MTH-4105-1 C1-C4 ExponentsRadicals_Layout 1 10-10-18 13:55 Page 1
ERadicals
MTH-4105-1
xponents
and
MTH-4105-1
EXPONENTS
AND
RADICALS
f
SO
AD
Société de formation à distance
des commissions scolaires du Québec
Project Coordinator: Jean-Paul Groleau
Author: Nicole Perreault
Update: Éric Lacroix
Content Revision: Jean-Paul Groleau
Line Régis
Translator: Claudia de Fulviis
Linguistic Revision: Johanne St-Martin
Electronic publishing: Productions P.P.I. inc.
Cover Page: Daniel Rémy
Printed: 2005
Reprint: 2006
© Société de formation à distance des commissions scolaires du Québec
All rights for translation and adaptation, in whole or in part, reserved for all countries.
Any reproduction by mechanical or electronic means, including microreproduction, is
forbidden without the written permission of a duly authorized representative of the
Société de formation à distance des commissions scolaires du Québec (SOFAD).
Legal Deposit: 2005
Bibliothèque et Archives nationales du Québec
Bibliothèques et Archives Canada
ISBN 978-2-89493-283-4
050331
MTH-4105-1
Answer Key
Exponents and Radicals
TABLE OF CONTENTS
Introduction to the Program Flowchart ................................................... 0.4
The Program Flowchart ............................................................................ 0.5
How to Use this Guide .............................................................................. 0.6
General Introduction ................................................................................. 0.9
Intermediate and Terminal Objectives of the Module ............................ 0.10
Diagnostic Test on the Prerequisites ....................................................... 0.13
Answer Key for the Diagnostic Test on the Prerequisites ...................... 0.17
Analysis of the Diagnostic Test Results ................................................... 0.19
Information for Distance Education Students ......................................... 0.21
UNITS
1. The Laws of Exponents ............................................................................. 1.1
2. Simplifying Algebraic or Numerical Expressions Written in
Exponential Form...................................................................................... 2.1
3. Converting an Expression Containing a Radical to Exponential Form
and Vice Versa ........................................................................................... 3.1
4. The Sum, Difference, Product and Quotient of Numerical Expressions
Containing Square Roots .......................................................................... 4.1
5. Operations on Polynomials Containing Square Roots ............................ 5.1
Final Review .............................................................................................. 6.1
Answer Key for the Final Review ............................................................. 6.4
Terminal Objectives .................................................................................. 6.5
Self-Evaluation Test.................................................................................. 6.7
Answer Key for the Self-Evaluation Test ................................................ 6.15
Answer Key for the Self-Evaluation Test Results ................................... 6.19
Final Evaluation........................................................................................ 6.20
Answer Key for the Exercises ................................................................... 6.21
Glossary ..................................................................................................... 6.59
List of Symbols .......................................................................................... 6.63
Bibliography .............................................................................................. 6.64
Review Activities ....................................................................................... 7.1
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INTRODUCTION TO THE PROGRAM FLOWCHART
Welcome to the World of Mathematics!
This mathematics program has been developed for the adult students of the
Adult Education Services of school boards and distance education. The learning
activities have been designed for individualized learning. If you encounter
difficulties, do not hesitate to consult your teacher or to telephone the resource
person assigned to you. The following flowchart shows where this module fits
into the overall program. It allows you to see how far you have progressed and
how much you still have to do to achieve your vocational goal. There are several
possible paths you can take, depending on your chosen goal.
The first path consists of modules MTH-3003-2 (MTH-314) and MTH-4104-2
(MTH-416), and leads to a Diploma of Vocational Studies (DVS).
The second path consists of modules MTH-4109-1 (MTH-426), MTH-4111-2
(MTH-436) and MTH-5104-1 (MTH-514), and leads to a Secondary School
Diploma (SSD), which allows you to enroll in certain Cegep-level programs that
do not call for a knowledge of advanced mathematics.
The third path consists of modules MTH-5109-1 (MTH-526) and MTH-5111-2
(MTH-536), and leads to Cegep programs that call for a solid knowledge of
mathematics in addition to other abiliies.
If this is your first contact with this mathematics program, consult the flowchart
on the next page and then read the section “How to Use This Guide.” Otherwise,
go directly to the section entitled “General Introduction.” Enjoy your work!
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THE PROGRAM FLOWCHART
CEGEP
MTH-5112-1
MTH-5111-2
MTH-536
MTH-5104-1
MTH-5103-1
MTH-5110-1
Introduction to Vectors
MTH-5109-1
Geometry IV
Trigonometric Functions and Equations
MTH-5107-1
Exponential and Logarithmic Functions
and Equations
Optimization II
MTH-5106-1
Real Functions and Equations
Probability II
MTH-5105-1
Conics
MTH-5102-1
Statistics III
MTH-5101-1
Optimization I
MTH-4111-2
Trades
DVS
MTH-436
MTH-4110-1
MTH-4109-1
MTH-426
MTH-314
MTH-216
MTH-116
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The Four Operations on
Algebraic Fractions
Sets, Relations and Functions
MTH-4108-1
Quadratic Functions
Straight Lines II
MTH-4106-1
Factoring and Algebraic Functions
MTH-4105-1
Exponents and Radicals
MTH-4103-1
MTH-4102-1
MTH-4101-2
Complement and Synthesis I
MTH-4107-1
MTH-4104-2
MTH-416
Complement and Synthesis II
MTH-5108-1
MTH-526
MTH-514
Logic
You ar e h er e
Statistics II
Trigonometry I
Geometry III
Equations and Inequalities II
MTH-3003-2
Straight Lines I
MTH-3002-2
Geometry II
MTH-3001-2
The Four Operations on Polynomials
MTH-2008-2
Statistics and Probabilities I
MTH-2007-2
Geometry I
MTH-2006-2
Equations and Inequalities I
MTH-1007-2
Decimals and Percent
MTH-1006-2
The Four Operations on Fractions
MTH-1005-2
The Four Operations on Integers
0.5
25 hours
= 1 credit
50 hours
= 2 credits
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HOW TO USE THIS GUIDE
Hi! My name is Monica and I have been
asked to tell you about this math module.
What’s your name?
Whether you are
registered at an
adult education
center or pursuing distance
education, ...
Now, the module you have in your
hand is divided into three
sections. The first section is...
I’m Andy.
... you have probably taken a
placement test which tells you
exactly which module you
should start with.
... the entry activity, which
contains the test on the
prerequisites.
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You’ll see that with this method, math is
a real breeze!
My results on the test
indicate that I should begin
with this module.
By carefully correcting this test using the
corresponding answer key, and recording your results on the analysis sheet ...
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... you can tell if you’re well enough
prepared to do all the activities in the
module.
And if I’m not, if I need a little
review before moving on, what
happens then?
Exponents and Radicals
In that case, before you start the
activities in the module, the results
analysis chart refers you to a review
activity near the end of the module.
Good!
In this way, I can be sure I
have all the prerequisites
for starting.
START
The starting line
shows where the
learning activities
begin.
Exactly! The second section
contains the learning activities. It’s
the main part of the module.
?
The little white question mark indicates the questions
for which answers are given in the text.
The target precedes the
objective to be met.
The memo pad signals a brief reminder of
concepts which you have already studied.
?
Look closely at the box to
the right. It explains the
symbols used to identify the
various activities.
The boldface question mark
indicates practice exercises
which allow you to try out what
you have just learned.
The calculator symbol reminds you that
you will need to use your calculator.
?
The sheaf of wheat indicates a review designed to
reinforce what you have just learned. A row of
sheaves near the end of the module indicates the
final review, which helps you to interrelate all the
learning activities in the module.
FINISH
Lastly, the finish line indicates
that it is time to go on to the self-evaluation
test to verify how well you have understood
the learning activities.
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There are also many fun things
in this module. For example,
when you see the drawing of a
sage, it introduces a “Did you
know that...”
It’s the same for the “math whiz”
pages, which are designed especially for those who love math.
