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History of Math
for the Liberal Arts
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CHAPTER 6
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Two Great Achievements:
Logarithms & Cubic
Equations
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Lawrence Morales
Seattle Central
Community College
2001, Lawrence Morales; MAT107 Chapter 6 – Page 1
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TABLE OF CONTENTS
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TABLE OF CONTENTS ............................................................................................................. 2
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PART 1: Introduction and Non-Western Math, and Evolving Calculating Methods............ 4
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Historical Background................................................................................................................ 4
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Multiplication from India............................................................................................................ 5
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Gelosia Multiplication ................................................................................................................ 6
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PART 2: The Solution of the Cubic and Quartic Equations .................................................. 10
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Historical and Mathematical Background................................................................................ 10
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Cardano and His Gang of Italian Algebraists.......................................................................... 11
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Ars Magna and the Solution of the Cubic................................................................................. 13
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Solving the General Cubic Equation ........................................................................................ 18
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The Solution of the Quartic Equation and Beyond ................................................................... 21
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PART 3: Modern Calculations .................................................................................................. 22
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Decimal Fractions .................................................................................................................... 22
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PART 4: Napier and The Emergence of Logarithms .............................................................. 25
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The Idea and Use of Logarithms............................................................................................... 27
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The Idea Behind Logs ............................................................................................................... 27
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Rule of Logs…One Step Closer to the Tables........................................................................... 28
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The Log Tables.......................................................................................................................... 34
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Larger Numbers and the Log Tables ........................................................................................ 36
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Logs of Very Large Numbers .................................................................................................... 38
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The Antilog Tables .................................................................................................................... 38
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The Evil Twins Meet Each Other…Finally............................................................................... 41
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Complex Calculations............................................................................................................... 41
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PART 5: New Calculating Devices ............................................................................................ 46
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The Slide Rule ........................................................................................................................... 46
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Napier and Other Calculating Devices..................................................................................... 46
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PART 6: Appendix A − Solution of the Quartic ...................................................................... 49
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PART 7: Appendix B – Log and Antilog Tables...................................................................... 51
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PART 8: Appendix C – Napier’s Rods ..................................................................................... 55
2001, Lawrence Morales; MAT107 Chapter 6 – Page 2
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PART 9: Homework ................................................................................................................... 57
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Multiplication with the Gelosia Grid System............................................................................ 57
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Checking Solutions of Cubic Equations.................................................................................... 57
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Using Cardano’s Formula on Depressed Cubics..................................................................... 57
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Using Cardano’s Formula on Non-Depressed Cubics............................................................. 58
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Stevin’s Notation....................................................................................................................... 58
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Basic Logarithm Rules.............................................................................................................. 59
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Using the Log Tables ................................................................................................................ 60
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Using the Antilog Tables........................................................................................................... 60
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Using Log and Antilog Tables to Do Calculations................................................................... 60
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Applications of Logarithms....................................................................................................... 61
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Writing ...................................................................................................................................... 62
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Blank Gelosia Grid ......................................................................Error! Bookmark not defined.
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PART 10: Endnotes .................................................................................................................... 63
2001, Lawrence Morales; MAT107 Chapter 6 – Page 3
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PART 1: Introduction and Non-Western Math, and Evolving
Calculating Methods
Historical Background
After the Greeks, a variety of cultures spread previous mathematical accomplishments and also
created many of their own. For example, in the last three centuries B.C.E., China worked on
ideas of square and cube roots as well as methods of solving systems of linear equations. They
also tackled mathematical surveying techniques during this period. In Egypt, Apollonius
developed a theory of conic sections. From 0 to 400 C.E., Ptolemy made inroads into astronomy
while Hypatia, the famous female mathematician, made a name for herself by commenting and
lecturing on the work of Apollonius and by being a gifted and inspiring teacher of mathematics
and philosophy.
From 400−800, the Italian, Boethius, contributed to the field by writing arithmetic books that
were used by Europe during a time when mathematics in that part of the world was in decline.
His book, Arithmetic taught others about Pythagorean number theory. On the other side of the
world, the Mayan numeration system was developed and used for astronomical purposes. And in
India, a major mathematical movement emerged: Aryabhata advanced trigonometry, while
Brahmagupta made major contributions in mathematics and astronomy. It was also during this
time that the Hindu−Arabic decimal place−value number system began to emerge and gain
popularity.
From 800−1000, India continued its work as it developed algebraic techniques. In what is
modern−day Iraq, Al−Khwarizmi wrote an influential text on algebra whose title actually gives
us the word “algebra.” (The title was “Hisab al-jabr w’al-muqabala.”) Other Islamic
mathematicians were also hard at work, not only preserving and translating ancient Greek texts,
but making their own advances, especially in the area of algebra. During this time, Spain became
a passageway for the Hindu−Arabic numbers into Europe.
From 1000−1200, Islamic mathematics continued to develop with work on what is now known
as Pascal’s triangle, found geometric solutions to certain cubic equations (important in this
chapter), and explore sums of powers. In India, Al−Biruni advanced spherical trigonometry,
while in China, Pascal’s triangle (as it is now called) was used to solve equations. In Spain,
Arabic works were translated into Latin, which would be important in the coming resurgence in
European mathematics. Toward that end, Leonardo of Pisa advocated the use of Hindu−Arabic
numbers. In Italy, the rich world of Islamic mathematics was introduced and began to spur
interest.
From 1200−1600, major contributions continued to flow from the Islamic world, China, and
India. In England, new algebra and trigonometry texts emerged. In France, Viète pushed a new
decimal fraction system. During this time, Copernicus proposed a new heliocentric theory that
would greatly affect how mathematics developed from that time onwards. In Italy, a group called
the “Italian Algebraists” conquered the problem of finding an equation for solving cubic
(third−degree) and quartic (fourth−degree) polynomial equations, which we will study in more
detail later in this chapter.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 4
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Many of these intermediate topics are worthy of their own independent treatment. Due to time
considerations, we will look at just one interesting contribution by Bhaskara. (Future versions of
this text will have more information on Islamic, Hindu, and Chinese mathematics). Later, we will
look at two major achievements in the history of mathematics. The first topic we will explore in
more detail is the algebraic solution of cubic and quartic equations. The second achievement we
will examine is the set of tools that were developed which greatly simplified the process of doing
long, complex computations. These tools would enable mathematicians and scientists to make
great inroads into their fields of studies during the 1600’s and 1700’s. Specifically, we will
focus on the invention of logarithms by John Napier and the dramatic impact they had on the
mathematical landscape.
Multiplication from India
In the 12th century, the Indian mathematician Bhaskara (also known as Bhasharacharaya in India)
represented the peak of mathematical knowledge. He was the head of an astronomy observatory
at Ujjain, which was the prominent center for mathematics in India at the time. He had a
thorough understanding of the number 0, negative numbers, and methods of solving equations −
centuries ahead of the Europeans. He established his reputation (in part) based on a work titled
Lilavati (“The Beautiful”), which had 13 chapters and covered topics in arithmetic, geometry,
and algebra.
One of the interesting
contributions he
included in his work was
a proof of the
Pythagorean theorem.
He started with a square
that was cut into four
triangles and one square,
and then rearranged
them to create two
squares that were
positioned immediately
next to each other. His
picture was accompanied by the simple phrase, “Behold!”
Think About It
Why is this a proof of
the Pythagorean
Theorem?
2001, Lawrence Morales; MAT107 Chapter 6 – Page 5
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This appears to have been inspired by the Chinese and
one of their own “proofs” of the Pythagorean Theorem.
Recall the hsuan−thu from the Pythagorean chapter.
You can see the corresponding inner−squares embedded
in the middle of the larger squares with side c. This
smaller square gets moved into the upper right hand
corner of the two larger squares that are positioned side
by side, above.
Gelosia Multiplication
In Lilavati, Bhaskara also provides five different
methods for multiplication. One of these is interesting
because it later emerged in the Middle Ages and was eventually adapted by the inventor of
logarithms, John Napier. It is now called the gelosia method of multiplication. The word gelosia
means “lattice” or “grating.”
In this method a grid is set up, several simple multiplications are placed into the grid, and then
161 diagonals on the grid are added together to get a final result. For
162 example, to do 286×734, a three−by−three grid would be drawn
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side.
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From the first picture (left), which looks a lot like a grid or lattice,
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you can see where the method gets its name. To fill in the grid, you
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take the number in a column and the number in a row and multiply
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intersect each other. For example, column 8 and row 3 give a
171 product of 24, so 24 goes in that square. Note that the tens digit
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goes to the left of the diagonal of that “cell”
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while the ones goes to the right of the
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diagonal. If the product of a column and
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row is less than ten, we put a 0 in the tens
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position…all spaces should be filled in to
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avoid confusion. The picture (left) shows
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all the cells filled in.
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This method allows you to do all the
multiplication at once…it’s only after this is done that we move to
addition. The addition is done along diagonals. Starting in the lower right corner, which
represents the ones place, we add up all numbers in that diagonal. If we get more than ten for the
sum, we carry any groups of ten into the next diagonal up. Hence, in the second diagonal, we
have 8+2+2=12, so we carry 1 up to the next diagonal and keep the 2 (just like when we add
vertically). The third diagonal has a sum of 1+8+3+4+1+2= 19. Again, that 1 will carry up and
the 9 is kept and recorded for that row. Continuing in this manner will produce a digit for each
2001, Lawrence Morales; MAT107 Chapter 6 – Page 6
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diagonal and we can read the final answer by starting from the upper left and reading down and
around the corner. In this case, we get the result that 286×734=209,924.
81916 Let’s compare this with a modern method of multiplication that is taught in
31924 U.S schools (but not everywhere). With this method, each digit in the first
×
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41934 number is multiplied with each digit in the second. However, carrying
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81940 must take place for each multiplication where more than 10 is produced.
2 0 0
01950 Any carrying leftovers must be added to the next product. Thus,
2 0 9
21964 multiplication, carrying, and addition are all intermingled during the
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multiplication process. Finally, everything can be added up, assuming that
zeros have been put into the proper locations so that all the place values line up correctly.
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Think About It
How are the two systems of multiplication alike? How are they
different? What are the advantages and disadvantages of each? If
you showed each to someone unfamiliar with either method,
which do you think would be easier to figure out?
Example 1
Use the gelosia method to calculate 5108×327
Solution:
For this product we need a 4 by 3 grid. Multiplying first and then adding
gives:
So our result is 1,670,316. ♦
2001, Lawrence Morales; MAT107 Chapter 6 – Page 7
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Check Point A
Multiply 726×4138
Solution:
See endnotes to check your answer.1
Example 2
Use the gelosia method to compute 2.35×46.7.
Solution:
This method works perfectly fine when we have numbers with fractional
parts. We simply multiply as usual, ignoring the decimal point for the time
being.
Think About It
Why does moving
the decimal point
three places “fix”
everything in this
example?
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Think About It
We get a result of 109,745, which is obviously too
large for 2.35×46.7. Since we have three decimal
places to compensate for, we simply move the decimal
point three places to the left. We get 109.745,which is
correct. ♦
How would you
do 355 ÷ 4 with
this method?
This method found it’s way into 15th century Europe (via Islamic mathematicians and their
arithmetic books) when Luca Pacioli (1445−1517) included it in his popular book on arithmetic.
His comment on the origin of the method’s name is interesting not only because it tells us where
the word may come from but also because of the insight it gives into a part of the culture of Italy
at the time.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 8
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By gelosia we understand the grating which is the custom to place at the windows of
houses where ladies and nuns reside, so that they cannot be early seen. Many such
abound in the noble city of Venice.2
The method did not survive for long after the fifteenth
century. When printing presses were invented it is likely
that the method, with all of its grids and lines, was too
demanding on the first printing presses, so it gave way to
other algorithms that were more easily typeset.
Think About It
Our word jealousy comes
from the word gelosia. Can
you think of a reason why
these two words would be
connected?
2001, Lawrence Morales; MAT107 Chapter 6 – Page 9
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PART 2: The Solution of the Cubic and Quartic Equations
Historical and Mathematical Background
For many hundreds of years, mathematicians had methods of solving quadratic equations.
Eventually, they developed techniques that are equivalent to the modern quadratic formula.
Today, we know that if we have an equation of the form ax 2 + bx + c = 0 , we can find the
solutions of this equation algebraically, if they exist, with the quadratic formula:
x=
− b ± b 2 − 4ac
2a
By “finding the solution algebraically,” we mean that we can find numbers that satisfy the
equation by using the only basic operations of addition, subtraction, multiplication, division,
powers and roots. Various mathematicians had some very creative solutions to quadratics that
were not algebraic solutions, but they were valid nonetheless. However, there was something
very alluring about finding the equivalent of a formula for these kinds of equations. This allure
led many mathematicians to look for solutions to other kinds of polynomial equations. In
particular, the third and fourth-degree polynomial equations and their solutions were pursued
long and hard by mathematicians.
