left to right. Therefore, the function decreases within its domain.

Example 4

Examine the following graph of an exponential function to determine the function’s domain, range, y-intercept, asymptote, and end behavior.

–4 –3 –2 –1 0 1 2 3 4

1

2

3

4

5

6

7

8

9

10


4.1.2

1. Analyze the values of x and y in the graph to determine the domain and range of the function.

The graph shows that the values of x extend infinitely in both the positive and negative directions, so the domain is all real numbers. The values of y extend infinitely in the positive direction, but they never reach 1, so the range is all real numbers greater than 1.


The graph crosses the y-axis at the point (0, 3), so the y-intercept is 3.


The graph approaches the line y = 1, so the line y = 1 is the asymptote.


Since we do not know the exact value of b in the function, look at the graph to see if it is rising or falling from left to right.

This graph shows the function rising from left to right. Therefore, the function increases within its domain.


Example 5

Juan scraped his knee while skateboarding and decided to graph the wound’s healing time as part of his science fair project. The area of his wound, measured in square millimeters, as a function of the amount of healing time in days can be modeled by the exponential function graphed as shown. What are the domain, range, y-intercept, asymptote, and end behavior of the function?

0 2 4 6 8 10 12 14 16 18

2

4

6

8

10

12

14

16

18

Are

a of

wou

nd (m

m2 )

Healing time (days)

1. Analyze the values of x and y in the graph to find the domain and range of the function.

The graph shows that the values of x start at 0, since the wound is created at 0 healing days, and extend infinitely in the positive direction, so the domain is all real numbers x such that x ≥ 0. The values of y start at 16, but they never reach 0, so the range is all real numbers y such that 0 < y ≤ 16. (Note that the exponential function is just a model; in the real world, y would reach 0 because the wound would completely heal.)


4.1.2


The graph touches the y-axis at the point (0, 16), so the y-intercept is 16.


The graph approaches the line y = 0, or the x-axis, so the x-axis is the asymptote.


Since we do not know the exact value of b in the function, look at the graph to see if it is rising or falling from left to right.

This graph shows the function falling from left to right. Therefore, the function decreases within its domain.

U4-28

PRACTICE



For problems 1–3, find the domain, range, y-intercept, and asymptote for the given exponential function. State whether the function increases or decreases within its domain.

1. a xx

( ) 61

914=

+

2. b(x) = 10(3)x + 7

3. c(x) = 9(0.71)x + 1

For problems 4–7, use the given information to determine the equation for the described exponential function.

4. The function’s base is 0.26, its graph crosses the y-axis at (0, 16), and it has an asymptote of y = 12.

5. The function’s base is 5, its graph crosses the y-axis at (0, 26), and it has an asymptote of y = 13.

6. The function’s base is 1

15, its graph crosses the y-axis at (0, 7), and it has an

asymptote of y = 2.

7. The function’s base is 18, its graph crosses the y-axis at (0, 17), and it has an asymptote of y = 6.

Practice 4.1.2: Properties of Exponential Functions

continued

U4-29

PRACTICE


Lesson 1: Exponential Functions 4.1.2

For problems 8–10, use the given information to find the domain, range, y-intercept, and asymptote of the given function. State whether the function increases or decreases within its domain.

8. Alexa drank a cup of tea containing 60 milligrams of caffeine. The number of milligrams of caffeine in her body x hours after drinking the tea can be modeled by the function h(x) = 60(0.87)x.

9. The temperature in degrees Fahrenheit of some apples that were placed in a refrigerator can be modeled by the function j(x) = 28(0.92)x + 40, where x is the number of minutes the apples have been in the refrigerator.

10. The value in dollars of a rare baseball card since it was purchased at a card show can be modeled by the function k(x) = 475(1.22)x, where x is the number of years since the card was purchased.

Lesson 2: Introducing Logarithmic Functions

UNIT 4 • EXPONENTIAL AND LOGARITHMIC FUNCTIONS



MCC9–12.F.IF.7e★

MCC9–12.F.IF.8b

MCC9–12.F.BF.5 (+)

Essential Questions

1. What is a logarithm?

2. How can a logarithm be written in exponential form and vice versa?

3. What is the difference between a common logarithm and a natural logarithm?

4. How does the graph of a logarithmic function change when the base changes?

5. How can logarithms that are not common be calculated on a calculator?

6. How are the properties of logarithms similar to those for exponents?

WORDS TO KNOW

argument the result of raising the base of a logarithm to the power of the logarithm, so that b is the argument of the logarithm loga b = c

asymptote an equation that represents sets of points that are not allowed by the conditions in a parent function or model; a line that a function gets closer and closer to, but never crosses or touches

change of base formula a formula that can be used to rewrite a logarithm so

that it has a base of 10: for logb x, where b is not equal

to 10, logb x is equal to x

b

log

log

common logarithm a base-10 logarithm which is usually written without the number 10, such as log x = log10 x

U4-31Lesson 2: Introducing Logarithmic Functions

4.2

compound interest formula

a formula used to calculate the balance on a loan or

investment for which interest earned on the principal

over time is added to the principal and also earns

interest: A Pr

n

nt

1= +

, where A is the ending

amount, P is the principal or initial amount, r is the

annual interest rate expressed as a decimal, n is the

number of times per year the interest is compounded,

and t is the time in years

inverse operation the operation that reverses the effect of another operation

logarithm a quantity that represents the power to which a base b must be raised in order to equal a quantity x; written logb x

logarithmic equation an equation involving logarithms. Given an exponential equation of the form x = by, the logarithmic equation is y = logb x, where y is the exponent, b is the base, and x is the argument.

logarithmic function the inverse of an exponential function; for the exponential function g(x) = 5x, the inverse logarithmic function is x = log5 g(x)

natural logarithm a logarithm whose base is the irrational number e; usually written in the form “ln,” which means “loge”



• Khan Academy. “Logarithm Properties.”


This series of videos demonstrates how to use logarithm properties. The examples are well explained and are accompanied by transcripts and answered questions from users of the site.

• MathIsFun.com. “Working with Exponents and Logarithms.”


This online lesson defines an exponent and a logarithm and gives examples. It also explains the inverse relationship between logarithms and exponents.

• Purplemath.com. “Logarithms: Introduction to ‘The Relationship’.”


This site defines logarithms and their relationship to exponents. There are examples with answers explained in addition to an animation of the inverse relationship of logarithms to exponents.


4.2.1

IntroductionExponential functions are often used to project population growth. In such cases, determining the number of years it would take for a population to reach a specific value requires solving the exponential equation for the exponent. We would need to find an operation to “undo” the exponent, or an operation that would yield the inverse of the exponent. An inverse operation is the operation that reverses the effect of another operation. Logarithms are numbers that allow you to reverse the exponent operation, thereby isolating the exponent. Specifically, for an exponential equation of the form x = by, where x is any quantity that is equal to a base b raised to the power y, the equivalent logarithmic equation is of the form logb x = y.

Key Concepts

• A logarithm is a quantity that represents the power to which a base b must be raised in order to equal a quantity x. Logarithms are written in the form logb x.

• For example, log10 100 = 2 because 10 raised to a power of 2 is 100. Notice the relationship between the logarithmic form and the exponential form. Substituting 100 for x, 10 for b, and 2 for y into the exponential form x = by yields 100 = 102; substituting these same values into the logarithmic form logb x = y results in log10 100 = 2.

• Similar to the inverse operations of addition and subtraction, a logarithm can be thought of as the inverse operation of an exponent.

• Common logarithms are logarithms with a base of 10. Common logarithms are usually written without the number 10, since it is understood to be included; that is, the general form log10 y would usually be written as log y. For example, x = log 15 is equivalent to x = log10 15; therefore, in either case, 10x = 15.

• Common logarithms can be easier to work with because they always deal with powers of 10.

• In a common logarithm of the form log x, the value x is the argument of the logarithm. The argument is the result of raising the base of a logarithm to the power of the logarithm, so that b is the argument of the logarithm logb x = y.

• A logarithmic equation is an equation involving logarithms. The general form of a logarithmic equation is y = logb x, where y is the exponent, b is the base, and x is the argument.

Lesson 4.2.1: Defining Logarithms


• Recall that natural logarithms are logarithms that have a base of e, an irrational number approximately equal to 2.71828. The abbreviation “ln” is used to denote natural logarithms. Therefore, “ln b” is read as “the natural logarithm of b.”

• Similarly, a natural exponential function is an exponential function with a base of e.

• Use the natural logarithm for solving problems with a base of e. Use logarithms for problems involving bases other than e.