For example. words in boldface italics appear in the
glossary at the end of the
module...
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Exponents and Radicals
A “Did you know that...”?
Yes, for example, short tidbits
on the history of mathematics
and fun puzzles. They are interesting and relieve tension at
the same time.
Must I memorize what the sage says?
No, it’s not part of the learning activity. It’s just there to
give you a breather.
And the whole module has
been arranged to make
learning easier.
They are so stimulating that
even if you don’t have to do
them, you’ll still want to.
... statements in boxes are important
points to remember, like definitions, formulas and rules. I’m telling you, the format makes everything much easier.
The third section contains the final review, which interrelates the different
parts of the module.
Great!
There is also a self-evaluation
test and answer key. They tell
you if you’re ready for the final
evaluation.
Thanks, Monica, you’ve been a
big help.
I’m glad! Now,
I’ve got to run.
See you!
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Later ...
This is great! I never thought that I would
like mathematics as much as this!
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GENERAL INTRODUCTION
EXPONENTS AND RADICALS: A WORLD TO DISCOVER
In science, it is common to use numbers expressed in exponential form or as
radicals. Chemists use numbers expressed in scientific notation to calculate
the equilibrium constant of an equation or to determine the degree of acidity of
a solution. Physicists make frequent use of radicals in mechanics or nuclear
science, and most of the constants they use in calculations are expressed in
scientific notation.
In this module, you will learn the laws of exponents and the rules for applying
them. These laws will allow you to simplify algebraic or numerical expressions
2
1
2
2
a
×
a
.
such as
a– 2
You will also learn the laws that apply to radicals. These laws will come in handy
for reducing polynomials such as 5 2 to simplest form.
2+ 3
You are already familiar with rational numbers (natural numbers, integers,
decimals and mixed numbers). In this module, you will learn about irrational
numbers which, historically speaking, had difficulty taking hold in
mathematics.
You will also learn to rationalize the denominator of an
expression, in other words, to make it rational.
To achieve the terminal objectives of this module, you will be required to solve
algebraic and numerical expressions in exponential form by applying the laws
of exponents. You will also be required to perform the four mathematical
operations on polynomials containing square roots, by following the rules of
priority of operations.
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INTERMEDIATE AND TERMINAL OBJECTIVES OF
THE MODULE
Module MTH-4105-1 consists of five units and requires 25 hours of study
distributed as follows. The terminal objectives appear in boldface.
Objectives
Number of hours*
% (Evaluation)
1 and 2
13
55%
3
4
15%
4 and 5
7
30%
* One hour is allocated for the final evaluation.
1. The laws of exponents
To calculate the numerical or algebraic value of a mathematical expression
containing powers of numbers or variables by applying one or more exponents
laws. The exponent laws are the following:
am × an = am + n
a m = am – n
an
a– m = 1m
a
ao = 1
(am)n = am × n
(abc)m = ambmcm
am
a m
= bm
b
where a, b, c are rational numbers or variables and m and n are rational
numbers.
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2. Simplifying algebraic or numerical expressions written in
exponential form
To reduce to simplest form an algebraic or numerical expression in
exponential form by applying the priority of operations and one or
more of the following exponent laws:
am × an = am + n
a m = am – n
an
a– m = 1m
a
ao = 1
(am)n = am × n
(abc)m = ambmcm
a
b
m
m
= am
b
where a, b, c are rational numbers or variables and m and n are
rational numbers.
3. Converting an expression containing a radical to exponential form
and vice versa
To convert a numerical or algebraic expression containing a radical
to an exponential expression with the lowest possible base.
The given expression is of the form am •
n
p
b , where a and b are
variables or positive rational numbers that are powers of like bases;
n and p are natural numbers; and m is a rational number.
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4. The sum, difference, product and quotient of expressions containing square
roots
To calculate the sum, difference, product and quotient of an arithmetic
expression containing a maximum of four radicals with the same root index.
When calculating a quotient, the denominator must be rationalized, if
necessary.
5. Operations on polynomials containing square roots
To perform operations on a numerical expression containing a
maximum of two polynomials and three square roots, and then
reducing it to simplest form.
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DIAGNOSTIC TEST ON THE PREREQUISITES
Instructions
1. Answer as many questions as you can.
2. You may use a calculator.
3. Write your answers on the test paper.
4. Do not waste any time. If you cannot answer a question, go on
to the next one immediately.
5. When you have answered as many questions as you can, correct
your answers using the answer key which follows the diagnostic
test.
6. To be considered correct, your answers must be identical to
those in the answer key. In addition, the various steps in your
solution should be equivalent to those shown in the answer key.
7. Copy your results onto the chart which follows the answer key.
This chart gives an analysis of the diagnostic test results.
8. Do only the review activities that apply to your incorrect
answers.
9. If all your answers are correct, you may begin working on this
module.
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1. Perform the following operations.
a) 3 × 1 = ........................................................................................................
2 4
b) 8.02 × 4.1 = ........................
c) 5.12 ÷ 0.2 = ........................
d) – 7 ÷ – 5 = ................................................................................................
2
8
2. Find all the factors of the numbers below.
a) 16: ................................................................................................................
b) 21: ................................................................................................................
c) 32: ................................................................................................................
d) 144: ..............................................................................................................
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3. Perform the following operations.
a) 2 + 7 = .......................................................................................................
3 8
b) – 3.56 – 2.49 = ..............................................................................................
4. Subtract the following polynomials. The resulting algebraic expression
must be in simplest form. Show all the steps in your solution.
2y – (8x – 7y – 3x)
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
5. Perform the required operations on the following polynomials.
The
resulting algebraic expression must be in simplest form. Show all the steps
in your solution.
a) 2a(7a – 5)
b) (5x + 4)(3x – 2)
...................................................
.......................................................
...................................................
.......................................................
...................................................
.......................................................
...................................................
.......................................................
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c) (5x – 3)(5x + 3)
Exponents and Radicals
d) (3y – 4)2
...................................................
.......................................................
...................................................
.......................................................
...................................................
.......................................................
...................................................
.......................................................
...................................................
.......................................................
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ANSWER KEY FOR THE DIAGNOSTIC TEST ON THE
PREREQUISITES
1. a) 3 × 1 = 3
2 4 8
b) 8.02 × 4.1 = 32.882
c) 5.12 ÷ 0.2 = 25.6
d) – 7 ÷ – 5 = – 7 × – 2 = 14 = 7
8
2
8
5
40 20
2. a) 16: 1, 2, 4, 8, 16
b) 21: 1, 3, 7, 21
c) 32: 1, 2, 4, 8, 16, 32
d) 144: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
3. a) 2 + 7 = 16 + 21 = 37 or 1 13
24
3 8 24 24 24
b) – 3.56 – 2.49 = – 6.05
4. 2y – (8x – 7y – 3x)
2y – (5x – 7y)
2y – 5x + 7y
9y – 5x
5. a) 2a(7a – 5)
b) (5x + 4)(3x – 2)
(2a × 7a) + (2a × (– 5))
5x(3x – 2) + 4(3x – 2)
14a2 – 10a
5x × 3x + 5x × (– 2) + 4 × 3x + 4 × (– 2)
15x2 – 10x + 12x – 8
15x2 + 2x – 8
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d) (3y – 4)2
c) (5x – 3)(5x + 3)
5x(5x + 3) – 3(5x + 3)
(3y – 4)(3y – 4)
5x × 5x + 5x × 3 – 3 × 5x – 3 × 3
3y(3y – 4) – 4(3y – 4)
25x2 + 15x – 15x – 9
3y × 3y + 3y × (– 4) – 4 × 3y + (–4)(–4)
25x2 – 9
9y2 – 12y – 12y + 16
9y2 – 24y + 16
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ANALYSIS OF THE DIAGNOSTIC TEST RESULTS
Answers
Questions Correct
Incorrect
1. a)
b)
c)
d)
2. a)
b)
c)
d)
3. a)
b)
4.