The algebraic solution of the cubic and quartic equations is one of the great highlights in all of
the history of mathematics. A cubic equation is one of the form:
ax 3 + bx 2 + cx + d = 0
The highest power of x is three. A quartic equation has four as the highest power of x and has
the form:
ax 4 + bx 3 + cx 2 + dx + e = 0
By 1500, algebraic solutions to third and fourth degree equations had not been found, however,
despite valiant attempts to do so. In the 12th century, al−Khayyami had a method of solving an
equation of the form x 3 + cx = d , but his method rested on seeing the equation as an equation
between solids. His solution was very geometric and today feels very “Greek” in nature. Al−
Khayyami actually examined 14 different kinds of cubic equations, described the physical
objects needed to solve each one, and then proved that each solution was correct.3 But this
approach was very different than the kind that was pursued later in history. Mathematicians
were not satisfied with a geometric approach; they wanted one that was more algebraic in nature.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 10
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Cardano and His Gang of Italian Algebraists
In 1494, Luca Pacioli, an Italian mathematician wrote a book called Summa de Arithmetica. In
this work, Pacioli discussed the solution of linear and quadratic equations. In this book, he
started using the word co, short for cosa (thing) to represent an unknown quantity. This was an
early version of our symbolic algebra system where a single letter, typically x, stands for an
unknown quantity. Pacioli also discussed the challenges of solving the cubic equation with an
algebraic approach and decided that it was probably NOT possible to do. In the next century,
however, many of his Italian counterparts did not share his opinion and attacked this problem
with great vigor.
At the University of Bologna, Scipione del Ferro (1465-1526) ignored Pacioli’s opinion and
discovered a formula for the solution to a cubic of the form:
x3 + mx = n
Pacioli did not publish his result. Instead, he kept it secret! To understand why he did this, it is
worthwhile to understand the Renaissance university of the time. This was a time when the
modern implementation of tenure did not exist. Jobs at universities were not secure like they are
today under tenure. To keep a post, you had to have not only political influence and the ability
to “shmooze” the appropriate people, but you also had to have the intellectual force to withstand
public challenges to your post.
These public challenges could occur at any time. In these scholarly battles, the current holder of
an academic post and his challenger would meet in public and match wits against each other. If
the challenger won, he would bring public humiliation to his counterpart, who would often have
to resign his post to his challenger. This was not a positive development in one’s career!
A new discovery like del Ferro had found was something to save in case of a challenge. If an
opponent appeared on the scene and had a list of problems to pose, del Ferro felt confident
enough that he could get at least some of them correct. On the other hand, if he was reasonably
sure that he was the only one with a solution to a cubic, he could counter his challenger’s list of
problems with a smattering of cubic equations to solve, thereby
giving him a decent chance of surviving the challenge.
As it turned out, del Ferro never needed to use his secret weapon. It is
reported that just before his death he passed on his solution to one of
his students, Antonio Fior. Unfortunately, Fior was not as prudent as
his master. Upon hearing that another mathematician, Niccolo
Fontana4, had boasted that he had found a solution to another form of
the cubic equation ( x3 + mx 2 = n) , Fior immediately challenged
Fontana to a public contest. (Fontana was and is currently also
known as Tartaglia – the stammerer – due to the fact that he was
unable to speak clearly. Fontana had suffered a sword wound to his
face in 1512 when a French soldier attacked his hometown.)
2001, Lawrence Morales; MAT107 Chapter 6 – Page 11
Tartaglia
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Tartaglia sent Fior a list of 30 varied problems. Fior, on the other hand, sent Tartaglia 30 cubic
equations, all of the form x3 + mx = n . As the story goes, Tartaglia worked furiously on these
problems, presumably with his previous knowledge of how to solve equations of the form
x3 + mx 2 = n available to help him. On February 13, 1535, Tartaglia cracked the solution and
was able to solve all 30 of Fior’s problems. Fior, of course, could not solve all of Tartaglia’s
problems as they were chosen more carefully. Gracefully, Tartaglia relieved Fior’s obligation as
the loser to shower Tartaglia with 30 banquets. However, the damage was done. Fior is no longer
remembered except for the foolishness he displayed during this challenge.
As word spread that Tartaglia had solved the solution to the cubic, it
eventually reached the ears of Gerolamo Cardano5 (1501-1576), one
of the most interesting characters in the entire history of mathematics.
Cardano was a doctor by trade, at one point serving the Pope, but
Cardano also worked extensively on mathematics. Cardano’s own
autobiography, De Vita Propria Liber (The Book of My Life), gives
us some idea of this man’s personality. He was a man who was
consumed with superstition and tragedy. There is much that could be
written about him, but readers are encouraged to do some basic
research and reading to get more information. (See Homework
Problem (85) for that opportunity.)
Cardano
Cardano wrote to Tartaglia and asked him for the solution to the cubic
equation. Tartaglia, of course, initially refused. Why would he give away such a valuable secret?
But Cardano did not give up. He continued to write Tartaglia until he wore him down. Finally,
on March 25, 1539, Tartaglia revealed the method to Cardano in the form of a coded cipher after
Cardano agreed to take the following oath:6
I swear to you by the Sacred Gospel, and on my faith as a gentleman,
not only never to publish your discoveries, if you tell them to me, but I
also promise and pledge my faith as a true Christian to put them down
in cipher so that after my death no one shall be able to understand
them.
Along with a student, Lodovico Ferrari (1522-1565), Cardano unraveled the secret of the cubic
equation in the form x 3 + mx = n , making significant progress. Not only did they master the
techniques given to them by Tartaglia, but they also extended his work to apply to cubic
equations of any form (a huge accomplishment) and used Tartaglia’s techniques to solve
polynomial equations of degree four! Cardano and Ferrari were bound by Cardano’s oath,
however. Even though they had pushed Tartaglia’s work far beyond where Tartaglia had ever
taken it, all of their progress was based on the oath-protected secrets of Tartaglia. Cardano,
though, was anxious to publish his results. He did not need to protect an academic post since he
was a doctor by trade.
Looking for a way out of this predicament, Cardano and Ferrari traveled to Bologna, where our
story began. There, they studied the private papers of del Ferro and saw the exact solution of
Tartaglia written in del Ferro’s own handwriting, but well before Tartaglia had discovered them
2001, Lawrence Morales; MAT107 Chapter 6 – Page 12
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independently for himself. As far as Cardano was concerned, he was no longer bound by his
oath. He would publish del Ferro’s findings, not Tartaglia’s, even though they were the same.
And that he did. In 1545, he published Ars Magna (Great Art), which is considered one of the
great masterpieces in mathematical history. (It’s still in print!) He prefaced his work with the
following attribution:
Scipio Ferro of Bologna well-nigh thirty years ago discovered this rule
and handed it on to Antonio Maria Fior of Venice, whose contest with
Niccolo Tartaglia of Brescia gave Niccolo occasion to discover it. He
gave it to me in response to my entreaties, though withholding the
demonstration. Armed with this assistance, I sought out its
demonstration in [various] forms. This was very difficult.7
Tartaglia was furious, of course. He accused Cardano of deceit, sending nasty letters to Cardano
with the charges clearly spelled out. Cardano basically ignored them. His student, however, was
more easily drawn into the debate. In 1548, Tartaglia and Ferrari squared off in a public,
mathematical challenge in Milan, Ferrari’s stomping grounds. Tartaglia lost, blaming his
performance on the “rowdiness and partisanship of the crowd.”8 (A rowdy crowd at a math
contest?) Tartaglia went back home, defeated once again, and Ferrari was proclaimed the winner.
Ars Magna and the Solution of the Cubic
In Ars Magna (1545), Cardano detailed del Ferro’s
and Tartaglia’s technique for finding the solution of a
cubic equation in the form:
x 3 + mx = n
This is called a depressed cubic because the x 2 term
is missing. Cardano’s solution was given entirely in
words. There were no modern algebraic symbols yet
present to help him express his results. In modern
notation, Cardano’s verbal solution of this kind of
cubic takes on the following modern form:
Cardano’s Formula
x=
3
n
n
+ R+3 − R
2
2
where R =
n 2 m3
+
4 27
2001, Lawrence Morales; MAT107 Chapter 6 – Page 13
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You can guess why this equation is not introduced into basic algebra courses. There is relatively
little that is “basic” about it. Prior to the advent of handheld calculators, using this formula
would prove to be almost impossible in most cases.
The demonstration of this formula is not an easy task to undertake.9 However, it is based on
basic algebraic principles and so if you are patient and diligent enough, you can follow how it
was derived. In this text, we will skip an explanation of where the equation comes from and
focus our efforts on learning how to use it.
(Note: In order to use this formula, you need to really know how to use your scientific or
graphing calculator. To take cube roots on your calculator you can check to see if your model has
a 3 x button. If it does not, you can take a cube root by raising a number to the power of 1/3. For
example, 81 / 3 = 2 . On a calculator, the power button looks usually like y x , x y , or the ^ symbol.
To do 81/3 would require a key sequence like: 8^(1/3), or 8 yx (1/3). Play with it on your model
until get 2 for 81/3 before you proceed.)
Before we do any specific examples, it might be helpful to propose some steps that can be
followed to use this formula correctly:
Steps in Using Cardano’s Formula
Step 1: Make sure that the equation given is in the form x 3 + mx = n .
Step 2: Identify the proper values of m and n.
Step 3: Calculate the value of R and also R . Simplify if possible.
Step 4: Plug in the value for R into Cardano’s equation and carefully compute the
final value.
We will start with the equation that Cardano used to illustrate his method. He stated his problem
in words (since he did not have variables to work with) in a form that might look like this:
Cube plus six times a number is twenty. What is the number?
A translation of this into modern algebraic notation would be the following:
x 3 + 6 x = 20 .
Cube plus six times a number is twenty. What is the number?
2001, Lawrence Morales; MAT107 Chapter 6 – Page 14
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Example 3
Use Cardano’s formula to solve x 3 + 6 x = 20 .
Solution:
Step 1:
Make sure that the equation given is in the form x 3 + mx = n . This
equation is already in this form so we can proceed.
Step 2:
Identify the proper values of m and n.
The value of m is 6 and the value of n is 20.
Step 3:
Calculate the value of R and also
n 2 m3
R=
+
4 27
202 63
=
+
4
27
400 216
=
+
4
27
= 100 + 8
= 108
481
482
483
484
485
486
487
488
This means that
Step 4:
R . Simplify if possible.
R = 108
Plug in the value for R into Cardano’s equation and carefully
compute the final value of x:
x=
3
n
n
+ R+3 − R
2
2
=
3
20
20
+ 108 + 3
− 108
2
2
= 3 10 + 108 + 3 10 − 108
489
490
491
492
This is what we could call the exact solution. If we want a decimal
approximation to this, we can use a calculator to do this:
x = 3 10 + 108 + 3 10 − 108
≈ 3 10 + 10.3923 + 3 10 − 10.3923
493
≈ 3 20.3923 + 3 −0.3923
≈ 2.7321 − 0.7320
≈ 2.0001
494
2001, Lawrence Morales; MAT107 Chapter 6 – Page 15
It’s amazing
that these two
are actually
equal.
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We get an estimate of 2.0001. This is pretty close to 2, so we’ll
round it off to 2 exactly and see how close we are. If we check this
in the original equation, we get:
x 3 + 6 x = 23 + 6 × 2
= 8 + 12
= 20
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We do get the correct value. ♦
Important Note
In order to be as accurate as possible with these calculations, it is best to keep as many
decimal places as possible. If you know how, it is best to try to do the entire sequence of
calculations without clearing our your calculator or using the clear button. (By not
using the clear button, my graphing calculator gives me the exact value of 2!) You
should experiment with your calculator and try to get an exact value without having to
use the clear button. If you cannot get it to work, use four decimal places throughout
your computations.
Note that this equation gives one solution for this
equation. We can look at the graphs of y = x 3 + 6 x and
y = 20 , to see where they intersect since the
intersection point(s) represent solutions of the original
equation. The graph here shows that there is indeed
one solution. You can see that when x = 2, the curve of
y = x 3 + 6 x crosses the line y = 20, just as we expect it
to. Obviously, Cardano did not have the advantage of
seeing such a graph…the x−y coordinate system was
not yet invented, but it does give us some insight into
what is going on here.
30
20
(2,20)
10
0
-2
0
2
4
-10
Example 4
Use the cubic formula to solve x 3 + 3 x = 10
Solution:
Step 1:
Make sure that the equation given is in the form x 3 + mx = n .
This equation is already in this form so we can proceed.
Step 2:
Identify the proper values of m and n.