• Logarithms that have bases other than 10 can be rewritten as common logarithms by using the change of base formula. This may be necessary if you have a calculator that only works with common logarithms.

• The change of base formula states that given logb x, where the base b is not

equal to 10, logb x is equal to x

b

log

log. For example, log2 10 is equal to

log 10

log 2.

This can be further reduced: log 10 is equal to 1 because 101 = 10 and log 2 is

approximately equal to 0.3 because 100.3 ≈ 2. Therefore, log2 10 = 1

0.3, which is

approximately equal to 3.3.

• Logarithms and exponents are frequently used in formulas in the banking

industry. One common formula is the compound interest formula, which is

used to calculate the balance on a loan or investment for which interest earned

on the principal over time is added to the principal and also earns interest. The

formula is A Pr

n

nt

1= +

, where A is the balance, P is the principal or initial

amount, r is the annual interest rate expressed as a decimal (that is, divided by

100), n is the number of times per year the interest is compounded, and t is

the time in years. (Note: n = 1 means interest is compounded once per year or

annually, n = 2 means it is compounded twice per year or semiannually, n = 4

means it is compounded four times per year or quarterly, and so on.)


4.2.1

• The following table can be useful when working with compound interest.

Compounded… n (number of times per year)Yearly/annually 1

Semiannually 2Quarterly 4Monthly 12Weekly 52Daily 365

• The compound interest formula is an exponential function because it involves exponents (n and t). Since it represents an exponential function, the compound interest formula could also be written as a logarithmic function.


Guided Practice 4.2.1Example 1

Rewrite the given exponential equations as logarithmic equations.

102 = 100

3t = 81

1. Write the general form of a logarithmic equation and assign variables.

The general form of a logarithmic equation is y = logb x, where b is the base and x is the argument.

2. Rewrite the equation 102 = 100 in the form y = logb x.

Define each part of the equation in terms of the base, the exponent, and the argument.

In the equation 102 = 100, the base is 10, the exponent is 2, and the argument is 100.

Therefore, 102 = 100 written as a logarithmic equation is 2 = log 100.

3. Rewrite the equation 3t = 81 in the form y = logb x.

Define each part of the equation in terms of the base, the exponent, and the argument.

In the equation 3t = 81, the base is 3, the exponent is t, and the argument is 81.

Therefore, 3t = 81 written as a logarithmic equation is t = log3 81.


4.2.1

Example 2

Given the logarithmic equation log4 64 = x, write the equivalent exponential equation.

1. Define each part of the given equation in terms of the base, the exponent, and the quantity that results when the base is raised to the exponent.

In the equation log4 64 = x, the base is 4, the exponent is x, and the number is 64.

2. Rewrite the logarithmic equation in the form y = bx.

The general form of an exponential equation is y = bx, where x is the exponent, b is the base, and y is the result of b raised to the power of x. Substitute each of the identified terms into the general form.

Thus, log4 64 = x written as an exponential equation is 4x = 64.

Example 3

Solve the logarithmic equation 2 = log5 (3a + 1) for a by using the definition of logarithms.

1. Determine values for b, x, and y from the logarithmic equation.

The general form of a logarithmic equation is y = logb x, where b is the base, x is the argument, and y is the logarithm of x to the base b.

In the function 2 = log5 (3a + 1), b = 5, x = 3a + 1, and y = 2.

2. Write the exponential equation in the form of by = x from the values in the original problem.

The logarithmic equation y = logb x is equivalent to the exponential equation by = x. Therefore, substitute the determined values for b, x, and y into the exponential form.

The logarithmic function 2 = log5 (3a + 1) is equivalent to the exponential equation 52 = 3a + 1.


3. Solve the exponential equation for a.

The exponential equation 52 = 3a + 1 can be solved for a using algebraic techniques.

52 = 3a + 1 Exponential equation

25 = 3a + 1 Square 5.

24 = 3a Subtract 1 from both sides.

8 = a Divide both sides by 3.

In the logarithmic equation 2 = log5 (3a + 1), a = 8.

Example 4

Solve the natural logarithm y = 4 ln 1 for y.

1. Rewrite the natural logarithm in exponential form.

The natural logarithm in exponential form has a base e raised to the power of x. Therefore, ln 1 can be rewritten as ex = 1.

2. Determine the value of x that would make ex = 1.

If ex = 1, then x must be 0 because the value of any number raised to the power of 0 is 1. Therefore, ln 1 = 0.

3. Solve the natural logarithm.

Let ln 1 = 0; substitute this value into the natural logarithm and solve.

y = 4 ln 1 Given natural logarithm

y = 4(0) Substitute 0 for ln 1.

y = 0 Simplify.

In the natural logarithm y = 4 ln 1, y = 0.


4.2.1

Example 5

Santiago has $1,500 that he would like to invest by opening a savings certificate of

deposit (CD). He compared the interest rates of the CDs he found at a few online

banks and two of them have caught his attention. The first CD option compounds

annually at 3.5%, and the second CD option compounds quarterly at 1.5%. Use the

compound interest formula, A Pr

n

nt

1= +

, to write a logarithmic equation for each

CD that Santiago can use to determine the number of years it would take for each CD

to yield a specific amount of money, A.

1. Write the specific compound interest formula for the first CD option.

Recall that in the compound interest formula, A Pr

n

nt

1= +

, A is

the balance, P is the initial amount, r is the interest rate expressed as

a decimal, n is the number of compounding periods per year, and t is

the number of years.

Determine the values for P, n, and r for the first savings CD.

It is given that the initial amount being deposited is $1,500, so P = 1500.

This CD compounds interest annually or once per year, so n = 1.

The rate is 3.5%. To express this rate as a decimal, divide it by 100: 3.5% ÷ 100 = 0.035. Thus, r = 0.035.

Substitute these values for P, n, and r into the compound interest formula and simplify.

A Pr

n

nt

1= +

Compound interest formula

At

(1500) 1(0.035)

(1)

(1)

= +

Substitute 1,500 for P, 0.035 for r, and 1 for n.

A = 1500(1 + 0.035)t Simplify the exponent and the fraction.

A = 1500(1.035)tSimplify the expression within the parentheses.

The compound interest formula for the first CD is A = 1500(1.035)t.


2. Write the specific compound interest formula for the second CD option.

Since the initial amount being deposited is unchanged, P = 1500.

The second CD compounds interest quarterly or 4 times per year, so n = 4.

The rate is 1.5%. To express this rate as a decimal, divide it by 100: 1.5% ÷ 100 = 0.015. Thus, r = 0.015.

Substitute these values for P, n, and r into the compound interest formula and simplify.

A Pr

n

nt

1= +


At

(1500) 1(0.015)

(4)

( 4)

= +

Substitute 1500 for P, 0.015 for r, and 4 for n.

A = 1500(1 + 0.00375)4t Simplify the exponent and the fraction.

A = 1500(1.00375)4t Simplify the expression within the parentheses.

The compound interest formula for the second CD is A = 1500(1.00375)4t.


4.2.1

3. Rewrite each compound interest formula as a logarithmic equation.

Use the formulas found in the previous steps to write the logarithmic equations.

For the first CD, A = 1500(1.035)t.

A = 1500(1.035)t Compound interest formula

At

15001.035= Divide both sides by 1500.

tA

log15001.035= Rewrite using the definition of logarithms.

The logarithmic function for the first CD is tA

log15001.035= .

For the second CD, A = 1500(1.00375)4t.

A = 1500(1.00375)4t Compound interest formula

At

15001.003754= Divide both sides by 1500.

tA1

4log

15001.00375= Rewrite using the definition of logarithms.

The logarithmic function for the second CD is

tA1

4log

15001.00375= .

U4-42

PRACTICE

UNIT 4 • EXPONENTIAL AND LOGARITHMIC FUNCTIONSLesson 2: Introducing Logarithmic Functions


Use what you have learned about logarithmic functions and equations to complete each problem.

1. Is log x equivalent to –log x? Use an example to support your answer.

2. Compare log x and log (x + 1). Is the value of log x always greater than or less than that of log (x + 1)? Explain your answer.

3. How would you write x

ylog1

= as an exponential equation?

4. Write a logarithmic function that models the values found in the following table.

s t0 11 22 43 8

5. Rewrite y 25= as a logarithmic equation.

6. Evaluate log10 x for x1

100= .

7. Explain why logb 1 always equals 0 regardless of the value of b.

8. Use a calculator to evaluate log 0.25 to the nearest hundredth.

9. Rewrite A = 3000(1.05)2t as a logarithmic equation.

10. What is the logarithmic equation for x1

327

= ?