5. a)
b)
c)
d)
Review
Section
7.1
7.1
7.1
7.1
7.2
7.2
7.2
7.2
7.3
7.3
7.4
7.5
7.5
7.5
7.5
Page
Before Going on to
7.4
7.4
7.4
7.4
7.13
7.13
7.13
7.13
7.18
7.18
7.28
7.31
7.31
7.31
7.31
Unit 1
Unit 1
Unit 1
Unit 1
Unit 4
Unit 4
Unit 4
Unit 4
Unit 4
Unit 4
Unit 4
Unit 5
Unit 5
Unit 5
Unit 5
• If all your answers are correct, you may begin working on this module.
• For each incorrect answer, find the related section listed in the Review
column. Do the review activities for that section before beginning the unit
listed in the right-hand column under the heading Before Going on to.
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INFORMATION
FOR
EDUCATION STUDENTS
DISTANCE
You now have the learning material for MTH-4105-1 and the relevant homework
assignments. Enclosed with this package is a letter of introduction from your
tutor, indicating the various ways in which you can communicate with him or her
(e.g. by letter or telephone), as well as the times when he or she is available. Your
tutor will correct your work and help you with your studies. Do not hesitate to
make use of his or her services if you have any questions.
DEVELOPING EFFECTIVE STUDY HABITS
Learning by correspondence is a process which offers considerable flexibility, but
which also requires active involvement on your part. It demands regular study
and sustained effort. Efficient study habits will simplify your task. To ensure
effective and continuous progress in your studies, it is strongly recommended
that you:
• draw up a study timetable that takes your work habits into account and is
compatible with your leisure and other activities;
• develop a habit of regular and concentrated study.
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The following guidelines concerning theory, examples, exercises and
assignments are designed to help you succeed in this mathematics course.
Theory
To make sure you grasp the theoretical concepts thoroughly:
1. Read the lesson carefully and underline the important points.
2. Memorize the definitions, formulas and procedures used to solve a given
problem; this will make the lesson much easier to understand.
3. At the end of the assignment, make a note of any points that you do not
understand using the sheets provided for this purpose. Your tutor will then
be able to give you pertinent explanations.
4. Try to continue studying even if you run into a problem. However, if a major
difficulty hinders your progress, contact your tutor before handing in your
assignment, using the procedures outlined in the letter of introduction.
Examples
The examples given throughout the course are applications of the theory you are
studying. They illustrate the steps involved in doing the exercises. Carefully
study the solutions given in the examples and redo the examples yourself before
starting the exercises.
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Exercises
The exercises in each unit are generally modeled on the examples provided. Here
are a few suggestions to help you complete these exercises.
1. Write up your solutions, using the examples in the unit as models. It is
important not to refer to the answer key found on the coloured pages at the
back of the module until you have completed the exercises.
2. Compare your solutions with those in the answer key only after having done
all the exercises. Careful! Examine the steps in your solutions carefully,
even if your answers are correct.
3. If you find a mistake in your answer or solution, review the concepts that you
did not understand, as well as the pertinent examples. Then redo the
exercise.
4. Make sure you have successfully completed all the exercises in a unit before
moving on to the next one.
Homework Assignments
Module MTH-4105-1 comprises three homework assignments. The first page of
each assignment indicates the units to which the questions refer.
The
assignments are designed to evaluate how well you have understood the
material studied. They also provide a means of communicating with your tutor.
When you have understood the material and have successfully completed the
pertinent exercises, do the corresponding assignment right away. Here are a few
suggestions:
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1. Do a rough draft first, and then, if necessary, revise your solutions before
writing out a clean copy of your answer.
2. Copy out your final answers or solutions in the blank spaces of the document
to be sent to your tutor. It is best to use a pencil.
3. Include a clear and detailed solution with the answer if the problem involves
several steps.
4. Mail only one homework assignment at a time.
After correcting the
assignment, your tutor will return it to you.
In the section “Student’s Questions,” write any questions which you wish to have
answered by your tutor. He or she will give you advice and guide you in your
studies, if necessary.
In this course
Homework Assignment 1 is based on units 1 and 2.
Homework Assignment 2 is based on units 3 to 5.
Homework Assignment 3 is based on units 1 to 5.
CERTIFICATION
When you have completed all your work, and provided you have maintained an
average of at least 60%, you will be eligible to write the examination for this
course.
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START
UNIT 1
THE LAWS OF EXPONENTS
1.1
LEARNING CONTEXT
Numbers That Defy the Imagination
While reading an article in a newspaper or science magazine, or a chemistry or
physics textbook, you may notice that the author used “astronomical” or
“infinitesimal” numbers to describe various facts.
Here are some examples.
• In 2002, the federal government debt was $421 000 000 000 (421 billion),
which is equivalent to $14 000 per inhabitant, regardless of age.
• The distance between our planet and the sun is 150 000 000 km, whereas
Alpha Centauri is located 40 400 000 000 000 000 000 km away from the sun.
• The human body contains about 15 000 000 000 000 blood corpuscles, also
called red blood cells; each of these cells has a diameter of 0.007 5 mm.
.
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• Each cubic centimetre of air (the equivalent of a thimbleful) we breathe
contains 27 000 000 000 000 000 000 molecules.
• The HTLV-III retrovirus ( a type of virus), the transmitting agent for AIDS,
has a diameter of 0.000 042 5 mm.
Fortunately, there is a method for writing such large numbers. This method is
called scientific notation. The table below shows the above numbers written
in scientific notation.
$
$421 000 000 000 = 4.21 × $1011
150 000 000 km = 1.5 × 108 km
CH3
OH
15 000 000 000 000 = 1.5 × 1013
0.007 5 mm = 7.5 × 10-3 mm
27 000 000 000 000 000 000 = 2.7 × 1019
0.000 042 5 mm
4.25 × 10-5 mm
Figure 1.1 From the extremely large to the extremely small
1.2
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Later in this unit we will look at how to express a number in scientific notation.
For the time being, let’s spend some time examining exponents, which are an
essential part of this type of notation.
An exponent is a numerical or algebraic expression that
indicates how many times a quantity is to be multiplied by itself.
Thus, in the expression 52 = 5 × 5, the exponent is 2.
We know that 22 (2 squared) = 4 because 2 × 2 = 4. We also know that
23 (2 cubed) = 8 because 2 × 2 × 2 = 8. In the expression 32 = 9, the base is 3 and
the exponent is 2.
In an algebraic or numerical expression, the base is the number
or variable with an exponent.
To calculate the power of a number, we apply the definition of exponents.
☞
{
an = a × a × ... × a
n times
Note
The base can be represented by any type of number: a natural number ( ), an
integer ( ), a fraction or a decimal ( ).
Example 1
2
5
3
= 2 × 2 × 2 = 8
125
5
5
5
(3.5)2 = (3.5) × (3.5) = 12.25
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Answer Key
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Exponents and Radicals
To achieve the objective of this unit, you must be able to solve various
forms of exponential expressions by applying the laws of exponents.
?
What is the value of (–2)2? ...............................................................................
?
What is the value of (–2)3? ...............................................................................
Your answers to the above questions should be 4 and – 8, since
(– 2)2 = (– 2) × (– 2) = 4 and (– 2)3 = (– 2) × (– 2) × (– 2) = 4 × (– 2) = – 8.
Law of signs in multiplication
(+)
×
(+)
=
+
(–)
×
(–)
=
+
(+)
×
(–)
=
–
(–)
×
(+)
=
–
Remember to pay particular attention to parentheses and signs when solving
expressions.
Example 2
23 = 2 × 2 × 2 = 8. The base is 2 and the exponent is 3.
– 23 = – (2 × 2 × 2) = – 8. The base is 2 and the exponent is 3.
(– 2)3 = (– 2) × (– 2) × (– 2) = – 8. The base is – 2 and the exponent is 3.
?
Calculate:
34 =
.....................................................................................
– 34 =
.....................................................................................
(– 3)4 =
.....................................................................................