The value of m is 3 and the value of n is 10.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 16
6
537
538
Step 3:
Calculate the value of R and also
R . Simplify if possible.
n 2 m3
+
4 27
102 33
=
+
4
27
100 27
=
+
4
27
= 25 + 1
= 26
R=
539
540
541
542
543
544
545
This means that
Step 4:
546
R = 26
Plug in the value for R into Cardano’s equation and carefully
compute the final value
x=
3
n
n
+ R+3 − R
2
2
=
3
10
10
+ 26 + 3
− 26
2
2
= 3 5 + 26 + 3 5 − 26
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548
549
550
This is we call the exact solution; the decimal approximation
is:
x = 3 5 + 26 − 3 − 5 + 26
≈ 3 5 + 5.09902 − 3 − 5 + 5.09902
≈ 3 10.09902 − 3 0.09902
≈ 2.161522 − .46263765
≈ 1.6989
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A check of this shows a close match, although the decimal
rounding causes a little error. ♦
Check Point B
Use the cubic formula to solve x 3 + 10 x = 57
Solution:
See endnote for an answer.10
2001, Lawrence Morales; MAT107 Chapter 6 – Page 17
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Check Point C
Use the cubic formula to solve x 3 + 5 x = 15
Solution:
Seen endnote for an answer.11
Solving the General Cubic Equation
Cardano next tackled the non−depressed equation. To do this, he came up with a clever way of
temporarily converting a non−depressed equation into one that is depressed. He then used his
formula on the depressed equation and then adjusted his result to reflect the fact that he had
changed the original equation.
The general method is as follows:
Solving the Non-Depressed Cubic Equation:
Given the equation ax 3 + bx 2 + cx + d = 0
b
Step1
Let x = y − .
3a
b
into the original equation to get an equation
3a
in y. The result should be a depressed cubic.
579
Step2
Substitute x = y −
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Step3
Use Cardano’s formula to solve the equation for y.
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Step4
Find the original, desired values of x using x = y −
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b
.
3a
Example 5
Solve 2 x3 − 30 x 2 + 162 x − 350 = 0
Solution:
b
− 30
= y−
= y + 5 . Remember this for later: x = y + 5
3a
3(2)
Step 2: Substitute x = y + 5 into the original equation to get a depressed cubic.
Step 1: Let x = y −
x=y+5
2 x 3 − 30 x 2 + 162 x − 350 = 2( y + 5)3 − 30( y + 5) 2 + 162( y + 5) − 350
594
= 2( y 3 + 15 y 2 + 75 y + 125) − 30( y 2 + 10 y + 25) + 162 y + 810 − 350
= 2 y 3 + 30 y 2 + 150 y + 250 − 30 y 2 − 300 y − 750 + 162 y + 810 − 350
= 2 y 3 + 12 y − 40
595
2001, Lawrence Morales; MAT107 Chapter 6 – Page 18
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597
Recall that this is equal to 0, so we can divide by 2 to get:
2 y 3 + 12 y − 40 = 0
y 3 + 6 y − 20 = 0
598
y 3 + 6 y = 20
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Step 3: This is a depressed cubic and we can now use Cardano’s formula on
it. But note that this is essentially the equation Cardano used to
illustrate the solution to the cubic. We saw in Example 3 that
x 3 + 6 x = 20 has a solution of 2, therefore y 3 + 6 y = 20 has a
solution of y=2.
Step 4: Since x = y + 5, and y = 2, we can see that x = 2 + 5 = 7. Thus, x = 7
is the solution to the original equation 2 x3 − 30 x 2 + 162 x − 350 = 0 .
Check:
2 ( 7 ) − 30 ( 7 ) + 162 ( 7 ) − 350 = 686 − 1470 + 1134 − 350
3
611
612
613
614
615
616
617
618
619
620
621
3
=0
This verifies the solution is correct. ♦
Using this process may seem cumbersome to some, but we should remember that Cardano did
not have variables or equations to work with. Cardano had to mainly use words to solve these
equations. Hence, the process that we have is actually a lot easier to use than Cardano’s.
Example 6
Solve x 3 − 6 x 2 + 15 x − 18 = 0
Solution:
(−6)
= y+2
3
622
Step1: Let x = y −
623
624
625
Step2: Substitute x in to the equation:
x 3 − 6 x 2 + 15 x − 18 = 0
626
( y + 2) 3 − 6( y + 2) 2 + 15( y + 2) − 18 = 0
y3 + 3y − 4 = 0
y3 + 3y = 4
627
628
629
This last equation is in the form we need to use Cardano’s formula.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 19
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Step3: We have m = 3, n = 4.
We first find the values of R and
R.
n 2 m3
+
4 27
42 33
=
+
4 27
= 4 +1
=5
R=
633
634
635
636
637
638
R = 5.
Thus,
Now find the value of y using Cardano’s formula.
639
y=
3
n
n
+ R+3 − R
2
2
=
3
4
4
+ 5+3 − 5
2
2
= 3 2+ 5 + 3 2− 5
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653
654
Using a calculator, we can get an estimate for this number:
3
2 + 5 + 3 2 − 5 ≈ 1.6180 + (−.6180)
=1
Therefore, y = 1. This means that y = 1 is a solution to y 3 + 3 y = 4 .
Keep in mind that our original equation was in terms of x and that we
used the substitution x = y + 2 . We now use this substitution again to
find the value of x that we are after.
Step4: y = 1, so x = y + 2 = 1+ 2 = 3. Hence, x = 3.
Check:
We let x = 3 and substitute in the original cubic equation.
x 3 − 6 x 2 + 15 x − 18 = 33 − 6(3) + 15(3) − 18
= 27 − 54 + 45 − 18
= 72 − 72
=0
2
655
656
657
This solution checks and we’re done.♦
2001, Lawrence Morales; MAT107 Chapter 6 – Page 20
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Check Point D
Use the cubic formula to solve x 3 + 12 x 2 + 4 x − 445 = 0
Solution:
See the endnotes for a solution.12
The Solution of the Quartic Equation and Beyond
With the cubic equation conquered, Cardano and his student, Ferrari, moved on to the quartic
equation of the form ax 4 + bx 3 + cx 2 + dx + e = 0 . Could a formula be found for that as well? The
answer was yes, as long as the equation could be reduced to a depressed cubic, in which case
they could use their previous discovery. It seems as though the cubic formula was more useful
than they originally thought. We won’t explore the solution of the fourth degree equation in
detail here due to its complexity. However, if you’re interested in seeing a modern
representation to the solution of the quartic, you can read Appendix A at the end of the chapter.
Fifteen years after Cardano and Ferrari did their work, Rafael Bombelli (1526−1572) wrote a
more systematic text to help students master these techniques. His book, Algebra, “marks the
high point of the Italian algebra of the Renaissance.”13 In this text, Bombelli introduced
equations and techniques that led to the establishment of what we call imaginary or complex
numbers and gave rules for working with them. Complex numbers include numbers of the form
a + bi where i = − 1 . That is, the square root of a negative number is allowed and is considered
a valid number. (These are not part of the real number system. In fact, the real numbers are
actually a subset of the complex numbers.) Complex numbers play important roles in higher
mathematics, physics, and engineering.
We make one last note about solving equations: It was only natural for Cardano and Ferrari to
pursue the quintic (fifth degree) equation and its solution. A quintic equation has the following
form:
ax 5 + bx 4 + cx 3 + dx 2 + ex + f = 0
But they were thwarted. They could not find an equation to solve a quintic
because there is none. It is impossible to express the solution of a fifth−degree
equation with a formula using roots, powers, and the four basic mathematical
operations. This was first proved by Niels H. Abel (1802−1829). Later
Evariste Galois (1811−1832) extended his work and gave general conditions
for when equations are solvable. Galois has a whole field of mathematics
named after his work…Galois Theory. Unfortunately, both Abel (top) and
Galois (bottom) died at very early ages. The story of Galois’ life and death is
fit for a television movie of the week.
As the 1500’s and 1600’s passed, mathematics was rediscovered in Europe
again. It stands on the shoulders of many civilizations that preceded them. By
the early 1600’s, Europe was ready for new techniques that would help them
tackle modern, complex calculations.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 21
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PART 3: Modern Calculations
The power and extent of modern calculations rest on at least three important developments.14
The introduction of the Hindu−Arabic numbers, the use and acceptance of numbers with
decimal/fractional parts, and the invention of logarithms all played important roles in how
mathematicians and scientists could do computations. The introduction of decimal fractions and
logarithms both took place in the early 1600’s and each of them came about because of efforts to
take very laborious computational techniques and simplify them. (Prior to this, very complicated
methods from trigonometry had been widely used instead.) The need for this kind of change was
motivated by the needs of research in astronomy, navigation, and commerce, to name a few.
Decimal Fractions
Simon Stevin15 (1548−1620) was one of the leading advocates of
decimal fractions. Just to clarify what we mean by “decimal
fractions,” we are talking about numbers like 34.657. These are
base−ten numbers where the whole part and fractional part of a
number are taken together and computations are done on them
without splitting up their individual parts. Prior to this time, the
whole and fractional part were often split apart and computed
separately. Not only that, but fractions were often expressed in base
60 rather than in base 10.
Think About It
Why Base 60?
Stevin worked in commercial, military, and administrative environments, and eventually taught
math at the Leiden School of Engineering. He was one of the first people to fully embrace the
emerging Copernican theory. He gains a role in this discussion because he published a pamphlet
called De Thiende (The Art of Tenths) where he introduced decimal fractions. His publication
came at a time when others had started taking advantage of decimal numbers, and so it was
received with open arms. It is interesting to see the difference between his notation and ours.
He used the symbols b,c,d,e,f… to represent descending powers of 10.
b Corresponds to the whole part of a number (ones place and higher, basically)
c Corresponds to the tenths place; (1/10)1 is tenths.
d Corresponds to the hundredths place; (1/10)2 is hundredths
e Corresponds to the thousandths place; (1/10)3 is thousandths, etc….
2001, Lawrence Morales; MAT107 Chapter 6 – Page 22
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However, instead of using “tenth,” “hundredth”, or “thousandth,” Stevin uses “prime,” “second,”
and “third,” respectively. If he wanted to express the number 34.657 it would look similar to this:
34b6c5d7e
Here’s a breakdown to clarify his notation:
34b6c5d7e
34 whole
6 in the tenths (prime) place 5 in the hundredths (second) place
7 in the thousandths (third) place
Example 7
Write 287.93 in Stevin notation
Solution:
287b9c3d
Check Point E
Write 3,456.28547 in Stevin notation
Solution:
See endnotes for solution.16
If Steven wanted to do arithmetic with decimal fractions, an addition
problem would look like the picture here, which is actually a piece of
Stevin’s publication on this subject.
Around this time, support for these numbers grew quickly17, despite their
clunky implementation. In 1592, Giovanni Antonio Magini, a mapmaker,
introduced the decimal point to separate the whole parts from the
fractional part of numbers. Despite its ongoing evolution and staunch
support for the system from people like Stevin, it would be another 200
years before the system was adopted for use in currency, weights, measures, etc.
It was not until the French Revolution that the “metric” system was adopted for
Think About It
such uses.
777
778 Another important name that paved the road for decimal What numbers are
being added here
779 fractions and logarithms to emerge was François Viète18
and what is the
780 (1540−1603). Viète, convinced that a geocentric model
result?
781 was more reliable than the emerging Copernican theory,
782 made great efforts to publish trigonometry calculations
783 that would help in the study of astronomy. For the most part, scientists
784 before him were used to using sexigesimal (base-60) fractions that were
785 a product of the Babylonian number system and their own work in
786 astronomy. Viète, along with more and more of his contemporaries, was
787 advocating a new system of decimal (base 10) fractions.
One way in which Viète and others tried to simplify multiplication and division was by reducing
them to addition and subtraction. The obvious reason for this was that adding and subtracting
numbers is generally much easier than multiplying or dividing them. This process of reduction is
2001, Lawrence Morales; MAT107 Chapter 6 – Page 23
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called prosthaphaersis.19 In his publication, Canon, Viète presents several equations from
trigonometry that accomplish this. For example, he gives the following two rules:
2 sin A × sin B = sin( A + B) + sin( A − B)
2 cos A × sin B = sin( A + B) − sin( A − B)
796
797
798
The “sin” and “cos” symbols are trigonometry functions that are actually abbreviations for sine
and cosine, respectively. The phrase cos( A + B ) does not mean cos “times” (A+B). Instead, it
799
means the cosine of (A+B). It’s sort of like a square root. The
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801
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itself…we have to take the square root of some particular number.
(usually).
symbol has no meaning by
A + B does have meaning
While we will not study these trigonometry functions in this course, what we will point out is
that each of these two equations reduces multiplication into addition (and/or subtraction). The
first equation takes two sines that are multiplied together and reduces the value to two sines that
are added together. With tables to give the values of these cosines and sines, computations could
(and did) go much more smoothly. We will see that this idea of reducing multiplication and
division to addition and subtraction is an important one in the next part of this chapter.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 24
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PART 4: Napier and The Emergence of Logarithms
The mere mention of the word “logarithm” can cause many algebra students’ eyes to gloss over.
Most do not understand these tools when they first see them and rarely remember them past the
final exam. I’m convinced that a big part of the reason that the rules and uses of logarithms are
so hard for students to grasp is that they are usually presented completely devoid of their
historical context. Furthermore, with the advent of technology, their general use has been
reduced significantly. In this section, we’ll try to establish that context and hopefully come to see
logarithms as an important and powerful development in the history of math (even if they have
been replaced by the modern calculator).