Practice 4.2.1: Defining Logarithms


4.2.2

IntroductionExponential and logarithmic functions are commonly used to model real-life problems. They are particularly useful with situations in which the quantities being observed increase or decrease at a fast rate. A logarithmic function is the inverse of an exponential function; for the exponential function g(x) = 5x, the inverse logarithmic function is x = log5 g(x). By evaluating logarithms, patterns that lead to solutions can be observed. Graphing logarithmic functions can generally give you a more efficient way to see patterns in the variables. In this lesson, you will learn how to graph logarithmic functions.

Key Concepts

• The graph of an exponential function in the form f(x) = bx appears as a slight curve when b > 1, such as in the graph of f(x) = 2x that follows.

–5 –4 –3 –2 –1 1 2 3 4 5

x

5

4

3

2

1

0

–1

–2

–3

–4

–5

y

f(x) = 2x

Lesson 4.2.2: Graphs of Logarithmic Functions


• When b < 1, the graph resembles the graph of the positive value of b, but is reflected about the x-axis. For example, observe the graph of f(x) = –2x, shown as follows. Notice that it is the reflection of the graph of f(x) = 2x. (Note: When b = 1, the graph is a horizontal line.)

–5 –4 –3 –2 –1 1 2 3 4 5

x

5

4

3

2

1

0

–1

–2

–3

–4

–5

y

f(x) = –2x

• The graph of an exponential function will approach an asymptote (a line that the graph of the function comes closer and closer to but never actually touches).

• In the graph of an exponential function, as the x-values increase, the y-values will either increase or decrease very rapidly.

• Logarithmic functions are inverses of exponential functions. Therefore, their graphs show similar behavior.

• The typical graph of a logarithmic function, f(x) = log x, is the mirror image of the exponential function. It has an asymptote as well.


4.2.2

• Furthermore, a logarithmic function also has a graph that shows a rapid increase or decrease in its y-values. For example, the graph that follows shows f(x) = log2 x.

–5 –4 –3 –2 –1 1 2 3 4 5

x

5

4

3

2

1

0

–1

–2

–3

–4

–5

y

f(x) = log2 x

Asymptote: x = 0

• Notice that this graph has an asymptote at x = 0. For this graph, the domain (the set of x-values) is all real numbers greater than 0, while the range (the set of all y-values) is all real numbers.



How do the graphs of f(x) = log x, g(x) = log2 x, and h(x) = ln x compare? How are they different? Describe any key features such as asymptotes, domain, and range for each graph. How do the different bases (i.e., base 10, base 2, and base e) affect the shape of each graph?

1. Graph each of the functions f(x) = log x, g(x) = log2 x, and h(x) = ln x on the same coordinate plane.

Graph the given functions using a graphing calculator. The result should resemble the following graph.

–50 –40 –30 –20 –10 10 20 30 40 50

25

20

15

10

5

–5

–10

–15

–20

–25

0

y

x

g(x)h(x)f(x)

2. Compare the shapes and end behavior of the graphs.

All three graphs are similar in that they all have the same shape and end behavior. Each graph has a slight curve and extends toward the right side of the x-axis; in other words, each graph approaches positive infinity on the x-axis.

Also, each graph approaches the y-axis but does not touch or intersect it.

The three graphs are different in terms of their proximity to the x-axis. As the graphs get farther away from the y-axis, their proximity to the x-axis changes. For example, near x = 1, the graphs are fairly close together. However, near x = 20, the graphs become more divergent and easier to distinguish from one another.

Thus, all of the graphs increase as the x-values approach positive infinity.


4.2.2

3. Identify the asymptote(s), intercept(s), domain, and range for each graph.

All three graphs have a vertical asymptote at x = 0.

All of the graphs have an x-intercept at (1, 0).

Since the x-values increase into positive infinity but never cross the y-axis, this means they are never negative. Thus, we can determine the domain of each graph is x > 0.

Since the y-values are not limited, we can determine the range of each graph is all real numbers.

4. State how the different bases (i.e., base 10, base 2, and base e) affect the shape of each graph.

In this situation, the overall shape and end behavior of each graph is the same regardless of the base number. However, the larger the base number, the closer the graph is to the x-axis. For instance, the graph of f(x) = log x, which has a base of 10, is much closer to the x-axis than the graph of g(x) = log2 x, which has a base of 2.


Example 2

Three pairs of logarithmic functions are provided. Graph each pair on one coordinate plane. Then, use the graphs to identify and compare the asymptotes, intercepts, domains, and ranges for each pair, as well as the shape and end behavior of each pair of graphs.

• f(x) = –log x and g(x) = log x

• h(x) = –log2 x and j(x) = log2 x

• m(x) = –ln x and p(x) = ln x

1. Graph the first pair of functions on the same coordinate plane.

Graph the given functions, f(x) = –log x and g(x) = log x, by hand or using a graphing calculator. The result should resemble the following.

–2 –1 1 2 3 4 5 6 7 8 9 10

4

3

2

1

0

–1

–2

–3

–4

y

x(1, 0)

g(x) = log x

f(x) = – log x


The shape and end behavior of the graph of f(x) = –log x are inverted from the shape and end behavior of the graph of g(x) = log x. In other words, the two graphs are reflections of each other. The line of reflection is the x-axis.

The graph of f(x) = –log x decreases as x approaches positive infinity. The graph of g(x) = log x increases as x approaches positive infinity.


4.2.2

3. Identify the two graphs’ asymptotes, intercepts, domains, and ranges.

Since neither graph crosses the y-axis, both graphs have a vertical asymptote at x = 0.

The graphs cross the x-axis at the x-intercept (1, 0).

Since the x-values increase into positive infinity but never cross the y-axis, this means they are never negative. Thus, we can determine the domain of both graphs is x > 0.

The range of both graphs is all real numbers.

4. Graph the second pair of functions together on a new coordinate plane.

The graph of j(x) = log2 x and h(x) = –log2 x is shown.

–2 –1 1 2 3 4 5 6 7 8 9 10

4

3

2

1

0

–1

–2

–3

–4

y

x(1, 0)

j(x) = log2 x

h(x) = – log2 x

5. Compare the shapes and end behavior of the second pair of graphs.

The shape and end behavior of the graph of h(x) = –log2 x are inverted from that of j(x) = log2 x. In other words, each graph is a reflection of the other over the x-axis.

The graph of h(x) = –log2 x decreases as x approaches positive infinity.

The graph of j(x) = log2 x increases as x approaches positive infinity.


6. Identify the two graphs’ asymptotes, intercepts, domains, and ranges.

There is a vertical asymptote at x = 0 for each graph.

Both graphs cross the x-axis at the x-intercept (1, 0).

The domain of both graphs is x > 0.

Since the y-values are not limited, we can determine the range of each graph is all real numbers.

7. Graph the third pair of functions together on a new coordinate plane. Compare the graphs’ shapes and end behavior.

The graph of p(x) = ln x and m(x) = –ln x is shown.

–2 –1 1 2 3 4 5 6 7 8 9 10

4

3

2

1

0

–1

–2

–3

–4

y

x(1, 0)

p(x) = ln x

m(x) = – ln x

The shape and end behavior of the graph of m(x) = –ln x are inverted from that of p(x) = ln x; therefore, the graphs are reflections of one another. The line of reflection is the x-axis.

The graph of m(x) = –ln x decreases as x approaches positive infinity. The graph of p(x) = ln x increases as x approaches positive infinity.


4.2.2

Example 3

Graph the logarithmic functions c(x) = log2 (x + 1) and d(x) = log2 x together on a coordinate plane. Use the graph to identify and compare the asymptotes, intercepts, domains, and ranges of the functions, as well as their shape and end behavior.

1. Graph the functions together on a coordinate plane.

Graph c(x) = log2 (x + 1) and d(x) = log2 x either by hand or using a graphing calculator. The result should resemble the following graph.

–2 –1 10 2 3 4 5 6 7 8 9 10

4

3

2

1

–1

–2

–3

–4

d(x) = log2 x

c(x) = log2 (x + 1)

y

x

8. Use the graphs to identify the asymptotes, intercepts, domains, and ranges of this third pair of functions.

There is a vertical asymptote at x = 0.

The graphs cross the x-axis at the x-intercept (1, 0).

The domain of both graphs is x > 0.


9. Summarize your findings.

In summary, all three pairs of graphs have the same vertical asymptote of x = 0. In every case, the x-intercept is (1, 0). The domain and range for all three pairs are identical. The three pairs of functions differ in the rate of increase or decrease as x increases.



The graph of c(x) = log2 (x + 1) has the same shape and end behavior as d(x) = log2 x, but the graph of c(x) is shifted to the left 1 unit.