You probably obtained 81, – 81 and 81 respectively, since:
34 = 3 × 3 × 3 × 3 = 81
– 34 = –( 3 × 3 × 3 × 3) = – 81
(– 3)4 = (– 3) × (– 3) × (– 3) × (– 3) = 81
1.4
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Exponents and Radicals
You no doubt noticed that the last result is positive even though the base is
negative. Indeed, any negative base with an even-numbered exponent yields
a positive result; however, a negative base with an odd-numbered exponent
yields a negative result.
• Any power with a positive base is positive.
• Any power with a negative base is positive if the exponent
is an even number.
• Any power with a negative base is negative if the exponent
is an odd number.
Now, try your hand at the following exercises to check your understanding of the
above concepts.
Exercise 1.1
1. Calculate the following expressions.
2
3
4
a) 23 =
...............................
b)
c) 13 =
...............................
d) (– 1.4)2 =
e) (– 5)3 =
...............................
=
..............................
..............................
2. Is the value of the following expressions positive or negative?
a)
1
2
3
:
.............................
c) (– 0.35)3: .............................
e) – 1.124:
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b) (– 0.5)2:
..............................
d) (– 1.12)4:
..............................
.............................
1.5
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Answer Key
Exponents and Radicals
Now let’s look at the laws of exponents. The first two laws were covered in a
previous course; however, we will review them here and examine them in more
depth.
First Law of Exponents
When two or more powers with like bases are multiplied, the
product has the same base and the exponents are added. In
general:
am × an = am + n
Example 3
Calculate t3 × t4.
We know that t3 = t × t × t and t4 = t × t × t × t.
Thus, t3 × t4 = t × t × t × t × t × t × t = t7.
By applying the first law of exponents, we get: t3 × t4 = t3 + 4 = t7.
Example 4
Calculate 22 × 23 × 2.
22 × 23 × 2 = 22 × 23 × 21 = 22 + 3 + 1 = 26 = 64
Example 5
Calculate (– y)5 × (– y)8.
(– y)5 × (– y)8 = (– y)5 + 8 = (– y)13 = – y13
1.6
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Exponents and Radicals
1
What is the product of x 2 × x 3 ? .......................................................................
The first law of exponents always applies, regardless of the nature of the
1
1
1
1
5
exponents. Thus, x 2 × x 3 = x 2 + 3 = x 6 .
Example 6
Calculate – 3y4 × 2y3.
First, we multiply the numerical coefficients of the terms. Then, we apply
the first law of exponents.
– 3y4 × 2y3 = – 6y4y3 = – 6y4 + 3 = – 6y7
Example 7
Calculate (– 2)3 × 25.
Since the bases are not identical, we must first determine the sign of (– 2)3. We
can then apply the first law of exponents.
(– 2)3 × 25 = – 23 × 25 = – 28 = – 256
Second Law of Exponents
When two powers with like bases are divided, the quotient
has the same base and the exponents are subtracted. In
general:
a m = am – n, a ≠ 0
an
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Exponents and Radicals
Example 8
5
Calculate a 3 .
a
a 5 = a × a × a × a × a = a × a = a2
a×a×a
1
a3
5
If we apply the second law of exponents we get: a 3 = a5 – 3 = a2.
a
?
3
What is the quotient of x 4 ?
x
?
2y – 2
What is the quotient of
? ...................................................................
4y3
...................................................................
2
a3
?
Calculate the quotient of
?
2
Calculate the quotient of 6 2 . ...................................................................
6
?
3y 4
Calculate the quotient of
. ..................................................................
6y3
1
. ...................................................................
a4
You probably obtained the following answers:
1 , 1 y – 5, a 125 , 1 and 1 y .
x 2
2
The last quotient is calculated by first simplifying the numerical coefficients.
3y 4 1y 4
=
. We then apply the second law of exponents:
6 y 3 2y 3
y
1y 4 1y 4 – 3 1y
=
=
or
or 1 y .
3
2
2
2
2
2y
We get:
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Exponents and Radicals
Until now, we have used numbers that are relatively simple to calculate.
However, this is not always the case in mathematics or science. These subjects
often involve operations on very small or very large numbers, making manual
calculations long and tedious. Fortunately, we have calculators to make our task
easier. In the following examples, you will learn how to use a calculator to
determine the numerical value of exponential expressions.
Example 9
37 × 32 = 37 + 2 = 39
To determine the value of this expression, key in:
3
yx
9
=
19683
Therefore: 39 =
19 683
Example 10
(– 2)14
(
2
+/–
)
yx
1
4
=
16 384
Example 11
– 214
We know that the value is negative. We key in: 2
yx
1
4
=
16 384
and write – 16 384.
N.B. Be careful if you are using a calculator to find – 214, as there are two
types of calculators.
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1.9
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• Conventional calculator: 2
+/–
yx
Exponents and Radicals
1
4
= .
• Direct algebraic logic (DAL) calculator:
+/–
2
yx
1
4
= .
Until now, we have looked at bases with positive exponents. But what happens
when the exponent is negative? In your opinion, what is the value of the
expression a– 2? This leads us to the third law of exponents.
Third Law of Exponents
When the base has a negative exponent, the base is inverted
and the exponent becomes positive.
a– m = 1m where m ∈
a
+
We can now answer the above question.
a–2 = 12
a
This law stems from the second law of exponents.
3
For example, a 5 = a3 – 5 = a– 2.
a
3
×a×a
1
1
But a 5 = a × a
a × a × a × a = a × a = a2
a
Therefore, a– 2 = 12 .
a
?
Calculate 3– 2. ....................................................................................................
Your answer should be 1 , since 3– 2 = 12 = 1 .
9
9
3
1.10
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3
Your
–4
1
3
Calculate
Exponents and Radicals
. ..............................................................................................
result
should
be
81,
since
1
3
–4
=
3 × 3 × 3 × 3 = 81 = 81.
1
1
1
1
1
1
1
3
4
= 3
1
4
=
Example 12
–3
Convert – 1
4
.
The base is – 1 and its reciprocal is – 4 .
4
1
Thus – 1
4
–3
= –4
1
3
= – 4 × – 4 × – 4 = – 64 = – 64.
1
1
1
1
Example 13
Convert 2
5
–3
.
The base is 2
5
Thus 2
5
–3
= 5
2
and its reciprocal is 5 .
2
3
= 5 × 5 × 5 = 125 .
2
2
2
8
Example 14
Convert 5a– 2.
The base is a and its reciprocal is 1
a.
Thus, 5a– 2 = 5 × 12 = 52 .
a
a
N.B. Here, the base with the negative exponent is a. The number 5 is the
numerical coefficient. It therefore remains in the numerator position.
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1.11
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Answer Key
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Exponents and Radicals
Can you identify which of the following statements is true?
a) 3– 1 = – 3: .................
d)
–5
4
–2
= – 16 : .......
25
b) (– 2)– 1 = 2: ................ c)
–2
5
e)
–3
7
3
–1
= – 3 : ...............
7
= – 125 : ....... f) (– 1)– 1 = 1: ..................
8
If you answered that only statement e) is true, you’re right! Indeed,
b) (– 2)– 1 = – 1
2
a) 3– 1 = 1
3
d)
–5
4
–2
= –4
5
2
= 16
25
e)
–2
5
–3
c)
= –5
2
3
= – 125
8
7
3
–1
= 3
7
f) (– 1)– 1 = – 1
The first two laws of exponents remain valid when the expressions involved have
negative exponents.
Example 15
Calculate
y–3
.
y–7
According to the second law of exponents,
y–3
= y–3 – (–7) = y–3 + 7 = y4.
y–7
Example 16
Calculate a × a– 3.
According to the first law of exponents, a × a– 3 = a1 × a– 3 = a1 + (– 3) = a– 2 = 12 .
a
We have seen what to do with negative exponents. But what about the exponent
0? In your opinion, what is the value of 50? To find out, let’s look at the fourth
law of exponents.
1.12
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Fourth Law of Exponents
Any expression other than 0 that has the exponent 0 is equal
to 1.
a0 = 1 where a ≠ 0
We can now answer the above question.