Keep in mind that we are considering a time in history when calculators
were not present and calculations all had to be done “by hand.” As work in
astronomy and other fields progressed, the need for an efficient method of
calculation had emerged. As we stated before, in some circles this was
done with prosthaphaersis, where multiplication and division were
reduced to addition and subtraction. John Napier20 (1550−1617) spent
about twenty years of his life developing a new and easy-to-use tool that
would use this approach to make computations easier to do. He gave it the
name logarithm.
Napier was an ardent Scottish Presbyterian who was strongly opposed to Catholicism. In his
book A Plaine Discovery of the Whole Revelation of Saint John: Set Downe in Two Treatises, he
identifies the Pope as the antichrist, urges King James to rid his house of “papists,” and predicts
the end of the world will take place between 1688 and 1700. He was also a resourceful
landowner, devising a hydraulic screw similar to that of Archimedes to control water levels in
coal pits. On the mathematical front, he used numerology to look for hidden prophecies in the
Bible and he looked for a way to simplify computations in trigonometry. To do this, he used a
well−known trigonometric equation of the time:
2 sin A × sin B = cos( A − B) − cos( A + B)
Although slightly different from the equations given
previously, we can see that the multiplication on the
left is reduced to addition and subtraction on the right.
Napier used this identity to build a table of logarithms
and “antilogarithms” that could be used to simplify
calculations. In 1614, he announced and published his
tables in Mirifici Logarithmorum Canonis Descriptio
(A Description of the Marvelous Laws of Logarithms).
Later, he published information on how he developed
logarithms and the theory behind them in Mirifici
logarithmorum canonis constructio (A Construction of
the Marvelous Laws of Logarithms.) The word logos
means ratio in this context and arithmos means
number; Napier merged them to form the word
2001, Lawrence Morales; MAT107 Chapter 6 – Page 25
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logarithm (ratio of number). The reason he chose this particular word has to do with how he
came up with the logarithm concept and later developed it.
It is important to point out that Napier’s logarithms are different than our own. It wasn’t until
after his tools had become widespread and had been refined and developed by others that they
took the form that they are in today. To Napier, logarithms were tied to physical motion and the
mathematics related to motion. Napier defines his logarithms as follows:21 Take a line segment
AB and an infinite ray DE, as shown below:
A
C
B
y
D
F
E
x
Let C be a point that starts at point A and moves along AB. Let F be a point that starts at point D
and moves along ray DE. Points C and F both start moving at the same speed. From that point
on, their speeds change according to the following rules:
Think About It
ŠPoint C always moves with speed equal to the distance CB = y.
Is point C slowing
ŠPoint F always moves at the same speed as when it started,
down or speeding
where x is the total distance F has moved from point D.
up as it moves
along
AB? Why?
According to Napier, x is the logarithm of y.
This definition looks very little like the modern definition of logarithms because the modern idea
of log didn’t come along until the time of Euler (1707-1783). We’ll explore the modern
definition more in a moment, but what we should try to keep in mind is that both versions of the
tool are trying to reduce multiplication and division to addition and subtraction. Using this (or
the modern) definition of logarithm, it is possible to build tables that will make these reductions
for us. Napier spent 20 years of his life painstakingly developing this theory and building his
tables to an incredible degree of accuracy. He made very few errors, which is amazing
considering the number and complexity of calculations that he undertook.
Astronomers and mathematicians alike were ecstatic about this new invention since it really did
do what Napier had intended. One of the most famous quotes about logs comes from
Pierre−Simon who stated that “[logarithms], by shortening labors [on computations], doubled the
life of astronomers.” The acceptance of Napier’s new tool was widespread and rapid. Rarely in
mathematics are inventions so quickly and universally adopted.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 26
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In 1616, Henry Briggs (1561−1631) worked with Napier to refine his invention so that they were
more convenient to use. Briggs is known for helping to construct tables where log 10 = 1, thus
creating what we today call the “common logarithm.” This is the log button you see on your
scientific or graphing calculator. Briggs published common log tables for values from 1 to 1000.
Later, others provided tables up to 100,000 to ten places of accuracy! These were in use all the
way until the early 1900’s when the tables were computed to twenty decimal places in Britain.
This was still before the age of the electronic calculator or computer!
Napier’s work had long−range implications for broader mathematics, navigation, astronomy,
banking, and number theory. In an indirect route, the logarithm eventually led down a path that
resulted in the final proof that the three great Greek construction problems could not be solved
by straightedge and compass alone.22
The Idea and Use of Logarithms
In this section we will explore basic ideas of how logarithms work and why they were useful in
the time of Napier. This will hopefully explain why they were so popular. Furthermore, it is my
hope that the “rules of logarithms” that may have perplexed you in the past will be placed into a
proper context and that you will see how and why they were useful.
Recall that Napier (and others after him) published tables of logarithms and antilogarithms that
were intended to make complex calculations easier. As we work our way through this section, try
to imagine yourself in a world where you cannot reach for a calculator to do any number
crunching for you. Furthermore, the modern algorithms for multiplication and division that you
learned as a child do not exist. With these restrictions, how would you do calculations like the
following?
(34,879.2945) × (1,998,736,882.812)
2.4 × 1012
57.2
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
These are certainly not computations we want to undertake without our little button−filled slabs
of plastic. But Napier and others in his time had no choice. Logarithms gave them hope of doing
these calculations more efficiently.
The Idea Behind Logs
Here’s the main idea behind logarithms and how Napier intended them to be used: We start with
a problem that involves complex arithmetic computations (like those given above). After
applying logarithm rules we get a new statement that we can look up on the logarithm tables. We
do some addition and subtraction operations on this statement (as needed) to get an intermediate
result. This intermediate result can be looked up in an antilog table to get us back to a result that
compensates for the fact that we reduced everything to addition and subtraction. Think of the
antilog table as the “evil twin” of the logarithm. What the logarithm does, its evil twin (the
2001, Lawrence Morales; MAT107 Chapter 6 – Page 27
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antilog) comes along and undoes. So while the log will reduce to addition and subtraction, the
antilog will covert back to multiplication and division. Here’s a diagram of what we mean:
Ugly
computation is
converted to a
statement using
logs
Via
log
table
Log rules/tables
are used so
addition and
subtraction yield
an intermediate
result
Via
anti-log
table
Antilog tables are
used to convert
the intermediate
result to base 10
numbers…the
final result
Rule of Logs…One Step Closer to the Tables
Well, we’ve put it off long enough. We can’t go any further until we talk about the rules of logs.
We will be working with modern versions of these rules that were unknown to Napier. But the
spirit of the task at hand remains the same.
The modern definition of the logarithm is defined in a way so that exponents are undone. For
example, if we have the equation x 2 = 25 , we know that we can take the square root of both
sides, as square roots undo powers of 2. If the equation looks like 2 x = 25 , taking a square root
no longer works since the exponent (x) is not a 2. In fact, it’s a variable! In order to undo this
exponent, we need a tool that will specifically deal with this kind of equation. Leonhard Euler
(1707−1783) is responsible for the definition of log that we will use:
log b n = p ⇔ b p = n
The ⇔ symbol here is used to imply interchangeability. Both sides of the double arrow lead to
each other and allow you to go back and forth between the two expressions. We often say that
the “log base b of a number n is the power to which b must be raised to produce n.” This
statement tells us what taking the log of a number (n) means. The number b is the base. The
number p is the result. You should observe that while the right side of the double arrow has a
power present, the left side does not. The logarithm has “undone” the power.
Some simple review examples are in order.
Example 8
Write 2 3 = 8 as a log statement.
Solution
2 = b, the base; 3 = p, the power; 8 = n, the number that is produced.
So, log 2 8 = 3
We say that the “log base 2 of 8 is 3.” That is, the power to which 2 must be
raised to get 8 is 3. ♦
2001, Lawrence Morales; MAT107 Chapter 6 – Page 28
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Example 9
Write 2 x = 25 as a log statement.
Solution
2 = b, the base; x is the power; 25 = n, the number.
So we have log 2 25 = x . ♦
Check Point F
Write 4 y = 16 as a log statement. What is y?
Solution
Example 10
Solution:
Check Point G
Solution
See the endnotes to check your answer.23
Write log 7 49 = x as a statement with exponents.
7 = b, the base; 49 = n, the number produced; x is the power required. So
we get 7 x = 49 . We can actually find the value of x here, since we know that
72 = 49. Thus x = 2. ♦
Write log y z = w as a statement with exponents
See the endnotes to check your answer.24
So, what does this have to do with making multiplication and division easier? Good question.
Unless otherwise specified, we will assume from now on that our base is 10. Hence we will be
working with the common logarithm that Briggs developed with Napier. If you don’t see a
base on a log, you can assume the base is 10. (By the way, that’s what your calculator log button
assumes.) Let’s take two numbers and multiply them. We’ll take two special numbers that will
help us demonstrate the first log rule, which reduces multiplication to addition.
Let A = 10 x and let B = 10 y
A and B are two numbers, which depend on x and y, of course. Let’s see what happens when we
multiply them:
2001, Lawrence Morales; MAT107 Chapter 6 – Page 29
Operation
AB = 10 x × 10 y
AB = 10 x + y
Comment
Multiply our numbers
Recall that by rules of exponents, when we
multiply the same base with different exponents,
AB = 10( x + y )
we add the exponents and keep the base. For
example x 2 ⋅ x 4 = x 2+ 4 = x 6 . The parentheses
around x + y emphasize that we can consider this
all one exponent.
We
simply apply the modern definition of the log,
log10 AB = ( x + y )
where 10 is the base, (x+y) is the exponent, and
AB is the number/result.
If we knew what x and y were, we could proceed. Although we don’t know what values
they take on, we can use the modern log definition to find expressions for each.
Apply the modern definition of the log to A and B.
A = 10 x ⇔ log10 A = x
B = 10 y ⇔ log10 B = y
Now comes the cool part…
log10 AB = x + y
= log10 A + log10 B
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1024
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1038
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Substitute our values for x and y into the new
equation.
This gives the first famous rule of logs:
log AB = log A + log B
We’ve omitted the base since this rule is true for all valid bases. No doubt you’ve seen this
before, unaware of where it came from or why it even showed up in the first place. It’s often
paraphrased as “The log of a product is the sum of the logs.” With a historical context in place,
we can see its importance. The left side is the log of two numbers that are multiplied. On the
right side, we have reduced it to two logs being added together! Aha. There it is again. We’ve
reduced multiplication to addition and subtraction, just like Napier was seeking to do.
We can look at a simple example to see how this law works:
Example 11
Solution
Write log(35 × 44) as the sum of logs.
log(35 × 44) = log 34 + log 44
If we knew the value of log 34 and the value of log 44, we could add them to
get the value of log(35×44). ♦
2001, Lawrence Morales; MAT107 Chapter 6 – Page 30
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Check Point H
Solution
Write log(1298 × 3982) as the sum of logs.
See the endnotes for solution.25
These examples so far don’t truly allow us to peer into the power of this rule. Here’s an example
that will begin to do that.
Example 12
What is z = 1654×3298?
Solution
Here we have a multiplication problem. It’s not hard. We can do it in a snap.
But let’s just pretend that these two numbers are very hard to multiply. Let’s
let x = 1654 and let y = 3298.
Operation
z = 1654 × 3298
log z = log(1654 × 3298)
log z = log1654 + log 3298
log z = 3.2185 + 3.5182
= 6.7367
log z = 6.7367
z = 5453809.951
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1065
1066
1067
Comments
This is what we want to compute, but we’re
assuming it’s too hard or too timeconsuming to do “by hand.”
We take the log of both sides. In this step we
are taking an “ugly computation and
converting it to a statement about logs.” (See
diagram above).
Apply the first log rule to the right side. “The
log of a product is the sum of the logs.”
Use the log tables to look up log1654 and
log3298. We then easily add up the results.
What tables, you ask? You’ll meet them soon
enough. We’ll finish this first.
Here is our final result, in “logland.”
Use an antilog table to get z. Remember, the
antilog is the evil twin of the log, so it undoes
the what the log does. In this case, it removes
the log from the z and returns the value of z.
(Again, the antilog tables are on their way.)
When we check our answer, we are just slightly off, but we only worked with
about four decimal places. Recall that the British had tables of twenty places
in place in the early 1900’s. ♦
2001, Lawrence Morales; MAT107 Chapter 6 – Page 31
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Ugly
computation is
converted to a
statement using
logs
log
1102
A
= log A − log B
B
Example 13
 95.2 
Simplify log 
.
 8 
Solution
 95.2 
log 
 = log 95.2 − log8 . Hence, if we can compute log95.2 and log8, we
 8 
can subtract them to get the value of the original log. ♦
Example 14
 85.3 
Simplify log 
.
 x 
1100
1101
Via
anti-log
table
This is the rule that reduces division to subtraction. It is related to the familiar rule of exponents
xm
that says n = x m − n . That is, when we divide exponents with the same base, we subtract the
x
x8
exponents. This is why 2 = x 6 instead of x4. We don’t divide the exponents. The proof of this
x
law is left as an exercise.