Both graphs increase as x approaches positive infinity.

3. Identify the graphs’ asymptotes, intercepts, domains, and ranges.

The graph of d(x) = log2 x has a vertical asymptote at x = 0, whereas the graph of c(x) = log2 (x + 1) has a vertical asymptote at x = –1.

The graph of d(x) = log2 x crosses the x-axis at the x-intercept (1, 0), whereas the graph of c(x) = log2 (x + 1) crosses the x-axis at (0, 0).

While the domain of d(x) = log2 x is x > 0, the domain of c(x) = log2 (x + 1) is x > –1.


U4-53

PRACTICE


Lesson 2: Introducing Logarithmic Functions 4.2.2

continued

Use what you have learned about the graphs of logarithmic functions to complete each problem.

1. Show by graphing that r(x) = log2 x and u(x) = –log2 x are reflections of each other.

2. What is the equation of the logarithmic function shown in the graph?

1 2 3 4 5 6 7 8 9 10

4

3

2

1

0

–1

–2

–3

–4

x

y

3. How does f(x) = log (x + 5) differ from g(x) = log x in terms of its asymptotes and intercepts?

4. What is the logarithmic equation of the growth rate in years of an investment that would begin with $1,000 and earn interest semiannually at a rate of 2%?

5. Describe the changes that happen to r(x) = log x to create the graph of v(x) = log (x + 4). Graph both functions to illustrate the changes.

Practice 4.2.2: Graphs of Logarithmic Functions

U4-54

PRACTICE



6. Graph f(x) = –ln x and then use the graph to evaluate f1

2

.

7. Give the domain of x-values in which the graph of f(x) = log (x + 1) is increasing and/or decreasing.

8. Show that the graph of f(x) = log3 x contains the point (3, 1).

9. Consider the graph of f(x) = log (x + 2). How would the domain and range change as the base decreases?

10. Does the function f(x) = ln x have a domain of all real numbers? Why or why not?


4.2.3

Introduction

Previously, you learned how to graph logarithmic equations with bases other than 10. It may be necessary to convert other bases to common logarithms if you are using a calculator that does not automatically do this. This process involves applying a property of logarithms in order to convert logarithms with different bases to common logarithms. Properties of logarithms allow you to solve problems containing logarithms by making the logarithms easier to work with. Applying logarithmic properties often allows for simpler and fewer calculations in a problem.

Key Concepts

• Recall that a logarithm is a quantity that represents the power to which a base b must be raised in order to equal a quantity x. It is typically written in the form logb x.

• Recall that when logarithms have bases other than 10, you can rewrite them as

common logarithms by using the change of base formula, xx

bbloglog

log= .

• Since logarithms are the inverses of exponents, they have similar properties. Like exponents, logarithms can be added, subtracted, multiplied, and divided.

• Performing operations with logarithms is the same as performing operations on exponents. For example, log (ab) = log a + log b is the same as multiplying two numbers raised to powers, as in xaxb = xa + b.

• The following table includes properties of logarithms and corresponding properties of exponents.

Exponent property Logarithm propertyLogarithm

property name

ax • ay = ax + y loga (x • y) = loga x + loga y Product Property of Logarithms

a

aa

x

yx y= −

x

yx ya a alog log log

= − Quotient Property of Logarithms

(ax)y = ax • y loga xy = y • loga x Power Property of Logarithms

Lesson 4.2.3: Properties of Logarithms



Let M = loga x and N = loga y. Use the properties of exponents to show that if loga x = loga y, then x = y. Verify this property mathematically.

1. Write the equivalent exponential equation for M.

As stated in the problem, M = loga x.

Use the definition of logarithms to rewrite the logarithmic equation as an exponential equation.

M = loga x is equivalent to aM = x.

2. Write the equivalent exponential equation for N.

As stated in the problem, N = loga y.

Use the definition of logarithms to rewrite the logarithmic equation as an exponential equation.

N = loga y is equivalent to aN = y.

3. Compare the exponential equations.

The exponential equations are aM = x and aN = y.

It follows that if M = N, then x = y since the exponential expressions have the same base, a, and are raised to equal exponents (M and N).


4.2.3

4. Verify your results using a specific example.

Show that if loga x = loga y, then x = y.

Let a = 10 and x = 10.

loga x = loga y Set the two logarithms equal.

log(10) (10) = log(10) y Substitute 10 for a and 10 for x.

1 = log(10) y Simplify.

Write the equivalent exponential equation.

101 = y Equivalent exponential equation

10 = y Simplify.

Since y = 10 and x = 10, y = x.

Example 2

Use the Product Property of Logarithms to show that loga (xy) = loga x + loga y. Verify this property mathematically.

1. Write the equivalent exponential equation for loga x.

To write the equivalent exponential equation, let M = loga x.

By the definition of logarithms, M = loga x is equivalent to ax = M.

2. Write the equivalent exponential equation for loga y.

To write the equivalent exponential equation, let N = loga y.

By the definition of logarithms, N = loga y is equivalent to ay = N.


3. Write the exponential form of the Product Property.

According to the Product Property, loga (xy) = M + N.

Therefore, the exponential form of the Product Property is a(M + N) = xy.

By the Product Property of Exponents, a(M + N) = (aM)(aN).

4. Substitute specific values for a, x, and y into the original equation to verify that the Product Property of Logarithms holds true for this problem.

The original equation was loga (xy) = loga x + loga y.

Start by substituting values into the left-hand side of the equation, loga (xy), and evaluating. Then, substitute the same values into loga x + loga y and solve to see if the result matches that for loga (xy).

For example, let a = 10, x = 3, and y = 5.

loga (xy) Original expression

= log(10) [(3)(5)] Substitute 10 for a, 3 for x, and 5 for y.

= log10 15 Multiply the numbers.

≈ 1.18 Use a calculator to compute log10 15.

Now evaluate the terms in the right-hand side of the equation, loga x + loga y, for the same values of a, x, and y.

loga x + loga y Original expression

= log(10) (3) + log(10) (5) Substitute 10 for a, 3 for x, and 5 for y.

≈ 0.48 + 0.70 Use a calculator to compute log10 3 and log10 5.

≈ 1.18 Add.

The results match: log10 15 ≈ 1.18 and log10 3 + log10 5 ≈ 1.18. Therefore, the Product Property of Logarithms holds true.


4.2.3

Example 3

Use the properties of logarithms to simplify the logarithmic function f(x) = 2 log x – log (x – 3).

1. Identify which logarithmic properties can be used to simplify the logarithmic function.

The given function contains the term 2 log x. This is the same as 2 • log x. This form is also seen in the Power Property of Logarithms, where loga xy = y • loga x.

Additionally, this function includes the subtraction of two

logarithmic terms. This form is also seen in the Quotient Property of

Logarithms, where x

yx ya a alog log log

= − .

Both the Power Property of Logarithms and the Quotient Property of Logarithms can be used to simplify the given function.

2. Simplify the logarithmic function by applying the identified properties.

Apply the Power Property of Logarithms to the term 2 log x from the original function. The base, a, is understood to be 10 and is omitted.

y log x = log xy Power Property of Logarithms

(2) log x = log x(2) Substitute 2 for y.

Thus, 2 log x = log x2.

Now apply the Quotient Property, log log logx

yx ya a a

= − , to the

original function. Use log x2 in place of 2 log x.

f(x) = 2 log x – log (x – 3) Original function

f(x) = (log x2) – log (x – 3) Substitute log x2 for 2 log x .

f xx

x( ) log

3

2

=−

Simplify using the Quotient Property of Logarithms.

The logarithmic function f(x) = 2 log x – log (x – 3)

simplifies to f xx

x( ) log

3

2

=−

.

U4-60

PRACTICE



Use what you have learned about the properties of logarithms to complete each problem.

1. Simplify x

log3 5

7

−.

2. Simplify x x1

3[log log ( 4)]2 2+ − .

3. Use the properties of logarithms to rewrite 2 log (x + 2).

4. Use log 4 and log 5 to rewrite log 200.

5. What is f x x( ) log ( 1)1

3

= + converted to a common logarithmic function?

6. Which property or properties of logarithms could be used to rewrite log2 (42 • 34)? What is the simplified expression?

7. Evaluate blog1

3 for logb 3 ≈ 0.57. Round the answer to the nearest hundredth.

8. Find the exact value of log 663 without a calculator. If this cannot be done,

explain why.