50 = 1
In fact, this law stems from the second law of exponents.
3
For example, 5 3 = 53 – 3 = 50.
5
3
But 5 3 = 5 × 5 × 5 = 125 = 1, therefore 50 = 1.
5 × 5 × 5 125
5
It’s now time to apply what you have learned. The following exercises will allow
you to practice working with the first four laws of exponents.
Exercise 1.2
1. Express the product of the following powers in exponential form and calculate the value of the resulting expression, if necessary. Your answers should
not contain any negative exponents.
a) 53 × 52 =
..............................................................................
b) 32 × 3– 1 =
..............................................................................
c)
7
4
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3
× 7
4
–5
=
..............................................................................
1.13
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1
Exponents and Radicals
d) x 4 × x 2 =
..............................................................................
e) 2y–2 × 3y3 =
..............................................................................
1
5
f) (– 4) 3 × (– 4) 3 =
..............................................................................
g) (– 2.5)3 × 2.5 =
..............................................................................
h) (– 3.25) × (– 3.25)2 =
..............................................................................
i) – 0.2x3 × 0.5x2 =
..............................................................................
j) 5x3 × x– 7 × 2x =
..............................................................................
k) 1 x 2 × – 1 x 3 × 2x – 4 = ..............................................................................
3
2
2. Express the quotients of the following powers in exponential form.
3
a) 3 2 =
3
b)
.........................................................................................
(– x)4
=
2
(– x)
.........................................................................................
3
2
c) 4a 1 =
2a 2
.........................................................................................
d) – 5t8 ÷ t6 =
.........................................................................................
4
4
e) 10 y 5 ÷ 5y 5 =
.........................................................................................
3.5a 3 =
0.7a – 3
.........................................................................................
f)
g) 1.5x2 ÷ (– 0.3x) = .........................................................................................
1.14
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3
h) – 3a3 =
6a
.........................................................................................
i) – 33y8 ÷ 11y9 =
.........................................................................................
j)
2 a 6 ÷ – 1 a 4 = .........................................................................................
3
2
Did you know that…
… the number 10 is the base of the “decimal” number
system? This means that all possible numbers can be
written with only 10 digits. Why was this base chosen
rather than another? It appears that the reason is one of
convenience. Almost all historians agree that this system was adopted
because humans have 10 fingers.
The Celts used a base 20 number system, very likely because they used
their fingers and toes to count. Traces of this number system were found
in the old British monetary system, in which 1 pound equalled 20
shillings.
Base 5 is also widely used. In many languages the words “five” and “hand”
are similar or have a common root. For example, “pentcha” means “hand”
in Persian and “pantche” means “five” in Sanskrit.
However, bases 5, 10 and 20 were not necessarily used worldwide. Certain
tribes in Africa, South America and Australia used the base 2 (binary)
system while others used the base 3 (tertiary) system because their
calculations were based on the 3 phalanxes of the fingers opposite the
thumb.
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1.15
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Exponents and Radicals
Base 4, while more rare, is still used by the Yiki Indians of California who
count by using the spaces between the fingers of one hand.
The most complicated number system is unquestionably that of the
Mesopotamian civilization. It is called the sexagesimal system because its
base is 60. Sixty different signs are required to represent the numbers
from 0 to 59. Despite its complexity, this system is still used to measure
the radian of an angle in degrees or the passing of time. There are 60
minutes in an hour and 60 seconds in a minute. It is obvious that if your
watch reads 5:07:09, then 5 hours, 7 minutes and 9 seconds have elapsed.
What is not so obvious is that in Mesopotamian notation, 5, 7, 9 mean:
5 × 602 + 7 × 601 + 9 × 600
or 18 429 in our decimal system. If 5 hours, 7 minutes and 9 seconds have
gone by, then 18 429 seconds have elapsed, starting from midnight.
SCIENTIFIC NOTATION
Before we examine the last three laws of exponents, let’s look at how to convert
a decimal number to scientific notation and vice versa.
• When we multiply a decimal number by a power of 10 (i.e. 10,
100, 1 000, 10 000, and so on), the decimal point is moved
to the right by as many places as there are zeros in the power
of 10.
Thus, – 3.29 × 10 = – 32.9
41.7 × 100 = 4 170
1.16
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Exponents and Radicals
• When we divide a decimal number by a power of 10, the
decimal point is moved to the left by as many places as there
are zeros in the power of 10.
Thus, 5.187 ÷ 1 000 = 0.005 187
8 ÷ 100 = 0.08
N.B. Zeros must be added where there are no digits.
Use your calculator to find 616.
6
yx
1
6
=
2.821109907 12
N.B. Some calculators cannot display more than 8 digits show a result of
2.8211099 12.
You may have guessed what these results mean. It is the answer written in
scientific notation. The number 2.821109907 12 stands for 2.821 109 907 × 1012.
With this notation, the calculator displays an approximation of the exact result,
which could not be displayed otherwise. The exact result is: 2 821 109 907 456.
N.B.
To display 2.821109907 × 1012 directly on your calculator, key in
2.821109907 and press EXP 1
2 .
In scientific notation, a number is always expressed in the
form a × 10n, where n is an integer and a is a decimal, such that
1 ≤ a < 10.
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1.17
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Answer Key
Exponents and Radicals
Let’s now look at how to convert a number expressed in scientific notation to a
decimal.
Example 17
Convert 7.547 8 × 102 to a decimal.
You need only move the decimal point two places to the right since the base
10 exponent is 2.
7.547 8 × 102 = 754.78
Indeed, 7.547 8 × 102 = 7.5478 × 100 = 754.78.
What happens if the base 10 exponent is negative?
Example 18
Convert 5.789 × 10– 4 to a decimal.
Simply move the decimal point four places to the left since the base 10
exponent is – 4.
0 0005.789 × 10– 4 = 0.000 578 9
1
Indeed, 5.789 × 10– 4 = 5.789 × 1 4 = 5.789 ×
= 5.789 ÷ 10 000 =
10
000
10
0.000 578 9.
1.18
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Exponents and Radicals
Let a be a decimal greater than or equal to 1 such that
1 ≤ a < 10, and let n be a natural number.
The number a × 10n can be converted to a decimal by moving
the decimal point in a to the right by n places.
The number a × 10–n can be converted to a decimal by moving
the decimal point in a to the left by n places.
?
Canada’s debt in 2002 can be expressed as $4.21 × 1011. Can you convert it
to a decimal?
...........................................................................................................................
If your answer is $421 000 000 000, you’re right!
The decimal point must be moved 11 places to the right by adding as many zeros
as necessary. Do you find it more encouraging to have a debt of $4.21 × 1011 or
a debt of $421 000 000 000?
?
The HTLV-III retrovirus has a diameter of 4.25 × 10– 5 mm. Can you write
this as a decimal?
...........................................................................................................................
If your answer is 0.000 042 5 mm, you’re right!
The decimal point must be moved 5 places to the left by adding as many zeros as
necessary.
Now, let’s do the reverse and express a decimal in scientific notation.
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1.19
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Exponents and Radicals
Example 19
Express 7 547.8 in scientific notation.
1. Place the decimal point after the first non-zero digit.
7 547.8
2. Count the number of places you moved the decimal point: 3 places to the
left.
7.547 8
3. Write 3 as the base 10 exponent to show the number of places the decimal
point was moved.
7.547 8 × 103
And that’s pretty much all there is to scientific notation!
The number 7 547.8 is equal to 7.547 8 × 103 since 7 547.8 × 1 000 = 7.547 8 × 103.
Example 20
Convert 0.000 647 to scientific notation.
1. Place the decimal point after the first non-zero digit.
0.000 6 47
2. Count the number of places you moved the decimal point: 4 places to the
right.
0 000 6.47
3. Write – 4 as the base 10 exponent to show the number of places the decimal
point was moved.
6.47 × 10– 4
1.20
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Exponents and Radicals
Simple, isn’t it?