1096
1097
1098
1099
Via
log
table
Antilog tables are
used to convert
the intermediate
result to base 10
numbers…the
final result
Hopefully you see from this example the process we described in the diagram above. We start by
creating a statement about logs. We then use log rules and tables to create results that are easily
added. Finally, we can use an antilog table to get us out of “logland” and to a final answer.
Before we introduce the log and antilog tables, we should mention two more log rules that are
important. The first of these is:
1094
1095
Log rules/tables
are used so
addition and
subtraction yield
an intermediate
result
Solution:
 85.3 
log 
 = log85.3 − log x ♦
 x 
1103
2001, Lawrence Morales; MAT107 Chapter 6 – Page 32
1104
Check Point I
 z 
Simplify log 
.
 3.5 
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Solution:
See endnotes for solution.26
The third log rule is also useful and is related to the familiar rule of exponents: (x m ) = x mn . For
n
example, (x 5 ) = x 3×5 = x15 . This rule states that when we raise an exponent to a power, we
multiply the powers. The related log rule is:
3
log b n c = c log b n
Many people remember this rule by observing that all it says is that “in a logarithm the power of
the number can move to the front of the log.”
Example 15
Simplify log x3 .
Solution
log x3 = 3log x ♦
Example 16
Simplify log x .
Solution
Recall that x = x1/ 2
1
log x = log x ♦
2
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Check Point J
Simplify log 3 x 2 .
Solution
See endnotes for solution.27
2001, Lawrence Morales; MAT107 Chapter 6 – Page 33
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Let’s prove this last log rule, log b n c = c log b n
Operation
Let x = log b n
bx = n
Consider log b nc
log b n c = log b (b x )c
= log b (b xc )
= xc
= cx
= c log b n
Comments
We start in this unusual place…but be patient
and you’ll see how things work out.
Use the modern definition of log. We file this
away for use later.
Now we look at the log we are seeking to
simplify.
Substitute n = b x from the step above.
Multiply exponents according to rules for
exponents.
This one is a little subtle. Remember that the
log base b of a number is the power that b
must be raised to get that number. In this case
our base is b and we want to raise that to some
power to get our number, b xc . Well, of course,
to get b xc we need to raise b to the power of
xc, so xc is the log.
Here we just substitute the value of x from the
very beginning to get our desired result.
So log b n c = c log b n
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The Log Tables
Most of you reading this have probably never seen a log table before. But before the relatively
recent appearance of the handheld calculator and computer, log tables and their more
sophisticated counterparts, slide rules, were the dominant tools for complex calculations.
Using a log table, we can do sophisticated problems of multiplication and division, as well as
powers and roots, without a calculator at our side. So for this section, you’ll want to put away
your calculator and journey back in time. For the most part, we will limit ourselves to
computations with numbers that have three or four decimal places. This will make using tables a
little easier than they might be otherwise. Also, we will be using base-10 logarithms for all of
our computations. You should have the log tables in front of you for the next series of examples.
They are in Appendix B.
There are four tables. Two are regular log tables. The other two are antilog tables. We’ll start
with the log tables. Looking up logs is relatively easy with these tables. To find the log of 4.25,
we look down the left side of the first table under the column labeled N. We locate the row that
says 4.2, as shown below.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 34
log10 N
N
4.1
4.2
4.3
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1
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4
5
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9
0.6128 0.6138 0.6149 0.6160 0.6170 0.6180 0.6191 0.6201 0.6212 0.6222
0.6232 0.6243 0.6253 0.6263 0.6274 0.6284 0.6294 0.6304 0.6314 0.6325
0.6335 0.6345 0.6355 0.6365 0.6375 0.6385 0.6395 0.6405 0.6415 0.6425
1
1
1
1
Ten Thousandth Parts
2 3 4 5 6 7 8
2 3 4 5 6 7 8
2 3 4 5 6 7 8
2 3 4 5 6 7 8
Then move along the row until you are in the column that is labeled with the 5. Ignore the “Ten
Thousandth Parts” of the table for now. You should see the number 0.6284. Thus, log10 4.25 =
0.6284.
What does this mean? We go back to the definition of log:
(0.6284 = log10 4.25) ⇔ (100.6284 = 4.25)
Hence, the log of 4.25 tells us to what power 10 must be raised to get 4.25. This makes sense
when we remember that 1 < 4.25 < 10. That is:
1 < 4.25 < 10
So… 100 < 4.25 < 101
The second inequality indicates that the power of 10 we require is somewhere between 0 and 1,
and 0.6284 certainly fits the bill.
Example 17
What is log10 9.32 ?
Solution
Looking at row 9.3 of the table, the number in column 2 is 0.9694. Hence,
100.9694 = 9.32, as desired. This makes sense since 9.32 is close to 10 = 101, so
our log (0.9694) should be close to 1. ♦
Example 18
What is log10 7.4 ?
Solution
We look up row 7.4 and inside column 0 (since 7.4 = 7.40) we find 0.8692. ♦
Check Point K
What is log10 8.55 ?
Solution:
See endnotes for an answer.28
2001, Lawrence Morales; MAT107 Chapter 6 – Page 35
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The Ten Thousandth Parts of the tables is there for when the number you are taking the log of
has three decimal places instead of two. To use this part of the table, you take the normal
two−decimal−place log and then add the ten thousandth part to the very end of the result. Here’s
an example:
Example 19
What is log 6.875?
Solution:
For this problem we first compute log 6.87 as before:
log10 N
N
6.7
6.8
6.9
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1234
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0
1
2
3
4
5
6
7
8
9
0.8261 0.8267 0.8274 0.8280 0.8287 0.8293 0.8299 0.8306 0.8312 0.8319
0.8325 0.8331 0.8338 0.8344 0.8351 0.8357 0.8363 0.8370 0.8376 0.8382
0.8388 0.8395 0.8401 0.8407 0.8414 0.8420 0.8426 0.8432 0.8439 0.8445
1
1
1
1
Ten Thousandths Parts
2 3 4 5 6 7 8
1 2 3 3 4 4 5
1 2 3 3 4 4 5
1 2 2 3 4 4 5
log 6.87 = 0.8370. Then, we look under the 5 column of the ten thousandth
parts of the table since the third decimal place of 6.875 is a 5. This gives a 3,
which really represents 0.0003 (That’s why it’s called a proportional part).
Take this and add it to the end of the previous result. So 0.8370+0.0003 =
0.8373.
Hence, log 6.875 = 0.8373. Since our calculators do not exist, we can’t check
this, but rest assured that it’s very close to the actual log of this number. ♦
Check Point L
Find log 4.638
Solution:
Check the table and make sure you can get 0.6663.
Note that the log requires that numbers strictly between 1 and 10 be used as its inputs.
Larger Numbers and the Log Tables
Okay, what about larger numbers? Suppose we want log 152. The tables don’t have a row with
152 in it, however. They only go up to 9.9. Here we have to rely on some common sense about
powers of 10. We know that 152 > 100, so therefore 152 > 102. Hence, our log must be bigger
than 2 since we need at least 100. We also know that 152 < 103, since 152 < 1000. Thus,
whatever power of 10 that gives us 152 must be between 2 and 3. Hence, log 152 must be
between 2 and 3.
To find how far past 2 we must go, we simply pretend that we are computing log 1.52. When we
look at row 1.5 of the table and move over to column 2, we find 0.1818. That’s how far past two
we must go. Hence, log 152 = 2.1818, since 2.1818 is between 2 and 3.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 36
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Here’s another way to think about it: We can rewrite 152 as 100×1.52. Thus,
log152 =
log(100×1.52) =
log100+log1.52. (This last step uses one of the basic log rules.)
We know that log 100 = 2 because 102 = 100. We can get log 1.52 from the table…it’s 0.1818.
So we have log 152=2+0.1818=2.1818
We call 2 (the integer part of the result) the characteristic and we call 0.1818 (the fraction part)
the mantissa. The term characteristic was first used by Briggs. The term mantissa is of Latin
origin, originally meaning an “addition” or “appendix.” It was probably first used by John Wallis
around 1693.
We can streamline some of this by trying to give a series of steps to follow when working with
the log tables.
Step 1 Move the decimal point of the given number so that you have a new number
that is strictly between 1 and 10.
Step 2 Count the number of places that you move the decimal point. This is the
characteristic.
Step 3 Look up the new number (that is between 1 and 10) in the log tables to get the
mantissa.
Step 4 Add the characteristic and the mantissa together to get the final result.
Example 20
Find log 4869.
Solution
If we write this as 4869.0, we note that need to move it 3 places to get 4.869.
Our characteristic, therefore, is 3. (We first note that 1,000<4869<10,000. Put
another way, 103 < 4869 < 104. Hence the power of ten required is more than
3 but less than 4.) To find the mantissa, we look up 4.869 on the table. We
look up row 4.8, column 6, and ten thousandth part 9. We get 0.6866+0.0008
= 0.6874. Combining our characteristic and mantissa gives log 4869 = 3.6874.
Hence, 103.6874 = 4869. ♦
Check Point M
Find log 35,630.
Solution
Check the endnote for an answer.29
2001, Lawrence Morales; MAT107 Chapter 6 – Page 37
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Logs of Very Large Numbers
You have to try to imagine what this means for doing calculations. With these tables, we can
now do sophisticated computations while at the same time getting three or four places of
accuracy without using a calculator (which they did not have at the time anyway). Here’s an
example, however, of why more than three places of accuracy might be needed.
Example 21
Find log 7,834,945.
Solution
We start by finding the characteristic. We need to move the decimal place 6
places to get 7.834945, so our characteristic is 6. To find the mantissa, we find
7.834945 on the log table. Looking up row 7.8, and down column 3 we see
0.8938. But what about the ten thousandth part? Since there are more than 3
decimal places we have to round 7.834945 to 7.835 and look in the 5 column
for the ten thousandth part, which gives us 3, representing 0.0003. Thus, our
mantissa is 0.8938 + 0.0003 = 0.8941. Finally, we get a log of 6+0.8941 =
6.8941. The actual value is something like 6.89403595. Even if we round this
number to the fourth decimal place, we do not get 6.8941. This is due to the
fact that we had to use an estimate to get the ten thousandth part of the
mantissa. ♦
This may not seem like a big deal to us. After all, three decimal places of accuracy are pretty
good for most applications. But remember that when working with very big or very small
numbers, like astronomers were at that time, losing accuracy in the fourth decimal place could
mean that your calculations end up being significantly off by the time you finish your
calculations.
Eventually, we want to use these tables to do complex calculations but we must first learn to use
the antilog tables.
The Antilog Tables
The analogy of the antilog as the evil twin of the log who undoes what the log does is helpful
here. Whereas the log tables tell us what power to raise 10 to in order to get a result, the antilog
tables tell us the opposite. Hence, the antilog tables are really just power of 10 tables, as you will
see here.
Example 22
If log N = 0.683, what is N?
Solution
Note that this is the opposite question than we had before. Here we know the
log but are looking for the original number N. By the modern definition of
logs, we have:
2001, Lawrence Morales; MAT107 Chapter 6 – Page 38
log N = 0.683
c
log10 N = 0.683
1322
c
10
1323
1324
1325
1326
1327
1328
1329
0.683
=N
Hence, we only need to compute 100.683. Without a calculator, we turn to the
antilog tables.
To compute 100.683, we find the 0.68 row and then move over into column
with 3 as its heading. Here is a piece of the antilog table that applies here:
ANTILOG TABLE: 10P = N
P
0
1
2
3
4
5
6
7
8
9
0.67 4.677 4.688 4.699 4.710 4.721 4.732 4.742 4.753 4.764 4.775
0.68 4.786 4.797 4.808 4.819 4.831 4.842 4.853 4.864 4.875 4.887
0.69 4.898 4.909 4.920 4.932 4.943 4.955 4.966 4.977 4.989 5.000
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2
2
2
2
Thousandth Parts
3 4 5 6 7
3 4 5 7 8
3 4 6 7 8
3 5 6 7 8
8 9
9 10
9 10
9 10
The table indicates that 100.683 = 4.819. This makes some sense when we
remember that since 0<0.683<1, then 100<100.683<101. Hence, we have
1<4.819<10, which is certainly true. ♦
To use the thousandths parts of the table, we follow the same basic idea as before.
Example 23
If log N = 0.8235, find N.
Solution
By the definition of log, we have 100.8235=N. We first find row 0.82 and then
move over to column 3. This takes us to 6.653.