9. Solve log x2 = 6 for x.

10. Compare the solutions of x = log5 625 and y = log2 32.

Practice 4.2.3: Properties of Logarithms

Lesson 3: Solving Exponential Equations Using Logarithms

UNIT 4 • EXPONENTIAL AND LOGARITHMIC FUNCTIONS

U4-61Lesson 3: Solving Exponential Equations Using Logarithms

4.3


MCC9–12.A.SSE.3c★

MCC9–12.F.IF.8b

MCC9–12.F.BF.5 (+)

MCC9–12.F.LE.4★

Essential Questions

1. Why is a logarithmic equation appropriate for modeling the magnitude of an earthquake based on the Richter scale?

2. If a natural logarithmic equation and a common logarithm equation each represent exponential decay, which equation model (natural or common) would represent a faster rate of decay?

3. For problems involving growth and decay, why is a graph sometimes a more useful tool in determining the growth/decay after many years?

4. How does the growth of an investment earning continuously compounded interest compare to that of an investment earning interest that is compounded quarterly?

WORDS TO KNOW

continuously compounded interest

compound interest that is being added to the balance at every instant

continuously compounded interest formula

a formula used to calculate the balance on a loan or investment for which compound interest is being added to the balance at every instant: A = Pert, where A is the ending amount, P is the principal or initial amount, e is a constant, r is the annual interest rate expressed as a decimal, and t is the time in years


e an irrational number with an approximate value of 2.71828

half-life the time it takes for a substance that is decaying exponentially to decrease to 50% of its original amount


• Monterey Institute. “Mathematical Modeling with Exponential and Logarithmic Functions.”


This site provides step-by-step worked examples of real-life problems involving exponential and logarithmic functions.

• PurpleMath.com. “The Common and Natural Logarithms.”


This site reviews the difference between common and natural logarithms. The natural logarithmic function is explained in relation to the natural exponential function. The site also explains how to rewrite logarithms as their inverse exponential expressions.

• PurpleMath.com. “Logarithmic Word Problems.”


This site explores several applications of logarithmic equations. Examples cover the intensity of sound, earthquake intensity, and compound interest.


4.3.1

Introduction

Exponential and logarithmic functions are great tools for modeling various real-life problems, especially those that deal with fast growth and decline rates. For example, in business, a logarithmic function can be used to calculate the amount of growth in an investment over the years. This information could then help a business owner forecast future trends in the growth of investments the business relies on to fund capital spending, debt payments, and payments to stockholders.

Key Concepts

• Recall that a logarithmic function is the inverse of an exponential function in the form f(x) = a(bx) + c, where a, b, and c are constants and b is greater than 0 but not equal to 1. For an exponential function, f(x) = bx, the inverse logarithmic function would be x = logb f(x). For example, the inverse logarithmic function of the exponential function g(x) = 5x is x = log5 g(x).

• Also recall that an inverse operation is the operation that reverses the effect of another operation. Applying a logarithmic operation to both sides of an exponential equation can serve as an inverse operation when solving equations with exponents.

• Common logarithms are often used when calculating compound interest.

• The compound interest formula can be used to calculate the interest from

the original balance (the principal) and the accrued interest. The formula is

A Pr

n

nt

1= +

, where A is the balance, P is the initial amount, r is the annual

interest rate expressed as a decimal, t is the amount of time in years, and n is

the number of compounding periods per year.

• When solving for n or t in the formula, you can apply logarithms as the inverse operation for the exponents in the formula.

• Common logarithms are also used frequently when working with pH problems.

• The pH scale is a base-10 scale that measures the acidity or alkalinity of a solution. A solution’s pH is found using the formula pH = –log [H+], where [H+] is the concentration of hydrogen ions in the solution measured in moles per liter (abbreviated as M). Generally speaking, the lower the [H+] concentration, the more acidic the solution is.

Lesson 4.3.1: Common Logarithms


• For instance, neutral solutions such as pure water have a pH of 7; pure water is neither acidic nor alkaline. Solutions that have a pH of less than 7 are acidic (such as vinegar, which has a pH of about 2.4), and solutions with a pH greater than 7 are basic or alkaline (i.e., bleach, which has a pH of about 12.6).

• The pH scale ranges from the highly acidic pH 0 (1 × 100 moles per liter) to the very alkaline pH 14 (1 × 10–14 moles per liter).


4.3.1


The magnitude of an earthquake on the Richter scale is calculated using the logarithmic equation r = 0.67 log E – 7.6, where r is the Richter scale number and E is the energy in ergs released by the earthquake. Use properties of logarithms to rewrite the equation in terms of E. Then, use the rewritten formula to calculate the energy in ergs released by an earthquake that measures 3.5 on the Richter scale.

1. Solve the Richter scale formula for E.

The Richter scale formula is given in the problem statement as r = 0.67 log E – 7.6; rearrange the formula so that it is solved for E, the earthquake energy in ergs.

r = 0.67 log E – 7.6 Richter scale formula

r + 7.6 = 0.67 log E Add 7.6 to both sides.r

E7.6

0.67log

+= Divide both sides by 0.67.

Er

107.6

0.67 =+

Rewrite the result as an exponential equation.

Solved for E, the Richter scale formula, r = 0.67 log E – 7.6, can be

rewritten as Er

107.6

0.67 =+

.


2. Use the rewritten formula to determine the energy released by an earthquake measuring 3.5 on the Richter scale.

Use the rewritten formula, Er

107.6

0.67 =+

, to determine the value of E

when r is 3.5.

Er

107.6

0.67 =+

Rewritten Richter scale formula

E10(3.5) 7.6

0.67 =+

Substitute 3.5 for r.

E1011.1

0.67 = Simplify the numerator of the exponent.

1016 ≈ EDivide to simplify the exponent, rounding to the nearest whole number.

4.0 × 1016 ≈ EUse a calculator to rewrite the result in scientific notation.

An earthquake with a magnitude of 3.5 releases approximately 4.0 × 1016 ergs of energy.

Example 2

Roger recently inherited some money. He plans to invest his inheritance in a mutual fund

earning 8.5% interest compounded annually. How long will it take for him to double

his inheritance? Round your final calculation to the nearest tenth. Use the compound

interest formula, A Pr

n

nt

1= +

, where A represents the fund’s balance, P is the

principal amount, r is the annual interest rate expressed as a decimal, n is the number of

times the balance is compounded each year, and t is the time in years.

1. Identify the known quantities.

The annual interest rate, r, is 8.5% or, expressed as a decimal, 0.085.

The number of times the balance is compounded each year, n, is 1 because the interest is compounded annually, or one time per year.

The fund’s balance, A, is 2P because Roger’s goal is to achieve a balance that is double the principal, P.


4.3.1

2. Determine how long it will take for Roger to double his inheritance.

Use the identified values and the compound interest formula,

A Pr

n

nt

1= +

, to solve for t, the time.

A Pr

n

nt

1= +


P Pt

(2 ) 1(0.085)

(1)

(1)

= +

Substitute 2P for A, 0.085 for r, and 1 for n.

2P = P(1.085)t Simplify.

2 = 1.085t Divide both sides by P.

log 2 = t log 1.085Rewrite the exponential form using logarithms.

tlog 2

log 1.085=

Divide both sides by log 1.085 and apply the Symmetric Property of Equality.

t ≈ 8.5 Simplify the logarithmic terms using a calculator.

It will take Roger approximately 8.5 years to double his inheritance.


Example 3

The acidity (or alkalinity) of a solution can be determined using the formula pH = –log [H+], where [H+] is the concentration in moles per liter of hydrogen ions in the solution. Acidic solutions have a pH of less than 7, neutral solutions have a pH of 7, and alkaline solutions have a pH of greater than 7. If the [H+] value of lemon juice is approximately 0.005 moles per liter (M), what is its pH? Given that unsweetened green tea is alkaline and thus less acidic than lemon juice, determine how the unknown pH of green tea compares with the pH of lemon juice. Explain your answer.

1. Determine the pH of lemon juice given the [H+] value.

The concentration of hydrogen ions in lemon juice is given as 0.005 M. Substitute this value for [H+] in the pH formula, pH = –log [H+], and solve.

pH = –log [H+] Formula for pH

pH = –log (0.005) Substitute 0.005 for [H+].

pH ≈ 2.3 Simplify using a calculator.

Lemon juice has a pH of approximately 2.3.

2. Determine what the pH of lemon juice means in terms of its acidity.

A pH of 2.3 is less than 7 (neutral), so lemon juice has an acidic pH.

3. Determine how the unknown pH of green tea compares with the pH of lemon juice.

Recall that an alkaline solution has a pH that is greater than 7.

We’ve determined that lemon juice is acidic. Given that green tea is alkaline, it should have a pH that is both higher than that of the lemon juice and higher than 7, which is neutral.