The number 0.000 647 is indeed equal to 6.47 × 10 – 4 since
1
0.000 647 = 6.47 = 6.47 ×
= 6.47 × 1 4 = 6.47 × 10– 4.
10 000
10 000
10
?
Canada’s debt in 2002 was $421 000 000 000. Can you express this in
scientific notation?
...........................................................................................................................
If you answered 4.21 × 1011, you’re right!
You need only move the decimal point 11 places to the left (4.21 000 000 000) and
then write 11 as the base 10 exponent: 4.21 × 1011.
This is indeed the amount of Canada’s debt since: $421 000 000 000 =
4.21 × $100 000 000 000 = $4.21 × 1011.
?
Now try converting the diameter of the HTLV-III retrovirus to scientific
notation: 0.000 042 5 mm?
...........................................................................................................................
If you answered 4.25 × 10– 5, you’re right!
You simply move the decimal point 5 places to the right (0000 04.25) and then
write – 5 as the base 10 exponent: 4.25 × 10– 5.
This is, in fact, the diameter of the HTLV-III retrovirus since 0.000 042 5 mm =
4.25 mm = 4.25 ×
1
mm = 4.25 × 1 5 mm = 4.25 × 10– 5 mm.
100 000
100 000
10
?
Can you now explain why your calculator displays 2.821109907 12 when you
calculate 616?
...........................................................................................................................
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1.21
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Answer Key
Exponents and Radicals
In fact, 2.821109907 12 stands for 2.821109907 × 1012. When we convert this
result to a decimal, we get 2 821 109 907 000, since the decimal point was moved
12 places to the right. This result is an approximation of the exact result:
2 821 109 907 456.
N.B. This integer has 13 digits since we moved the decimal point 12 places to the
right.
To express a number in scientific notation:
1. Place the decimal to the right of the first non-zero digit.
2. Count the number of places the decimal point was moved.
3. Write this number as the base 10 exponent:
a) this exponent is positive if the decimal point was
moved to the left;
b) this exponent is negative if the decimal point was
moved to the right.
Exercise 1.3
1. Convert the following numbers to scientific notation.
a) 142 857 = .................................... b) 1 230 000 000 = .............................
c) 0.054 = ........................................ d) 0.007 21 = ......................................
1.22
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2. Convert the following numbers to decimals.
a) 5.1 × 105 = ................................... b) 7.654 3 × 103 = ...............................
c) 8.193 × 10– 4 = .............................. d) 4 × 10– 7 =........................................
3. Which number is larger: 0.000 007 2 or 7.1 × 10– 5?
...........................................................................................................................
...........................................................................................................................
4. The speed of sound in dry air at standard temperature and pressure is about
331.4 m/s. Convert this number to scientific notation.
...........................................................................................................................
5. The speed of light in a vacuum is about 2.997 924 6 × 108 m/s. Convert this
number to a decimal.
...........................................................................................................................
6. Express 532 in scientific notation by keeping three digits after the decimal
point.
...........................................................................................................................
7. How many digits are in 2131?
...........................................................................................................................
...........................................................................................................................
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Exponents and Radicals
Now let’s look at the last three laws of exponents.
You are now ready to tackle a greater level of difficulty. Problem-solving in
mathematics or science sometimes requires mastery of another law of exponents. For example, how would you solve an expression of the form (x2)3?
We know that x2 = x × x.
If this expression is itself cubed (power of three), then we get:
x2
×
x2
×
x2
(x × x)
×
(x × x)
×
(x × x)
which is equivalent to x6. Therefore, (x2)3 = x2 × 3 = x6
We have just discovered the fifth law of exponents.
Fifth Law of Exponents
To raise a power m with base a to a power n, multiply the
exponents. In general:
(am)n = am × n
Example 21
(24)3 = 24 × 3 = 212
To determine the value of this expression, key in:
2
yx
1
2
=
4096
Therefore: (24)3 = 212 = 4 096
1.24
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Example 22
– 3(a2)6 = – 3a2 × 6 = – 3a12
☞
Here, the negative sign applies only to the numerical coefficient of
base a. We must therefore keep the negative sign throughout the
calculations.
Example 23
((– 3)2)– 2 = (32)– 2 = (3)2 × (– 2) = (3)– 4 = 14 = 1
81
3
☞
Here, base – 3 is negative. As the power of a negative base is positive
when the exponent is an even number, the result is therefore positive.
• The power of a positive base is positive.
• The power of a negative base is positive if the exponent is
an even number.
• The power of a negative base is negative if the exponent is
an odd number.
Example 24
–(32)2 = –(32 × 2) = –(34) = – 81
☞
Here the base is positive. Be careful: –(32)2 ≠ [(– 3)2]2. The base is positive
in the first expression, and negative in the second.
To assign an exponent to several variables or numbers, place
these between parentheses. Otherwise, the exponent will
apply only to the closest variable or number.
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1.25
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Exponents and Radicals
?
What is the value of (xyz)2?
.....................................................................
?
Solve (a2y3)2 .
.....................................................................
These problems are slightly more complex. But don’t worry, solving them is
child’s play – or almost!
To determine the value of the first expression:
(xyz)2 = (x × y × z) × (x × y × z) = x2 y2 z2
The second problem is solved as follows:
(a2y3)2 = (a × a × y × y × y) × (a × a × y × y × y) = a4y6
As you may already have guessed, there is yet another law of exponents that can
be used to solve this type of problem.
Sixth Law of Exponents
To raise a product in exponential form to a power n, multiply
the exponents of each of the factors by n. In general:
(abc)m = am bm cm
where a, b et c represent exponential expressions.
Example 25
1
2
1
2
1
2
1
2
3a 2b 2
3a 2b 2
3a 2b 2
3a 2b 2
1
2
2
= 3 2 × (a ) × b 2
2
• 6th law of exponents
1 ×2
= 32 × a2 × 2 × b 2
• 5th law of exponents
= 9 × a4 × b
= 9a4b
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Example 26
(x2y– 2)3 = (x2)3 × (y– 2)3
• 6th law of exponents
(x2y– 2)3 = x2 × 3 × y– 2 × 3
• 5th law of exponents
(x2y– 2)3 = x6 × y– 6
6
(x2y– 2)3 = x 6
y
• 3rd law of exponents
N.B. In future, we will apply the fifth and sixth laws at the same time. We
will therefore go directly from (x2y– 2)3 to x2 × 3 × y– 2 × 3.
Example 27
(– x2 y)3 = (– 1)3 × x2 × 3 × y1 × 3 = – x6 y3
Example 28
(– a3 b2)2 = (– 1)2 × a3 × 2 × b2 × 2 = a6 b4
☞
Be careful!
Always remember that an expression of the type (2abc)2 stands for
(2 × a × b × c)2. By applying the sixth law of exponents, we get
(22 × a2 × b2 × c2). The expression 2(abc)2 is equivalent to 2 × a2 × b2 × c2.
Watch out for the parentheses!
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Example 29
(2x3 3y– 2 4z2)– 2 = 2– 2 × x3 × (– 2) × 3– 2 × y– 2 × (– 2) × 4– 2 × z2 × (– 2)
(2x3 3y– 2 4z2)– 2 = 12 × x – 6 × 12 × y 4 × 12 × z – 4
2
3
4
(2x3 3y– 2 4z2)– 2 = 1 × 16 × 1 × y 4 × 1 × 14
4 x
9
16 z
(2x3 3y– 2 4z2)– 2 =
y4
576x 6 z 4
You will now learn a useful trick for simplifying some of your calculations. In
science and mathematics, we must often simplify numerical or algebraic
expressions. Thus (22 × 8)2 = 22 × 2 × 8. So far, no problem. However, 82 can also
be written (23)2 because 8 = 2 × 2 × 2. The purpose of this type of conversion is
to obtain a single base in order to make your calculations easier. Therefore
22 × 2 × 82 = 24 × (23)2 = 24 × 26 = 210.
Convert each of the following expressions to expressions with a single base.
?
(3 × 272)3
.......................................................................................................
?
(16 × 82)2
.......................................................................................................