ANTILOG TABLE: 10P = N
P
0
1
2
3
4
5
6
7
8
9
6.486
6.501 6.516 6.531 6.546 6.561 6.577 6.592
0.81 6.457 6.471
0.82 6.607 6.622 6.637 6.653 6.668 6.683 6.699 6.714 6.730 6.745
0.83 6.761 6.776 6.792 6.808 6.823 6.839 6.855 6.871 6.887 6.902
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1
1
1
1
1
2
2
2
2
3
3
3
Thousandth Parts
3 4 5 6 7
5 6 8 9 11
5 6 8 9 11
5 6 8 10 11
8
12
12
13
9
14
14
14
Then, we look up the 5 in the thousandth parts of the table and see and 8 there,
which now represents 0.008. We add these, 6.653+0.008 = 6.661. Therefore,
we have 100.8235 = 6.661. ♦
Note that the antilog requires numbers between 0 and 1 as inputs.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 39
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Check Point N
If logN = 0.1834, find N. What does N represent as a power of 10?
Solution
See the endnote for an answer.30
So then what do we do about larger numbers?
Example 24
If logN = 3.874, what is N?
Solution
We rewrite this as 103.874 = N.
First we notice that 103.874 = 103 × 100.874. (Rules of exponents… x m + n = x m x n )
Thus, 103.874 = 1000 × 100.874.
The power of 10 on the end can be obtained from the table. 100.874 = 7.482.
So we have 1000×7.482 = 7482.
The actual value is 7481.69500511, so we are reasonably close (four places of
accuracy when rounded.). ♦
Example 25
If logN = 7.2159, find N.
Solution
First we write 107.2159 = N.
Next, we write N = 107 × 100.2159 = 10,000,000 × 100.2159
From the table we have: N = 10,000,000 × 1.644 = 16,440,000. ♦
There is an alternative way to approach antilog problems. Notice that in the previous few
examples, the whole number part of the number we are given does not play a role when we use
the tables. Only the fractional part has an impact. So, as a shortcut, we can look up the fractional
part in the antilog tables and then move the decimal point in the result. The number of places that
we move corresponds to the whole number part of the given number. In the previous example,
note that you had to move the decimal point 7 places which corresponds to the 7 in 7.2159.
Check Point O
If logN = 5.623, find N.
Solution
See the endnotes to check your answer.31
2001, Lawrence Morales; MAT107 Chapter 6 – Page 40
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The Evil Twins Meet Each Other…Finally
We have been saying that antilogs and logs undo each other’s work. Let’s see a pair of examples
that demonstrate that.
Example 26
What is the value of x = log 76,540?
Solution
We note that 104 < 76,540 < 105. This means our characteristic is 4 and we
need to find the mantissa from the table. Since the table requires an input
between 1 and 10, we use 7.654. From the log table, we get
0.8837+0.0002=0.8839. Hence, log 76,540 = 4.8839. This means that 104.8839
= 76540. ♦
Now we’ll do the opposite of this problem…it’s twin.
Example 27
If logN = 4.8839, what is N?
Solution
This is linked to the previous example…we should expect to get back 76,540.
(Why?) Because logN = 4.8838 means that 104.8838 = N, we can use the
antilog table to find N.
104.8838 = 104×100.8838 = 10,000×100.8838 = 10,000×(7.638+.014) =
10,000×7.652 = 76,520.
Okay….we’re off a little bit, but if we had more decimal places of accuracy in
our table, we would get an even better result. ♦
What is important to observe from the last two examples is that applying log and antilog tables
basically take you back to the original starting point. For this reason, we say that the log and
antilog tables are inverses of each other in the same way that square roots and powers of two are
inverses of each other.
Complex Calculations
Now imagine for a moment that you are living in the 1600’s and your astronomical calculations
require that you multiply 874,200,000×7,853,00. You have no calculator. You don’t have any
easy method of multiplying these two numbers…until Napier knocks on your door and
introduces you to his logs. With our three basic log rules, and our four handy log/antilog tables,
we can do this and many other nasty computations.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 41
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Recall:
Definition of log
log b n = p ⇔ b p = n
Basic log Rules
1.) log AB = log A + log B
A
2.) log = log A − log B
B
3.) log n c = c log n
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Example 28
Multiply 874,200,000 × 7,853,000
Solution
We’ll start all such problems by letting x equal the quantity we want to
compute.
Operation
x = 874,200,000×7,853,000
log x = log(874,200,000 × 7,853,000)
log x = log 874,200,000 + log 7,853,000
108 < 874,200,000 < 109, so
log 874,200,000 = 8 + log 8.742 = 8.9416
106 < 7,853,000 < 107, so
log 7,853,000 = 6 + log 7.853 = 6.8951
log x = 8.9416 + 6.8951 = 15.8367
log x = 15.8367 ⇔ 1015.8367 = x
x = 1015.8367 = 1015 ×10.8367 = 1015 × 6.866
x = 6.866×1015
Comments
This is our computation
Take the log of both sides. This is
going to transform our statement into
one with logs, taking us into
“logland.” It’s a key step!
Apply log rule #1 (see above)
Compute log 874,200,000 with the
table and save it for later. You can
also think of 874,200,000 as
8.742×108 to help find the
characteristic.
Compute log 7,853,000 with the table.
You can also think of 7,853,000 as
7.853×106 to help find the
characteristic.
Add these two logs together by hand.
Addition is easy! This is not our
answer, of course. We now need to
call in the evil twin to take us out of
logland.
Rewrite using the definition of logs
Use the antilog table to compute what
x is. Remember, x is the desired
quantity.
We’ll leave this in scientific notation.
1451
2001, Lawrence Morales; MAT107 Chapter 6 – Page 42
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If we cheat and take out our calculator, we’ll find that 874,200,000×7,853,00
= 6.8650926×1015. This means that our log table was good to three decimal
places, but missed the fourth. With tables that had more accuracy, we could
get a better answer. ♦
Check Point P
Multiply 357,600,000×4,591,000,000,000 using the log and antilog tables.
Solution
Example 29
Compute x =
1465
1466
1467
1468
1469
See the endnotes to check your answer.32
893.5
with the log tables.
2.028
Solution
Although these numbers are not very ugly, they will demonstrate the process
just fine.
Operation
893.5
x=
2.028
Comments
 893.5 
log x = log

 2.028 
log x = log 893.5 − log 2.028
log x = 2.9511 − 0.3071 = 2.644
Take the log of both sides
log x = 2.644 ⇔ 102.644 = x
x = 10 2.644 = 10 2 ×10 0.644 = 100 ×10 0.644
= 10 × 4.406
x = 440.6
1470
1471
This is our final answer. A calculator gives 440.555 as
the final answer. Not too bad (but not that great,
either).♦
Check Point Q
Compute x =
1472
1473
1474
1475
Apply log rule #2
Use the log tables to carefully find the two logs on the
right. You should verify these before you proceed. Note
that the characteristic of log893.5 is 2 since
102<893.5<103. The characteristic of 2.028 is 0. (Why?)
Definition of log
Rewrite and use the antilog table to compute the value of
x.
Solution
235.60
with the log tables.
17.22
See endnotes for a solution.33
2001, Lawrence Morales; MAT107 Chapter 6 – Page 43
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Example 30
Find the value of y = 84.658
Solution
Many complex important calculations require that numbers be raised to
powers like this. The tables can help us here as well.
Operation
y = 84.658
log y = log 84.658
log y = 8 log 84.65
Comments
Take the log of both sides.
log y = 8 × (1 + 0.9277) = 8 × 1.9277
log y = 15.4216
log y = 15.4216 ⇔ 1015.4216 = 1015 × 10 0.4216
log y = 1015 × 2.640 = 2.64 × 1015
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Use log rule #3 to move the power in front of the
log.
Take the log of 84.65. Note that the characteristic is
1 since 101<84.65<102.
At this point we have to do 8×1.9277. While we
could use a log table to do this (nested logs, yuck),
we observe that a simple multiplication like this is
probably something Napier’s contemporaries could
have done…they were concerned about the nasty
ones.
Multiply by hand. 8×1.9277. Now we know what log
y is. We can undo the log by using the antilog table.
Apply the log definition and simplify so that we can
use the antilog table.
Use the antilog table to do 100.4216 and we have a
final answer. A calculator gives the result to be
2.63643×1015.
This last example truly shows the power of logarithms. In the 1600’s, it was
not easy at all to do this last kind of computation. By converting to logs,
working with addition and subtraction (and perhaps some easy multiplication),
and then converting back from logs to regular numbers, they could raise a
decimal fraction to a power with relatively little work. ♦
Check Point R
Solution
Compute 278.410 with the log tables.
See endnotes for a solution.34
We can even take roots of numbers with logs. We simply need to remember a basic rule of
exponents:
xm/ n = n xm
2001, Lawrence Morales; MAT107 Chapter 6 – Page 44
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This tells us how to express roots with fractional powers. Here are two common identities:
x = x1 / 2
3
x = x1 / 3
Example 31
Find the value of x = 789.2
Solution
Operation
1/ 2
x = 789.2 = (789.2)
Comments
Use the exponent rule to convert.
log x = log(789.2)
1
log x = log 789.2
2
1
log x = (2.8972)
2
log x = 1.4486
Take the log of both sides.
1/ 2
Use log rule #3 to move the power in front.
Use the log table to compute log 789.2
x = 101.446 = 101 × 10 0.4486 = 10 × 2.809 = 28.09
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Take half of the result by hand…this is easy
enough to do without invoking another layer
of logs.
Rewrite x and then use the antilog table to
compute 100.4486. The final answer of 28.09
compares with a calculator result of
28.0931. ♦
Check Point S
Find the value of 3 1,342,000 using the log tables
Solution:
See endnotes for a solution.35
With this capability in hand, astronomers, mathematicians, and other scientists could speed up
their calculations and accelerate their discoveries. Of course, they still had to learn the basic
skills of arithmetic, but that wasn’t a major problem. It is analogous to when the handheld
calculator infiltrated the classroom. They made calculations much faster and easier, but they are
not the answer to learning math. Students of math still need to learn the basics so that they can
check the reasonableness of answers and so that when confronted with a relatively simple
problem, they don’t have to go searching for a calculator to pound on. Unfortunately, it seems as
though we are overly dependent on calculators. For some, it slows them down since every
computation must be done with that wretched piece of plastic. (I’ve even seen students reach for
their calculator when asked what 2×3 is!). For most, it leads to a false sense of security about
“answers” because their addiction to the calculating machine leaves them powerless to question
whether or not their answers even make sense.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 45
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PART 5: New Calculating Devices
The Slide Rule
As word of the logarithm spread, mathematicians took the idea and not only used it to help them
with their work, but also extended it. Edmund Gunter, a faculty member at Gresham College
where Briggs was a professor of geometry, invented a tool with log, tangent, and sine scales
which was used by navigators. In 1621, William Oughtred invented a slide rule that helped with
computations. The slide rule slowly evolved and became an important computing device until the
handheld calculator and desktop computer emerged as the computing workhorses of the modern
era.
Napier and Other Calculating Devices
Napier is also known for creating another, less−powerful computing device called “Napier’s
Bones.” They are multiplication devices based on the gelosia method of multiplication that we
discussed earlier. Napier noted that the columns on the gelosia grid were merely multiples of
numbers at the top of the columns. He took the columns and placed them on vertical “rods”
which could then be moved around easily and used as a mobile computing device. He called this
invention “Rabdologia” which means “a collection of rods” in Greek. They eventually picked up
the nickname “Napier’s Rods,” or “Napier’s Bones.” There is a full set of the rods in the
appendix that you can cut out if you would like. We can see how these rods
4
8
6
work with an example. Let’s compute 486×7. To do this:
0
1.
2.
3.
We take the 4, 8, and 6 rods and place them in order next to each other,
and we place the “index rod” on the right of these.
To find the product 486×7, read along the row marked with a 7 on the
index rod.
Add the numbers along the diagonals, just as before, to get 3402.
Ignore all other rows, as they are not needed (since we are multiplying
by 7).
0
0
4
2
5
8
4
1
1
1
4
3
2
4
0
4
2
8
5
3
6
6
4
7
5
5
0
1
0
3
5
6
4
6
2
4
0
3
7
8
5
5
5
6
4
4
6
4
0
0
4
7
3
4
5
3
2001, Lawrence Morales; MAT107 Chapter 6 – Page 46
2
3
3
8
8
2
2
9
1
2
5
2
Here we have two steps to take. First multiply 569 by 7.
Next, multiply 569 by 2 tens. Essentially, we are thinking
of this problem as (569)×(7×1)+569×(2×10)
0
6
1
1
1
2
3
4
5
6
7
8
9
9
0
0
Solution
4
6
1
Use the rods to find the product 569×27
8
2
5
Example 32
2
4
6
As you can imagine, this eliminates the need to write the grid down because
you can rearrange rods and read off results rather quickly.
6
4
2
3
0
3
4
8
2
4
3
0
2
8
2
6
2
2
1
2
1
6
6
2
4
6
0
8
8
2
3
0
4
2
8
4
1
1
2
3
4
5
6
7
8
9
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1600
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569×(7×ones) =
569×(2×tens) =
(569×7)×1
(569×2)×10
=
=
3983×1
1138×10
Result
= 03983
= 11380
= 15363
We are once again thinking of multiplication in terms of its relationship to
addition. By adding together all the appropriate places, we get to a result.♦
Example 33
Use the rods to compute 631×854.