4.3.1

Example 4

The sound intensity level of a noise, expressed as a number of decibels (dB), can be

found using the formula I

10 log10 12β =

− , where I is the intensity or “power” of

the sound in watts per square meter. The lowercase Greek letter beta, β, represents

the sound intensity level. If the sound intensity of a large orchestra was measured at

6.3 × 10–3 watts per square meter, determine the number of decibels the orchestra

produced. Then, find the number of decibels produced by a sound measured at a rock

concert (I = 1 × 10–1) and the number of decibels produced by an average conversation

(I = 1 × 10–6). Compare all three results.

1. Determine the number of decibels produced by the large orchestra.

Substitute the sound intensity for the orchestra, I = 6.3 × 10–3, into the given formula and then solve.

I10 log

10 12β =

− Given formula

10 log6.3 10

10

3

12β( )

=×

−

− Substitute 6.3 × 10–3 for I.

β = 10 log [6.3 × 10–3 – (–12)] Apply the rules of exponents.

β = 10 log (6.3 × 109) Simplify the exponent.

β ≈ 98 Evaluate the logarithm using a calculator.

The orchestra produced a sound intensity level of approximately 98 decibels.


2. Determine the number of decibels produced by the rock concert.

Substitute the sound intensity for the rock concert, I = 1 × 10–1, into the given formula and then solve.

I10 log

10 12β =

− Given formula

10 log1 10

10

1

12β( )

=×

−

− Substitute 1 × 10–1 for I.

β = 10 log [1 × 10–1 – (–12)] Apply the rules of exponents.

β = 10 log (1 × 1011) Simplify the exponent.

β = 110 Evaluate the logarithm using a calculator.

The rock concert produced a sound intensity level of 110 decibels.


4.3.1

3. Determine the number of decibels produced by the average conversation.

Substitute the sound intensity for the average conversation, I = 1 × 10–6, into the given formula and then solve.

I10 log

10 12β =

− Given formula

10 log1 10

10

6

12β( )

=×

−

− Substitute 1 × 10–6 for I.

β = 10 log [1 × 10–6 – (–12)] Apply the rules of exponents.

β = 10 log (1 × 106) Simplify the exponent.

β = 10 log 106 Simplify.

β = 60 Evaluate the logarithm using a calculator.

The average conversation produced a sound intensity level of 60 decibels.

4. Compare the three sounds by sound intensity level.

The previous calculations show that the rock concert has the highest sound intensity level at 110 dB, followed by the orchestra performance at approximately 98 dB, and then the average conversation at 60 dB.

U4-72

PRACTICE

UNIT 4 • EXPONENTIAL AND LOGARITHMIC FUNCTIONSLesson 3: Solving Exponential Equations Using Logarithms


Use the given information and what you have learned about common logarithms to complete each problem.

1. What is the value of P = –7.52 log (t) + 53 when t = 39? Round your answer to the nearest hundredth.

2. Use the pH formula, pH = –log [H+], to calculate the concentration of hydrogen ions, [H+], in moles per liter (M) for a solution that has a pH of 3.2. Write your answer in scientific notation.

3. Jeremiah invested $1,000 in an account that compounded interest annually. If his balance was $1,771.56 after 6 years, what was the annual interest rate on the account?

4. The temperature of a cooling liquid can be modeled by the exponential model T – 21 = 54.4(0.964)t, where t is the time in minutes and T is the temperature in degrees Celsius of the cooling liquid. If t runs from 0 to 30, prove or disprove that the liquid’s temperature is 51.2°C when the time is 15 minutes.

5. Rewrite x

5000(1 0.005)

0.005250,000

+

= as a logarithmic equation.

6. Determine the sound intensity level of an alarm in decibels if the alarm has

an intensity I of 5.8 × 10–9 watt per square meter. Use the formula for sound

intensity level, I

10 log10 12β =

− .

Practice 4.3.1: Common Logarithms

continued

U4-73

PRACTICE


Lesson 3: Solving Exponential Equations Using Logarithms 4.3.1

7. Determine the concentration of hydrogen ions in moles per liter (M) for a sample of coffee with a pH of 5.3. Use the pH formula, pH = –log [H+].

8. A particular citrus fruit has a pH of 2.2 and an antacid pill has a pH of 10.1. How much higher is the concentration of hydrogen ions, [H+], in moles per liter for the fruit compared that of the antacid? Use the pH formula, pH = –log [H+].

9. The sound of a slamming car door has an intensity I of 10–4 watt per square

meter, a loud car horn has an intensity of 10–6 watt per square meter, and an

ambulance siren has an intensity of 10–3 watt per square meter. Use the formula I

10 log10 12β =

− , where β is the sound intensity level expressed as a number

of decibels and I is the intensity of the sound, to rank each sound from highest

sound intensity level to lowest.

10. Kaylee deposited $750 in an account earning 5% interest, compounded annually.

If she withdrew her money once the balance reached $1,500, how long was her

money in the account? Use the formula A Pr

n

nt

1= +

, where A is the balance,

P is the initial amount, r is the annual interest rate expressed as a decimal, t is the

amount of time in years, and n is the number of compounding periods per year.

Round your answer to the nearest whole number.


Introduction

Logarithms can be used to solve exponential equations that have a variable as an

exponent. In compound interest problems that use the formula A Pr

n

nt

1= +

,

logarithms can be used to solve for t, which indicates the amount of time an account or

investment takes to mature. A financial advisor at a business can use this information

to forecast and plan for the company’s future. In natural exponential equations

(those with the variable e), a natural logarithm can be used to solve for a variable that

exists as an exponent, since ln x = loge x. So, natural logarithms can be used to model

situations, graphed as functions, and written as exponents. Additionally, they can be

evaluated by hand using the properties of logarithms (e.g., ln 1 = 0 because e0 = 1),

and with a calculator when properties of logarithms cannot be applied. For example,

ln 8 ≈ 2.07 because e2.07 ≈ 8.

Key Concepts

• Recall that natural logarithms consist of a number with a base of e. As with common logarithms, natural logarithms have an inverse relationship with exponential functions. That is, the natural logarithm function ln b = a can also be rewritten as the exponential function, ea = b, where e is an irrational number with an approximate value of 2.71828.

• Natural logarithms differ from common logarithms because of their base.

Common logarithms have a base of 10 whereas natural logarithms have a base

of e. For convenience, natural logarithms can be converted to base 10 by using

the change of base formula, xx

bbloglog

log= .

• The same properties that apply to common logarithms also apply to natural logarithms.

• One application for natural logarithms involves calculating interest on money that is compounded continuously, or at every instant.

Lesson 4.3.2: Natural Logarithms


4.3.2

• The continuously compounded interest formula is A = Pert, where A is the

ending amount, P is the principal or initial amount, e is a constant, r is the

annual interest rate expressed as a decimal, and t is the time in years. Note this

formula’s similarity to the compound interest formula, A Pr

n

nt

1= +

.

• The continuously compounded interest formula, A = Pert, is a natural exponential function because it is in the form f(x) = ex and involves a base of e.

• Other uses of natural logarithms include situations of exponential growth or decay, which involve rapid increase (growth) or decrease (decay).

• One formula for modeling the exponential growth or decay of a substance or population is P = P0ekt. In this formula, P represents the amount of a substance or population after the time, t, has elapsed, P0 is the initial amount, e is a constant, and k is the rate of growth or decay.

• Notice this formula is similar to the continuously compounded interest formula, but instead of modeling an amount of money over time, this formula models the amount of a substance or population over time. When k > 0, the formula models growth; when k < 0, the formula models decay. Since this formula uses the base e, it represents a natural exponential function.

• A common application of exponential decay is finding the half-life of a substance. Half-life is the time it takes for a substance that is decaying exponentially to decrease to 50% of its original amount.



Cheyenne deposited $200 in a bank account earning continuously compounded interest. After 10 years, she closed the account and withdrew the entire balance, which totaled $364.42. What was her annual interest rate? Rounded to the nearest dollar, how much would Cheyenne have received if she had left the money in the account for 15 years? 20 years? Use the continuously compounded interest formula, A = Pert, where A is the ending amount, P is the principal or initial amount, e is a constant, r is the annual interest rate expressed as a decimal, and t is the time in years.

1. Determine values for the continuously compounded interest formula.

The continuously compounded interest formula is A = Pert.

The ending amount, A, is the account balance when Cheyenne withdrew the money, $364.42.

The principal, P, is $200.

The time, t, is 10 years.

Therefore, let A = 364.42, P = 200, and t = 10.

2. Substitute the known values into the formula and solve for the annual interest rate, r.

A = Pert Continuously compounded interest formula

(364.42) = (200)er (10) Substitute 364.42 for A, 200 for P, and 10 for t.