The first expression can be converted to [3 × (33)2]3 because 27 = 3 × 3 × 3 = 33.
[3 × (33)2]3 = (3 × 36)3 = 33 × 318 = 321 or (3 × 36)3 = (37)3 = 321.
To simplify the second expression, you must convert it using base 2.
(16 × 82)2 = [24 × (23)2]2 = (24 × 26)2 = 28 × 212 = 220 or (24 × 26)2 = (210)2 = 220.
As you can see, it’s quite simple. Pay close attention to the parentheses and the
signs of the bases, the numerical coefficients and the exponents, and presto! Now
let’s see if you have understood the concepts you have learned so far.
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Exercise 1.4
1. Simplify the following expressions.
a) (x2)3 =
.........................
b) (y– 3)– 2 =
c) 3(a6)2 =
.........................
d) –(x 3) 3 =
e) (– y2)2 =
.........................
.............................
1
.............................
2. Calculate the value of the following expressions by applying the appropriate
law of exponents.
a) (– 23)– 1 = .........................
b) (– 42)3 =
c) (3– 4)– 1 =
d)
e)
1
22
.........................
1
44
...........................
4
=
...........................
–2
= .........................
3. The following products are in exponential form. Raise them to the required
power.
a) (x2y3)2 = ....................................
c)
m 3n 6 p 3
1
3
= ...........................
1
b) –2(a 8b 4) 2 = ..............................
d) (a– 1b)– 3 = ..................................
e) (x– 1yz2)– 2 = ...............................
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4. Calculate the value of the following expressions by applying the appropriate
law of exponents.
a) (42 × 33)2 = ....................................................................................................
b)
1
4 2
(– 1.5) × (– 2.2)
2
= ...................................................................................
c) (– 34 × 52)2 = ..................................................................................................
d)
1
35 – 1 × 7 2
–2
= .............................................................................................
5. Simplify and convert the following expressions to exponential form.
a) (22 × 4)3 =......................................................................................................
b) (33 × 92)–1 = ...................................................................................................
c) (4 × 162 × 33)2 = ............................................................................................
d)
e)
1
2
1
2
2
× 1
8
8 × 16
1
3
3
= ..............................................................................................
1
2
= ..............................................................................................
There is one last case involving both a quotient and exponents. In your opinion,
2 3
what is the quotient of a 3 ? To answer this question, you need to know the
b
seventh and last law of exponents.
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Seventh Law of Exponents
To raise a quotient of two exponential expressions to a
power n, multiply the exponents of each expression by n. In
general:
a
b
m
m
= a m where b ≠ 0
b
Determine the quotient of each of the following expressions.
?
x
y2
?
3a 2
4b 3
3
= .............................................................................................................
2
= ...........................................................................................................
To obtain the first result, proceed as follows:
x
y2
3
3
3
= x2 × 3 = x 6
y
y
To determine the second quotient, proceed as follows:
3a 2
4b 3
2
2
4
2 ×2
= 3 2 × a 3 × 2 = 9a 6
4 ×b
16b
Obviously, we can always make things more difficult! Here are some examples.
Example 30
x–2
y3
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– 2 × (– 3)
6
6
y9
= x 3 × (– 3) = x–9 = x = x 6 ×
= x6y9
1
1
y
y
y9
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Example 31
a2
3
(– b)
3
a 2 × 3 = a 6 = a 6 = – a 6 (because the exponent is an odd
3×3
9
9
9
(– b)
(– b)
–b
b
=
number)
Since there’s nothing like practice, try your hand at a few exercises involving the
seventh law of exponents.
Exercise 1.5
1. The quotients below are in exponential form. Raise them to the required
power.
a)
m
n
c)
y–2
x3
e)
4
=
............................
b)
–2
x2
y3
1
=
............................
d)
a2
1
2
=
..............................
=
..............................
3
b4
3
x4
3
(– y)
= ............................
2. Calculate the following expressions by applying the appropriate exponent
law of exponents.
3
a)
3
2
b)
(– 6)3
2
(– 4)
c)
3– 2
43
=
................................................................................................
3
=
2
=
2
(– 2)
d) –
(– 5)3
e)
2.5 4
0.5 2
................................................................................................
1
2
................................................................................................
2
= ................................................................................................
=
................................................................................................
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Before you continue, here is a summary of the laws of exponents and the rules
for applying them.
Laws of Exponents
1. am × an = am + n
2.
am
an
= am – n
3. a–m = 1
a
m
where 1m
a
4. a0 = 1 where a ≠ 0
5. (am)n = am × n
6. (abc)m = ambmcm
7.
a
b
m
m
= a m where b ≠ 0
b
Rules for Applying the Laws of Exponents
1. The power of a positive base is positive.
2. The power of a negative base is positive if the exponent is
an even number.
3. The power of a negative base is negative if the exponent is
an odd number.
4. When a base has a negative exponent, invert the base and
change the sign of the exponent to obtain an equivalent
expression in which the exponent is positive.
5. When an expression preceded by a negative sign has an
exponent, each term must be raised to this power,
including the digit – 1.
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It can never be repeated enough: signs merit special attention. For example, they
can radically change the result of a chemistry experiment or a financial
statement! The law of signs, while simple, applies and will always apply
regardless of the complexity of the problems to be solved. Speaking of signs…
Did you know that…
…around the year 500 CE, a bank employee of East Indian
origin came up with the ingenious idea of using a sign to
distinguish between assets and liabilities in his clients’
accounts.
All the rules regarding how these numbers are handled were established
and codified by the mathematicians Brahmagupta in the VII century and
Bhaskara in the XII century.
East Indians, lovers of poetry and
mathematics, translated the law of signs into sayings. These sayings are
still used today:
“The friends of my friends are my friends.”
“The enemies of my friends are my enemies.”
“The friends of my enemies are my enemies.”
“The enemies of my enemies are my friends.”
Substitute the + sign for the word “friends” and the – sign for the word
“enemies” and you get the current law of signs for multiplication and
division.
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At the beginning of this unit, we studied the sign of the power of a number. We
will now go a little further in our analysis by describing the value of am depending
on the sign and value of a as well as that of its exponent m.
Example 32
Describe the sign and the value of am knowing that – 1 < a < 0 and m is a
positive even integer.
An example of this is – 4
5
–4
5
2
2
.
= – 4 × – 4 = 16
25
5
5
16 has a positive sign and is between 0 and 1.
25
We can therefore conclude that am has a positive sign and 0 < am < 1.
Example 33
Let’s describe the sign and the value of am knowing that a < – 1 and m is an
odd negative integer.
An example of this situation is – 5
4
–3
.
According to the third law of exponents, – 5
4
–4
5
3
–3
= –4
5
3
.
= – 4 × – 4 × – 4 = – 64
125
5
5
5
– 64 has a negative sign and is between – 1 and 0.
125
We can therefore deduce that am has a negative sign and that – 1 < am < 0.
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Can you describe the sign and value of am knowing that – 1 < a < 0 and that
m is an odd negative integer?
...........................................................................................................................
If you answered am < – 1, you’re right!
A numerical example of this situation is – 4
5
–3
.
–3
3
= –5 .
According to the third law of exponents, – 4
4
5
3
– 5 = – 5 × – 5 × – 5 = – 125 and – 125 is smaller than – 1.
64
4
4
4
4
64
We can therefore conclude that am < – 1.
This method of solving problems is relatively simple; however, sometimes it is
necessary to generalize. There are several ways of doing this.
We know that any base with an even-numbered positive exponent yields a
positive result. Thus, if a > 1 or a < – 1, then am > 1. For example, 52 = 25 and
(– 5)2 = 25. Likewise, if 0 < a < 1 or – 1 < a < 0, then a < am < 1. For example,
2
2
1 = 1 and – 1 = 1 .
5
25
5
25
This, of course, is a fairly easy example.
?
In your opinion, what is the value of am if m > 0 and odd and that a > 1 or that
a < – 1?
...........................................................................................................................
?
Now what is the value of am if m > 0 and odd and that 0 < a < 1 or that
– 1 < a < 0?