Solution
We think of this as 631×(4×1) + 631×(5×10) +
631×(8×100)
631×(4×1)
=
(631×4)×1
=
2524
631×(5×10) =
(631×5)×10 =
31550
631×(8×100) =
(631×8)×100 =
504800
Result
=
538,874
6
3
0
0
6
1
0
3
0
2
1
0
0
2
0
1
3
0
2
1
0
4
0
5
1
6
2
9
4
3
1
6
8
3
A calculator check will show this to be true.♦
1
5
0
8
6
1
2
3
4
5
6
4
2
0
7
Since addition was (and is) much easier than multiplication, this method
2
1
7
4
2
0
provided a nice way to be able to compute “on the fly.” In a time when
8
merchants and other professionals in fields such as commerce were required 5 8 2 4 0 8
9
4
7
9
to do increasingly complication computations, time−saving devices like
these were often cherished. It is reported that some people even “jealousy” guarded their secrets
from others so that they could have a computational advantage.
Check Point T
Use your own set of the rods to find 542×8, and 986×36
Solution
You can use a calculator to check your answers.
Think About It
Could you adapt
the rods so that
they would
compute
785×65.3? How?
2001, Lawrence Morales; MAT107 Chapter 6 – Page 47
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The rods were eventually manufactured and carried around so that they could be used to
“mobile” calculating. Here are some sets of Napier rods that show how they looked.36
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2001, Lawrence Morales; MAT107 Chapter 6 – Page 48
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PART 6: Appendix A − Solution of the Quartic37
A quartic equation of the form:
x 4 + ax 3 + bx 2 + cx + d = 0
has a related equation (its resolvent cubic equation) of the form:
y 3 − by 2 + (ac − 4d ) y − a 2 d + 4bd − c 2 = 0
Let y be any solution of this equation and let
R=
1622
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1624
1625
1626
If R is not 0, then let
3a 2
4ab − 8c − a 3
2
− R − 2b +
D=
4
4R
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and
E=
1629
1630
1631
4ab − 8c − a 3
3a
− R 2 − 2b −
4R
4
2
If R = 0 then let
D=
1632
1633
E=
1634
1635
1636
1637
3a 2
− 2b + 2 y 2 − 4d
4
and
3a 2
− 2b − 2 y 2 − 4d
4
Then the four solutions of the original equation are given by:
a R
x=− + ±
4 2
a R
x=− − ±
4 2
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a2
−b+ y
4
D
2
E
2
Holy Cow!!!
2001, Lawrence Morales; MAT107 Chapter 6 – Page 49
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2001, Lawrence Morales; MAT107 Chapter 6 – Page 50
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PART 7: Appendix B – Log and Antilog Tables
log10 N
N
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
0
0.0000
0.0414
0.0792
0.1139
0.1461
0.1761
0.2041
0.2304
0.2553
0.2788
0.3010
0.3222
0.3424
0.3617
0.3802
0.3979
0.4150
0.4314
0.4472
0.4624
0.4771
0.4914
0.5051
0.5185
0.5315
0.5441
0.5563
0.5682
0.5798
0.5911
0.6021
0.6128
0.6232
0.6335
0.6435
0.6532
0.6628
0.6721
0.6812
0.6902
0.6990
1
0.0043
0.0453
0.0828
0.1173
0.1492
0.1790
0.2068
0.2330
0.2577
0.2810
0.3032
0.3243
0.3444
0.3636
0.3820
0.3997
0.4166
0.4330
0.4487
0.4639
0.4786
0.4928
0.5065
0.5198
0.5328
0.5453
0.5575
0.5694
0.5809
0.5922
0.6031
0.6138
0.6243
0.6345
0.6444
0.6542
0.6637
0.6730
0.6821
0.6911
0.6998
2
0.0086
0.0492
0.0864
0.1206
0.1523
0.1818
0.2095
0.2355
0.2601
0.2833
0.3054
0.3263
0.3464
0.3655
0.3838
0.4014
0.4183
0.4346
0.4502
0.4654
0.4800
0.4942
0.5079
0.5211
0.5340
0.5465
0.5587
0.5705
0.5821
0.5933
0.6042
0.6149
0.6253
0.6355
0.6454
0.6551
0.6646
0.6739
0.6830
0.6920
0.7007
3
0.0128
0.0531
0.0899
0.1239
0.1553
0.1847
0.2122
0.2380
0.2625
0.2856
0.3075
0.3284
0.3483
0.3674
0.3856
0.4031
0.4200
0.4362
0.4518
0.4669
0.4814
0.4955
0.5092
0.5224
0.5353
0.5478
0.5599
0.5717
0.5832
0.5944
0.6053
0.6160
0.6263
0.6365
0.6464
0.6561
0.6656
0.6749
0.6839
0.6928
0.7016
4
0.0170
0.0569
0.0934
0.1271
0.1584
0.1875
0.2148
0.2405
0.2648
0.2878
0.3096
0.3304
0.3502
0.3692
0.3874
0.4048
0.4216
0.4378
0.4533
0.4683
0.4829
0.4969
0.5105
0.5237
0.5366
0.5490
0.5611
0.5729
0.5843
0.5955
0.6064
0.6170
0.6274
0.6375
0.6474
0.6571
0.6665
0.6758
0.6848
0.6937
0.7024
5
0.0212
0.0607
0.0969
0.1303
0.1614
0.1903
0.2175
0.2430
0.2672
0.2900
0.3118
0.3324
0.3522
0.3711
0.3892
0.4065
0.4232
0.4393
0.4548
0.4698
0.4843
0.4983
0.5119
0.5250
0.5378
0.5502
0.5623
0.5740
0.5855
0.5966
0.6075
0.6180
0.6284
0.6385
0.6484
0.6580
0.6675
0.6767
0.6857
0.6946
0.7033
6
0.0253
0.0645
0.1004
0.1335
0.1644
0.1931
0.2201
0.2455
0.2695
0.2923
0.3139
0.3345
0.3541
0.3729
0.3909
0.4082
0.4249
0.4409
0.4564
0.4713
0.4857
0.4997
0.5132
0.5263
0.5391
0.5514
0.5635
0.5752
0.5866
0.5977
0.6085
0.6191
0.6294
0.6395
0.6493
0.6590
0.6684
0.6776
0.6866
0.6955
0.7042
7
0.0294
0.0682
0.1038
0.1367
0.1673
0.1959
0.2227
0.2480
0.2718
0.2945
0.3160
0.3365
0.3560
0.3747
0.3927
0.4099
0.4265
0.4425
0.4579
0.4728
0.4871
0.5011
0.5145
0.5276
0.5403
0.5527
0.5647
0.5763
0.5877
0.5988
0.6096
0.6201
0.6304
0.6405
0.6503
0.6599
0.6693
0.6785
0.6875
0.6964
0.7050
8
0.0334
0.0719
0.1072
0.1399
0.1703
0.1987
0.2253
0.2504
0.2742
0.2967
0.3181
0.3385
0.3579
0.3766
0.3945
0.4116
0.4281
0.4440
0.4594
0.4742
0.4886
0.5024
0.5159
0.5289
0.5416
0.5539
0.5658
0.5775
0.5888
0.5999
0.6107
0.6212
0.6314
0.6415
0.6513
0.6609
0.6702
0.6794
0.6884
0.6972
0.7059
9
0.0374
0.0755
0.1106
0.1430
0.1732
0.2014
0.2279
0.2529
0.2765
0.2989
0.3201
0.3404
0.3598
0.3784
0.3962
0.4133
0.4298
0.4456
0.4609
0.4757
0.4900
0.5038
0.5172
0.5302
0.5428
0.5551
0.5670
0.5786
0.5899
0.6010
0.6117
0.6222
0.6325
0.6425
0.6522
0.6618
0.6712
0.6803
0.6893
0.6981
0.7067
1644
2001, Lawrence Morales; MAT107 Chapter 6 – Page 51
1
4
4
3
3
3
3
3
2
2
2
2
2
2
2
2
2
2
2
2
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Ten Thousandth Parts
2 3 4 5 6 7 8 9
8 12 16 20 24 28 32 36
7 11 15 18 22 26 29 33
7 10 14 17 20 24 27 30
6 9 13 16 19 22 25 28
6 9 12 15 18 20 23 26
5 8 11 14 16 19 22 25
5 8 10 13 15 18 21 23
5 7 10 12 15 17 19 22
5 7 9 12 14 16 18 21
4 7 9 11 13 15 18 20
4 6 8 10 12 15 17 19
4 6 8 10 12 14 16 18
4 6 8 10 11 13 15 17
4 5 7 9 11 13 15 16
3 5 7 9 10 12 14 16
3 5 7 8 10 12 13 15
3 5 6 8 10 11 13 15
3 5 6 8 9 11 12 14
3 5 6 8 9 11 12 14
3 4 6 7 9 10 12 13
3 4 6 7 8 10 11 13
3 4 5 7 8 10 11 12
3 4 5 7 8 9 11 12
3 4 5 6 8 9 10 12
2 4 5 6 7 9 10 11
2 4 5 6 7 8 10 11
2 4 5 6 7 8 9 11
2 3 5 6 7 8 9 10
2 3 4 6 7 8 9 10
2 3 4 5 7 8 9 10
2 3 4 5 6 7 9 10
2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9
2 3 4 5 6 6 7 8
2 3 4 5 5 6 7 8
2 3 4 4 5 6 7 8
2 3 3 4 5 6 7 8
2 3 3 4 5 6 7 8
1645
log10 N
N
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
7.0
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
8.0
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
9.0
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
0
0.7076
0.7160
0.7243
0.7324
0.7404
0.7482
0.7559
0.7634
0.7709
0.7782
0.7853
0.7924
0.7993
0.8062
0.8129
0.8195
0.8261
0.8325
0.8388
0.8451
0.8513
0.8573
0.8633
0.8692
0.8751
0.8808
0.8865
0.8921
0.8976
0.9031
0.9085
0.9138
0.9191
0.9243
0.9294
0.9345
0.9395
0.9445
0.9494
0.9542
0.9590
0.9638
0.9685
0.9731
0.9777
0.9823
0.9868
0.9912
0.9956
1
0.7084
0.7168
0.7251
0.7332
0.7412
0.7490
0.7566
0.7642
0.7716
0.7789
0.7860
0.7931
0.8000
0.8069
0.8136
0.8202
0.8267
0.8331
0.8395
0.8457
0.8519
0.8579
0.8639
0.8698
0.8756
0.8814
0.8871
0.8927
0.8982
0.9036
0.9090
0.9143
0.9196
0.9248
0.9299
0.9350
0.9400
0.9450
0.9499
0.9547
0.9595
0.9643
0.9689
0.9736
0.9782
0.9827
0.9872
0.9917
0.9961
2
0.7093
0.7177
0.7259
0.7340
0.7419
0.7497
0.7574
0.7649
0.7723
0.7796
0.7868
0.7938
0.8007
0.8075
0.8142
0.8209
0.8274
0.8338
0.8401
0.8463
0.8525
0.8585
0.8645
0.8704
0.8762
0.8820
0.8876
0.8932
0.8987
0.9042
0.9096
0.9149
0.9201
0.9253
0.9304
0.9355
0.9405
0.9455
0.9504
0.9552
0.9600
0.9647
0.9694
0.9741
0.9786
0.9832
0.9877
0.9921
0.9965
3
0.7101
0.7185
0.7267
0.7348
0.7427
0.7505
0.7582
0.7657
0.7731
0.7803
0.7875
0.7945
0.8014
0.8082
0.8149
0.8215
0.8280
0.8344
0.8407
0.8470
0.8531
0.8591
0.8651
0.8710
0.8768
0.8825
0.8882
0.8938
0.8993
0.9047
0.9101
0.9154
0.9206
0.9258
0.9309
0.9360
0.9410
0.9460
0.9509
0.9557
0.9605
0.9652
0.9699
0.9745
0.9791
0.9836
0.9881
0.9926
0.9969
4
0.7110
0.7193
0.7275
0.7356
0.7435
0.7513
0.7589
0.7664
0.7738
0.7810
0.7882
0.7952
0.8021
0.8089
0.8156
0.8222
0.8287
0.8351
0.8414
0.8476
0.8537
0.8597
0.8657
0.8716
0.8774
0.8831
0.8887
0.8943
0.8998
0.9053
0.9106
0.9159
0.9212
0.9263
0.9315
0.9365
0.9415
0.9465
0.9513