1.82 = e10rDivide both sides by 200 and simplify the exponent.

ln 1.82 = ln e10r Rewrite each side as a natural logarithm.

ln 1.82 = 10r ln eApply the Power Property for natural logarithms.

re

ln 1.82

10 ln= Divide to isolate r and then apply the Symmetric

Property of Equality.

r ≈ 0.06Evaluate each natural logarithm using a calculator.

The value of r is about 0.06; therefore, the annual interest rate was 6%.


4.3.2

3. Determine how much money Cheyenne would have received if she had left the money in the account for 15 years.

Use the continuously compounded interest formula, A = Pert.

The values of P and r remain the same.

The value of t is 15.

Substitute these values into the formula and solve for the ending amount, A.


A = (200)e(0.06)(15) Substitute 200 for P, 0.06 for r, and 15 for t.

A = 200e0.9 Simplify the exponent.

A ≈ 491.92 Evaluate using a calculator.

If Cheyenne had left her money in the account for 15 years, she would have received about $491.92.

4. Determine how much money Cheyenne would have received if she had left the money in the account for 20 years.

Once again, use the continuously compounded interest formula, A = Pert.

The values of P and r remain the same.

The value of t is 20.

Substitute these values into the formula and solve for the ending amount, A.


A = (200)e(0.06)(20) Substitute 200 for P, 0.06 for r, and 20 for t.

A = 200e1.2 Simplify the exponent.

A ≈ 664.02 Evaluate using a calculator.

If Cheyenne had left her money in the account for 20 years, she would have received about $664.02.


Example 2

A new father has decided to invest $20,000 in a savings account for his child’s college education. Which savings account option (detailed in the following table) will earn the most money after 18 years?

Account Annual interest rate Compounding scheduleX 3.7% ContinuouslyY 3.75% AnnuallyZ 3.5% Quarterly

1. Determine which formula to use to calculate the balance after 18 years for each account.

Account X compounds continuously; therefore, use the continuously compounded interest formula, A = Pert.

Accounts Y and Z compound a set number of times each year instead

of continuously (annually and quarterly, respectively). Therefore, use

the compound interest formula, A Pr

n

nt

1= +

.

In both formulas, A is the account balance, P is the principal amount, r is the annual interest rate expressed as a decimal, and t is the time in years.

In the compound interest formula, n is the number of compounding periods.


4.3.2

2. Determine the balance after 18 years for Account X.

The principal, P, is $20,000.

The interest rate, r, is 3.7% or 0.037.

The time in years, t, is 18.

Therefore, let P = 20,000, r = 0.037, and t = 18.

Substitute these values into the continuously compounded interest formula and solve for the ending amount.

A = PertContinuously compounded interest formula

(AX) = (20,000)e(0.037)(18)Substitute AX for A, 20,000 for P, 0.037 for r, and 18 for t.

AX ≈ 20,000e0.67 Simplify the exponent.

AX ≈ 39,084.75 Evaluate using a calculator.

After 18 years, Account X would have a balance of approximately $39,084.75.


3. Determine the balance after 18 years for Account Y.


The interest is compounded annually, or once per year, so n is 1.

The principal, P, is $20,000.

The time in years, t, is 18.

Therefore, let P = 20,000, r = 0.0375, n = 1, and t = 18.

Substitute these values into the compound interest formula and solve for the ending amount.

A Pr

n

nt

1= +


A (20,000) 1(0.0375)

(1)Y

(1)(18)

( )= +

Substitute AY for A, 20,000 for P, 0.0375 for r, 1 for n, and 18 for t.

AY = 20,000(1.0375)18Simplify within the brackets and simplify the exponent.

AY ≈ 38,798.59 Evaluate using a calculator.

After 18 years, Account Y would have a balance of approximately $38,798.59.


4.3.2

4. Determine the balance after 18 years for Account Z.

The values for P and t remain the same as in the calculations for the two previous accounts.


The interest is compounded quarterly, or 4 times per year, so n = 4.

Therefore, let P = 20,000, r = 0.035, n = 4, and t = 18.

Substitute these values into the compound interest formula and solve for the ending amount.

A Pr

n

nt

1= +


A (20,000) 1(0.035)

(4)Z

( 4)(18)

( )= +

Substitute AZ for A, 20,000 for P, 0.035 for r, 4 for n, and 18 for t.

AZ = 20,000(1.00875)72Simplify within the brackets and simplify the exponent.

AZ ≈ 37,449.45 Evaluate using a calculator.

After 18 years, Account Z would have a balance of approximately $37,449.45.

5. Determine which account would yield the most money after 18 years.

Account X yields approximately $39,084.75, Account Y yields approximately $38,798.59, and Account Z yields approximately $37,449.45. The account with the greatest amount of money is Account X, so it would yield the most money after 18 years.


Example 3

Studies have shown that the number of bacteria on a public restroom sink can grow exponentially from 5,000 bacteria to 12,000 bacteria in 10 hours. Write the natural logarithmic equation that would represent how the number of bacteria would grow over the given time period. Use the exponential growth/decay formula, P = P0ekt, where P represents the number of bacteria after t hours, P0 is the initial number of bacteria, and k is the rate of growth or decay. Given the rate of bacterial growth, how many bacteria would there be after 24 hours? How many bacteria would you expect to be present after 48 hours? Verify your answers algebraically.

1. Determine the growth rate, k, using properties of logarithms.

Begin by identifying the known values.

Let P, the final population of the bacteria, be 12,000.

Let P0, the initial population of the bacteria, be 5,000.

Let t, the time in hours, be 10.

Substitute these values into the given formula and solve for k.

P = P0ekt Exponential growth/decay formula

(12,000) = (5000)ek(10)Substitute 12,000 for P, 5,000 for P0, and 10 for t.

2.4 = e10kDivide both sides by 5,000 and simplify the exponent.

ln 2.4 = ln e10k Rewrite each side as a natural logarithm.

ln 2.4 = 10k ln eApply the Power Property of natural logarithms.

ke

ln 2.4

10 ln= Divide to isolate k and then apply the

Symmetric Property of Equality.

k ≈ 0.088 Evaluate using a calculator.

The growth rate, k, is approximately 0.088.


4.3.2

2. Determine the number of bacteria after 24 hours.

Use the exponential growth/decay formula, P = P0ekt.

The value for P0 remains the same (5,000).

Let the growth rate, k, be 0.088.

Let the time, t, be 24.

Substitute these values into the given formula and solve for P, the final population.


P = (5000)e(0.088)(24) Substitute 5,000 for P0, 0.088 for k, and 24 for t.

P = 5000e2.112 Simplify the exponent.

P ≈ 41,324 Evaluate using a calculator.

After 24 hours, there will be approximately 41,324 bacteria on the sink.

3. Determine the number of bacteria after 48 hours.


The values for P0 and k remain the same (5,000 and 0.088, respectively).

Let the time, t, be 48.

Substitute these values into the given formula and solve for P, the final population.


P = (5000)e(0.088)(48) Substitute 5,000 for P0, 0.088 for k, and 48 for t.

P = 5000e4.224 Simplify the exponent.

P ≈ 341,531 Evaluate using a calculator.

After 48 hours, there will be approximately 341,531 bacteria on the sink.


Example 4

When Rebekah left her hometown for college, 300 people lived in her hometown. When she moved back to her hometown 25 years later to take a job as the town planner, the population had grown to 6,000 people. Rebekah wants to create a long-term plan to accommodate the growing population, which she estimates could grow to 50,000 someday. Use the formula P = P0ekt to write an equation that will help Rebekah predict how many years it will take for the town’s population to reach 50,000 people based on the growth over the past 25 years.

1. Determine the growth rate, k, using properties of logarithms.

Begin by identifying the known values.

Let P, the final (current) population, be 6,000.

Let P0, the initial population, be 300.

Let t, the time in years, be 25.

Substitute these values into the given formula and solve for k.


(6000) = (300)ek(25) Substitute 6,000 for P, 300 for P0, and 25 for t.

20 = e25k Divide both sides by 300.

ln 20 = ln e25k Rewrite each side as a natural logarithm.

ln 20 = 25k ln e Apply the Power Property of natural logarithms.

ke

ln 20

25 ln= Divide to isolate k and then apply the Symmetric

Property of Equality.

k ≈ 0.12 Evaluate using a calculator.

The growth rate, k, is approximately 0.12.


4.3.2

2. Determine when the population will grow to 50,000 people.


The value for P0 remains the same (300).

Let P, the final (projected) population, be 50,000.

Let the growth rate, k, be 0.12.