...........................................................................................................................
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If you answered am > 1 or am < – 1 to the first question, you’re right. Indeed,
53 = 125 or (– 5)3 = – 125. Likewise, 0 < am < 1 or – 1 < am < 0 in the second case.
3
3
You will be convinced of this if you consider that 1 = 1 and – 1 = – 1 .
5
125
5
125
Is everything clear up to now? Yes? Then we can go one step further and look
at what happens when m < 0.
?
What do we know when m < 0? In other words, what rule do we follow?
...........................................................................................................................
We must invert the base and then use the same reasoning as for the preceding
cases.
Let m be an even-numbered negative exponent. If a > 1 or a < – 1, then
0 < am < 1. If you are not convinced, then consider this: 5– 2 = 12 = 1 and
25
5
(– 5)2 = 1 2 = 1 .
25
(– 5)
However, if m is an odd-numbered negative exponent, 0 < am < 1 or – 1 < am < 0:
5– 3 = 13 = 1 and (– 5)3 = 1 3 = – 1 .
125
125
5
(– 5)
?
What is the value of am if m < 0 and even and if 0 < a < 1 or – 1 < a < 0?
...........................................................................................................................
?
And if m is an odd number?
...........................................................................................................................
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Answer
In the first case, am > 1
1
5
–2
= 5
1
2
= 25
and
–1
5
in the second case a m > 1 or a m < – 1
–1
5
–3
1
5
–2
= –5
1
–3
= 5
1
2
= 25 , while
3
= 125
and
=(– 5)3 = –125 .
Until now, we have proved each statement with a numerical example. Try doing
the following exercise without looking at the above examples.
Exercise 1.6
1. Describe the sign and value of am knowing that:
a) a < – 1 and m is an even positive integer. ..................................................
b) a < – 1 and m is an even negative integer. ................................................
c) – 1 < a < 0 and m is an odd negative integer. ............................................
d) 0 < a < 1 and m is an odd negative integer................................................
e) a < – 1 and m is an odd positive integer.....................................................
The time has come to put what you have learned to the test. The following
practice exercises will tell you whether you have understood the concepts in this
unit. If you have difficulty doing the exercises, reread the relevant passages. You
should have a firm grasp of the content of this unit since it forms the basis for the
material in later units.
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PRACTICE EXERCISES
1. Determine the numerical or algebraic value of the following expressions by
applying the appropriate law of exponents.
N.B. Round off your answers to the nearest thousandth.
a) 0.7– 2 = ...........................................................................................................
b) – 3y2 × y– 1 = ..................................................................................................
1
2
c) 2 2 × – (2) = .................................................................................................
d) 3y2 ÷ 2y = .....................................................................................................
2
e) – 8.1c2 = ......................................................................................................
0.9c
f)
(– 5)
1
2 2
= ....................................................................................................
1
g) (x 2) 3 = .........................................................................................................
h)
32
23
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= ........................................................................................................
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i) (1.43 × 2.2– 1)2 = .............................................................................................
j)
–2
1
m 3n 2 p 2
= ...............................................................................................
k) – 1.6– 3 = ........................................................................................................
l) (– 3)– 2 = .........................................................................................................
m) 3x2 × 2x– 1 = ...................................................................................................
n) 2 y 2 ÷ 1 y 3 = ................................................................................................
3
6
o) 1 y – 2 × 1 y 2 = ..............................................................................................
3
8
–4
p) 3a – 5 = .........................................................................................................
2a
q) 6x3 ÷ 2x3 =.....................................................................................................
r)
1.22
0.4 3
–2
=....................................................................................................
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s) 1.5 × 103 × 0.4 × 10– 2 =.................................................................................
1
3
t)
u)
3
× 1
2
1
8–3 × 2 2
2
× 1
4
4
–1
= ............................................................................
–2
= ...............................................................................................
2. a) In 1988, the federal government debt was about $310 000 000 000.
Express this number in scientific notation.
.......................................................................................................................
b) The mass of the moon is about 7.36 × 1022 kg. Express this number as a
decimal.
.......................................................................................................................
c) The
mass
of
an
electron
at
rest
is
0.000 000 000 000 000 000 000 000 000 000 911 kg. Express this number
in scientific notation.
.......................................................................................................................
d) How many digits are in 345.
.......................................................................................................................
3. What is the sign and the value of am, knowing that:
a) a > 1 and m is an even negative integer. ...................................................
b) – 1 < a < 0 and m is an odd positive integer. .............................................
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4. Which of the following expressions is equal to (5x6)– 2?
A) 252 ......... B) 56 .........
x
x
C) 25x4......... D)
1 .........
25x 12
E) 5x4.........
12
1 .........
F) x ......... G) 112 ......... H) 1 4 ......... I)
25
5x
5x
25x 4
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REVIEW ACTIVITIES
1. In your own words, describe the seven laws of exponents and give an example
of each one.
1. .....................................................................................................................
.....................................................................................................................
.....................................................................................................................
Example: ......................................................................................................
2. .....................................................................................................................
.....................................................................................................................
.....................................................................................................................
Example: ......................................................................................................
3. .....................................................................................................................
.....................................................................................................................
.....................................................................................................................
Example: ......................................................................................................
4. .....................................................................................................................
.....................................................................................................................
.....................................................................................................................
Example: ......................................................................................................
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5. .....................................................................................................................
.....................................................................................................................
.....................................................................................................................
Example: ......................................................................................................
6. .....................................................................................................................
.....................................................................................................................
.....................................................................................................................
Example: ......................................................................................................
7. .....................................................................................................................
.....................................................................................................................
.....................................................................................................................
Example: ......................................................................................................
2. Complete the following sentences by inserting the missing terms or
expressions in the blanks.
a) The power of a positive base is ..................................... . The power of a
negative base is .......................... if the exponent is an even number. The
power of a ............................ base is negative if the exponent is an odd
number.
b) When a base has a negative exponent, simply .........................
........................ the base and change the ........................ of the exponent to
obtain an equivalent expression in which the exponent is ...................... .
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c) Let a be a decimal greater than or equal to 1 and smaller than 10 and let
n be a natural number. The number a × 10n is converted to a decimal by
moving the decimal point in a to the ....................... by n places. The
number a × 10– n is converted to a decimal by moving the decimal point in
a to the ....................... by n places.
d) To express a number in scientific notation:
1. Place the decimal point to the ....................... of the first non-zero digit.
2. Count the number of places the decimal point was moved.
3. Write this number as the base 10 exponent:
a) this exponent is ....................... if the decimal point was moved to
the left;
b) this exponent is ....................... if the decimal point was moved to
the right.
e) Let exponent m be a positive integer:
1. if a > 1, then am ....... 1;
2. if ....................... , then 0 < am < 1.
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THE MATH WHIZ PAGE
Spotlight on Scientific Notation
Convert the numbers in the problems below to scientific notation and
perform the calculations using the same notation. Watch those units!
1. Each second, Niagara Falls releases more than 840 000 000 000
drops of water.
Each of these drops of water contains
1 700 000 000 000 000 000 000 molecules. Calculate the number
of water molecules that fall each second.
..........................................................................................................
..........................................................................................................
2. The earth’s circumference is approximately 40 000 km.
a) If sound travels at a speed of 340 m/s, how much time will it
take a sound to circle the earth?
.....................................................................................................
.....................................................................................................
.....................................................................................................
...............................................................................................
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b) The speed of light is 300 000 km/s. How many times will light
circle the earth in one minute?
.....................................................................................................
.....................................................................................................
.....................................................................................................
.....................................................................................................
.....................................................................................................
.....................................................................................................
.....................................................................................................
.....................................................................................................
.....................................................................................................
.....................................................................................................
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c) We say that an airplane flies at Mach 2 when its speed is double
that of sound. How much time would it take such an airplane
to circle the earth?
.....................................................................................................
.....................................................................................................
.....................................................................................................
.....................................................................................................
.....................................................................................................
.....................................................................................................
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