Substitute these values into the given formula and solve for t.


(50,000) = (300)e(0.12)tSubstitute 50,000 for P, 300 for P0, and 0.12 for k.

166.67 = e0.12t Divide both sides by 300.

ln 166.67 = ln e0.12t Rewrite each side as a natural logarithm.

ln 166.67 = 0.12t ln eApply the Power Property of natural logarithms.

te

ln 166.67

0.12 ln= Divide to isolate t and then apply the

Symmetric Property of Equality.

t ≈ 42.5 Evaluate using a calculator.

If the population growth continues at the current rate, it will take approximately 42.5 years for the town’s population to grow to 50,000 people.


Example 5

Carbon-14 is a radioactive isotope used to determine the age of once-living organisms by studying the amount of carbon left behind as the organism decays. Carbon-14 has a half-life of 5,730 years, meaning that it takes 5,730 years for half of the carbon-14 to decay from the material in which it is found. If an archaeologist finds an animal bone, how long ago did the animal die if the bone contains only 70% of its original amount of carbon-14?

1. Determine the decay rate, k, using properties of logarithms.

Let P, the amount of remaining carbon-14, be P1

2 0 , because we’re

looking for the decay rate relative to the half-life of carbon-14.

Let t, the time in years, be 5,730.

Substitute these known values into P = P0ekt and then solve for k.


P P e k1

2 0 0(5730)

= Substitute P

1

2 0 for P and 5,730 for t.

e k1

25730= Divide both sides by P0.

e kln1

2ln 5370= Rewrite each side as a natural logarithm.

k eln1

25730 ln= Apply the Power Property of natural logarithms.

ke

ln1

25730 ln

=Divide to isolate k and then apply the Symmetric Property of Equality.

k ≈ –0.00012 Evaluate using a calculator.

The decay rate, k, is –0.00012.


4.3.2

2. Determine how old the bone is if it contains only 70% of its original carbon-14.

Create an equation using k and the information from the problem statement to rewrite the equation in terms of t.

Let P be 0.70P0.

Let k, the decay rate, be 0.00012.


(0.70P0) = P0(–0.00012)t

Substitute 0.70P0 for P0 and –0.00012 for k.

0.70 = e–0.00012t Divide both sides by P0.

ln 0.70 = ln e–0.00012t Rewrite each side as a natural logarithm.

ln 0.70 = –0.00012t ln eApply the Power Property of natural logarithms.

te

ln 0.70

0.00012 ln=−

Divide to isolate t and then apply the Symmetric Property of Equality.

t ≈ 2972.29 Evaluate using a calculator.

The bone is approximately 2,972.29 years old.

U4-88

PRACTICE



Use the given information and what you have learned about natural logarithms to solve each of the problems.

1. How long will it take your money to double if you deposit $500 in an account that pays 6.75% interest, compounded continuously? Use the formula A = Pert, where A is the balance, P is the initial amount, r is the annual interest rate expressed as a decimal, and t is the time in years.

2. For the years from 1985 to 2004, the average salary (in thousands of dollars) for public school teachers in a particular district for the year t can be modeled by the function f(t) = –1.562 + 14.584 ln t, for t = 5 to t = 24. In what year during this time period did the average salary for public school teachers reach $44,000?

3. The timber yield V (in millions of cubic feet per acre) for a forest at age t years is

given by V e t6.748.1

=−

. Solve for t in the given equation.

4. The exponential growth equation P = 5400e0.0118t represents the world population in billions from t = 8 to t = 17 years. For what value of t did the world population reach 6.8 billion? (Hint: Use 6,800 for 6.8 billion.)

5. Carrie hopes to make $10,000 from an investment over the next 10 years. If her account earns 4.5% continuously compounded interest, will she reach her desired amount if she deposits $5,000 into the bank now? If not, how many years would it take to reach a balance of $10,000? Use the formula for continuously compounded interest, A = Pert.

Practice 4.3.2: Natural Logarithms

continued

U4-89

PRACTICE


Lesson 3: Solving Exponential Equations Using Logarithms 4.3.2

6. The population of a country grew from approximately 65.7 million people in the year 2000 to about 73.3 million people in 2010. Which function best reflects this change: f(x) = aekt, or f(x) = ae–kt?

7. If $1 were invested in an account, which option would yield a balance of

$50 faster: 7.5% interest compounded annually or 7% interest compounded

continuously? How long would it take this account to reach $50? Use the

formulas A Pr

n

nt

1= +

and A = Pert.

8. The equation P = 134.0ekt represents the population of a town in Nevada, with t = 0 corresponding to the year 1990. If the population increased to 180,000 residents in the year 2000, find the value of k in the equation. Based on the same growth rate, k, did the population increase to more than 200,000 residents in the year 2010? Explain.

9. Use logarithms to solve for x in the equation 100e0.005x = 125,000.

10. How much time would it take to triple $10,000 at 3.5% interest compounded continuously? What would be the balance of the account in 10 years? Use the formula A = Pert.

AK-1Answer Key

Answer KeyLesson 1: Exponential Functions

Practice 4.1.1: Rewriting Exponential Functions, pp. U4-14–U4-15

1. f xx

( ) 5 2=2.

3. 2.84%4.

5. 0.19%6.

7. 2.99%8.

9. 1.64%

Practice 4.1.2: Properties of Exponential Functions, pp. U4-28–U4-291. domain:allrealnumbers;range:allrealnumbersgreaterthan14;y-intercept:20;asymptote:y=14.The

functiondecreaseswithinitsdomain.2.

3. domain:allrealnumbers;range:allrealnumbersgreaterthan1;y-intercept:10;asymptote:y=1.Thefunctiondecreaseswithinitsdomain.4.

5. f(x)=13(5)x+136.

7. f(x)=11(18)x+68.

9. domain:allrealnumbersxsuchthatx≥0;range:allrealnumbersysuchthat40<y≤68;y-intercept:68;asymptote:y=40.Thefunctiondecreaseswithinitsdomain.

Lesson 2: Introducing Logarithmic Functions

Practice 4.2.1: Defining Logarithms, p. U4-421. No,itisnot.Forexample,log10=1while–log10=–1.

2.

3. 10y=1

x4.

5. log25

y =1

26.

7. Becauseanynumberraisedtothepowerof0isequalto18.

9.1

2log 30001.05=t

Practice 4.2.2: Graphs of Logarithmic Functions, pp. U4-53–U4-541. Thegraphsarereflectionsofeachother.

–8 –6 –4 –2 2 4 6 8

5

4

3

2

1

0

–1

–2

–3

–4

–5

u(x)

r(x)

y

x

2.

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AK-2Answer Key

3. Theasymptoteforf(x)=log(x+5)isx=–5,whereastheasymptoteforg(x)=logxisx=0.Theinterceptsforf(x)=log(x+5)are(–4,0)and(0,0.7),whereastheinterceptforg(x)=logxis(1,0).4.

5. Thegraphofr(x)=logxshifts4unitstotheleftalongthex-axistocreatethegraphofv(x)=log(x+4).r(x)hasinterceptsof(0,0.6)and(–3,0),wherev(x)hasonlyoneintercept,at(1,0).r(x)hasanasymptoteatx=–4,butv(x)hasanasymptoteatx=0.

–4 –2 2 4

4

3

2

1

0

–1

–2

–3

–4

y

xv(x)

r(x)

6.

7. Thegraphoff(x)=log(x+1)isincreasingforvaluesofx>–1.8.

9. Asthebaseoff(x)=log(x+2)decreases,thedomainwillremainx>–2buttherangevalueswillincrease.

Practice 4.2.3: Properties of Logarithms, p. U4-60

1.1

2log 3 – 5 – log 7( )x

2.

3. log(x+2)2

4.

5.log ( 1)

log1

3

+x

6.

7. logblog 6 log 6

1

36

36

1

3= = =–0.578.

9. x=1000

Lesson 3: Solving Exponential Equations Using Logarithms

Practice 4.3.1: Common Logarithms, pp. U4-72–U4-731. P=41.042.

3. 10%4.

5. x=log1.005

256.

7. 5×10–6molesperliter8.

9. ambulancesiren,slammingdoor,carhorn

Practice 4.3.2: Natural Logarithms, pp. U4-88–U4-89

1. t = ≈ln2

0.067510.27 years

2.

3. tV

=−

48.1

ln674.

5. Carriewillnotbeabletoreachabalanceof$10,000in10yearsatthegiveninterestrate.Itwilltakeapproximately15years.6.

7. Thebalanceoftheannuallycompoundedaccountwouldreach$50faster(in54.09years).8.

9. x=1426

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