0.9562
0.9609
0.9657
0.9703
0.9750
0.9795
0.9841
0.9886
0.9930
0.9974
5
0.7118
0.7202
0.7284
0.7364
0.7443
0.7520
0.7597
0.7672
0.7745
0.7818
0.7889
0.7959
0.8028
0.8096
0.8162
0.8228
0.8293
0.8357
0.8420
0.8482
0.8543
0.8603
0.8663
0.8722
0.8779
0.8837
0.8893
0.8949
0.9004
0.9058
0.9112
0.9165
0.9217
0.9269
0.9320
0.9370
0.9420
0.9469
0.9518
0.9566
0.9614
0.9661
0.9708
0.9754
0.9800
0.9845
0.9890
0.9934
0.9978
Ten Thousandth Parts
6
0.7126
0.7210
0.7292
0.7372
0.7451
0.7528
0.7604
0.7679
0.7752
0.7825
0.7896
0.7966
0.8035
0.8102
0.8169
0.8235
0.8299
0.8363
0.8426
0.8488
0.8549
0.8609
0.8669
0.8727
0.8785
0.8842
0.8899
0.8954
0.9009
0.9063
0.9117
0.9170
0.9222
0.9274
0.9325
0.9375
0.9425
0.9474
0.9523
0.9571
0.9619
0.9666
0.9713
0.9759
0.9805
0.9850
0.9894
0.9939
0.9983
7
0.7135
0.7218
0.7300
0.7380
0.7459
0.7536
0.7612
0.7686
0.7760
0.7832
0.7903
0.7973
0.8041
0.8109
0.8176
0.8241
0.8306
0.8370
0.8432
0.8494
0.8555
0.8615
0.8675
0.8733
0.8791
0.8848
0.8904
0.8960
0.9015
0.9069
0.9122
0.9175
0.9227
0.9279
0.9330
0.9380
0.9430
0.9479
0.9528
0.9576
0.9624
0.9671
0.9717
0.9763
0.9809
0.9854
0.9899
0.9943
0.9987
8
0.7143
0.7226
0.7308
0.7388
0.7466
0.7543
0.7619
0.7694
0.7767
0.7839
0.7910
0.7980
0.8048
0.8116
0.8182
0.8248
0.8312
0.8376
0.8439
0.8500
0.8561
0.8621
0.8681
0.8739
0.8797
0.8854
0.8910
0.8965
0.9020
0.9074
0.9128
0.9180
0.9232
0.9284
0.9335
0.9385
0.9435
0.9484
0.9533
0.9581
0.9628
0.9675
0.9722
0.9768
0.9814
0.9859
0.9903
0.9948
0.9991
9
0.7152
0.7235
0.7316
0.7396
0.7474
0.7551
0.7627
0.7701
0.7774
0.7846
0.7917
0.7987
0.8055
0.8122
0.8189
0.8254
0.8319
0.8382
0.8445
0.8506
0.8567
0.8627
0.8686
0.8745
0.8802
0.8859
0.8915
0.8971
0.9025
0.9079
0.9133
0.9186
0.9238
0.9289
0.9340
0.9390
0.9440
0.9489
0.9538
0.9586
0.9633
0.9680
0.9727
0.9773
0.9818
0.9863
0.9908
0.9952
0.9996
1646
2001, Lawrence Morales; MAT107 Chapter 6 – Page 52
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
2
2
2
2
2
2
2
2
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
1
1
1
1
1
1
1
1
1
1
1
1
4
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
5
4
4
4
4
4
4
4
4
4
4
4
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
6
5
5
5
5
5
5
5
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
7
6
6
6
6
5
5
5
5
5
5
5
5
5
5
5
5
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
3
3
3
3
3
3
3
3
3
3
3
3
3
8
7
7
6
6
6
6
6
6
6
6
6
6
5
5
5
5
5
5
5
5
5
5
5
5
5
5
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
3
9
8
7
7
7
7
7
7
7
7
6
6
6
6
6
6
6
6
6
6
6
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
4
4
4
4
4
4
4
4
4
4
4
4
4
P
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.20
0.21
0.22
0.23
0.24
0.25
0.26
0.27
0.28
0.29
0.30
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.40
0.41
0.42
0.43
0.44
0.45
0.46
0.47
0.48
0.49
0.50
0
1.000
1.023
1.047
1.072
1.096
1.122
1.148
1.175
1.202
1.230
1.259
1.288
1.318
1.349
1.380
1.413
1.445
1.479
1.514
1.549
1.585
1.622
1.660
1.698
1.738
1.778
1.820
1.862
1.905
1.950
1.995
2.042
2.089
2.138
2.188
2.239
2.291
2.344
2.399
2.455
2.512
2.570
2.630
2.692
2.754
2.818
2.884
2.951
3.020
3.090
3.162
1
1.002
1.026
1.050
1.074
1.099
1.125
1.151
1.178
1.205
1.233
1.262
1.291
1.321
1.352
1.384
1.416
1.449
1.483
1.517
1.552
1.589
1.626
1.663
1.702
1.742
1.782
1.824
1.866
1.910
1.954
2.000
2.046
2.094
2.143
2.193
2.244
2.296
2.350
2.404
2.460
2.518
2.576
2.636
2.698
2.761
2.825
2.891
2.958
3.027
3.097
3.170
ANTILOG TABLE: 10P = N
2
3
4
5
6
1.005 1.007 1.009 1.012 1.014
1.028 1.030 1.033 1.035 1.038
1.052 1.054 1.057 1.059 1.062
1.076 1.079 1.081 1.084 1.086
1.102 1.104 1.107 1.109 1.112
1.127 1.130 1.132 1.135 1.138
1.153 1.156 1.159 1.161 1.164
1.180 1.183 1.186 1.189 1.191
1.208 1.211 1.213 1.216 1.219
1.236 1.239 1.242 1.245 1.247
1.265 1.268 1.271 1.274 1.276
1.294 1.297 1.300 1.303 1.306
1.324 1.327 1.330 1.334 1.337
1.355 1.358 1.361 1.365 1.368
1.387 1.390 1.393 1.396 1.400
1.419 1.422 1.426 1.429 1.432
1.452 1.455 1.459 1.462 1.466
1.486 1.489 1.493 1.496 1.500
1.521 1.524 1.528 1.531 1.535
1.556 1.560 1.563 1.567 1.570
1.592 1.596 1.600 1.603 1.607
1.629 1.633 1.637 1.641 1.644
1.667 1.671 1.675 1.679 1.683
1.706 1.710 1.714 1.718 1.722
1.746 1.750 1.754 1.758 1.762
1.786 1.791 1.795 1.799 1.803
1.828 1.832 1.837 1.841 1.845
1.871 1.875 1.879 1.884 1.888
1.914 1.919 1.923 1.928 1.932
1.959 1.963 1.968 1.972 1.977
2.004 2.009 2.014 2.018 2.023
2.051 2.056 2.061 2.065 2.070
2.099 2.104 2.109 2.113 2.118
2.148 2.153 2.158 2.163 2.168
2.198 2.203 2.208 2.213 2.218
2.249 2.254 2.259 2.265 2.270
2.301 2.307 2.312 2.317 2.323
2.355 2.360 2.366 2.371 2.377
2.410 2.415 2.421 2.427 2.432
2.466 2.472 2.477 2.483 2.489
2.523 2.529 2.535 2.541 2.547
2.582 2.588 2.594 2.600 2.606
2.642 2.649 2.655 2.661 2.667
2.704 2.710 2.716 2.723 2.729
2.767 2.773 2.780 2.786 2.793
2.831 2.838 2.844 2.851 2.858
2.897 2.904 2.911 2.917 2.924
2.965 2.972 2.979 2.985 2.992
3.034 3.041 3.048 3.055 3.062
3.105 3.112 3.119 3.126 3.133
3.177 3.184 3.192 3.199 3.206
7
1.016
1.040
1.064
1.089
1.114
1.140
1.167
1.194
1.222
1.250
1.279
1.309
1.340
1.371
1.403
1.435
1.469
1.503
1.538
1.574
1.611
1.648
1.687
1.726
1.766
1.807
1.849
1.892
1.936
1.982
2.028
2.075
2.123
2.173
2.223
2.275
2.328
2.382
2.438
2.495
2.553
2.612
2.673
2.735
2.799
2.864
2.931
2.999
3.069
3.141
3.214
8
1.019
1.042
1.067
1.091
1.117
1.143
1.169
1.197
1.225
1.253
1.282
1.312
1.343
1.374
1.406
1.439
1.472
1.507
1.542
1.578
1.614
1.652
1.690
1.730
1.770
1.811
1.854
1.897
1.941
1.986
2.032
2.080
2.128
2.178
2.228
2.280
2.333
2.388
2.443
2.500
2.559
2.618
2.679
2.742
2.805
2.871
2.938
3.006
3.076
3.148
3.221
9
1.021
1.045
1.069
1.094
1.119
1.146
1.172
1.199
1.227
1.256
1.285
1.315
1.346
1.377
1.409
1.442
1.476
1.510
1.545
1.581
1.618
1.656
1.694
1.734
1.774
1.816
1.858
1.901
1.945
1.991
2.037
2.084
2.133
2.183
2.234
2.286
2.339
2.393
2.449
2.506
2.564
2.624
2.685
2.748
2.812
2.877
2.944
3.013
3.083
3.155
3.228
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2001, Lawrence Morales; MAT107 Chapter 6 – Page 53
Thousandth Parts
3 4 5 6 7
1 1 1 1 2
1 1 1 1 2
1 1 1 1 2
1 1 1 2 2
1 1 1 2 2
1 1 1 2 2
1 1 1 2 2
1 1 1 2 2
1 1 1 2 2
1 1 1 2 2
1 1 1 2 2
1 1 2 2 2
1 1 2 2 2
1 1 2 2 2
1 1 2 2 2
1 1 2 2 2
1 1 2 2 2
1 1 2 2 2
1 1 2 2 2
1 1 2 2 3
1 1 2 2 3
1 2 2 2 3
1 2 2 2 3
1 2 2 2 3
1 2 2 2 3
1 2 2 3 3
1 2 2 3 3
1 2 2 3 3
1 2 2 3 3
1 2 2 3 3
1 2 2 3 3
1 2 2 3 3
1 2 2 3 3
2 2 3 3 4
2 2 3 3 4
2 2 3 3 4
2 2 3 3 4
2 2 3 3 4
2 2 3 3 4
2 2 3 3 4
2 2 3 4 4
2 2 3 4 4
2 2 3 4 4
2 3 3 4 4
2 3 3 4 5
2 3 3 4 5
2 3 3 4 5
2 3 3 4 5
2 3 4 4 5
2 3 4 4 5
2 3 4 4 5
8
2
2
2
2
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
4
4
4
4
4
4
4
4
4
4
5
5
5
5
5
5
5
5
5
6
6
6
6
9
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
3
3
3
3
3
4
4
4
4
4
4
4
4
4
4
4
5
5
5
5
5
5
5
5
5
6
6
6
6
6
6
6
7
7
1647
P
0.51
0.52
0.53
0.54
0.55
0.56
0.57
0.58
0.59
0.60
0.61
0.62
0.63
0.64
0.65
0.66
0.67
0.68
0.69
0.70
0.71
0.72
0.73
0.74
0.75
0.76
0.77
0.78
0.79
0.80
0.81
0.82
0.83
0.84
0.85
0.86
0.87
0.88
0.89
0.90
0.91
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
0
3.236
3.311
3.388
3.467
3.548
3.631
3.715
3.802
3.890
3.981
4.074
4.169
4.266
4.365
4.467
4.571
4.677
4.786
4.898
5.012
5.129
5.248
5.370
5.495
5.623
5.754
5.888
6.026
6.166
6.310
6.457
6.607
6.761
6.918
7.079
7.244
7.413
7.586
7.762
7.943
8.128
8.318
8.511
8.710
8.913
9.120
9.333
9.550
9.772
1
3.243
3.319
3.396
3.475
3.556
3.639
3.724
3.811
3.899
3.990
4.083
4.178
4.276
4.375
4.477
4.581
4.688
4.797
4.909
5.023
5.140
5.260
5.383
5.508
5.636
5.768
5.902
6.039
6.180
6.324
6.471
6.622
6.776
6.934
7.096
7.261
7.430
7.603
7.780
7.962
8.147
8.337
8.531
8.730
8.933
9.141
9.354
9.572
9.795
2
3.251
3.327
3.404
3.483
3.565
3.648
3.733
3.819
3.908
3.999
4.093
4.188
4.285
4.385
4.487
4.592
4.699
4.808
4.920
5.035
5.152
5.272
5.395
5.521
5.649
5.781
5.916
6.053
6.194
6.339
6.486
6.637
6.792
6.950
7.112
7.278
7.447
7.621
7.798
7.980
8.166
8.356
8.551
8.750
8.954
9.162
9.376
9.594
9.817
ANTILOG TABLE: 10P = N
3
4
5
6
3.258 3.266 3.273 3.281
3.334 3.342 3.350 3.357
3.412 3.420 3.428 3.436
3.491 3.499 3.508 3.516
3.573 3.581 3.589 3.597
3.656 3.664 3.673 3.681
3.741 3.750 3.758 3.767
3.828 3.837 3.846 3.855
3.917 3.926 3.936 3.945
4.009 4.018 4.027 4.036
4.102 4.111 4.121 4.130
4.198 4.207 4.217 4.227
4.295 4.305 4.315 4.325
4.395 4.406 4.416 4.426
4.498 4.508 4.519 4.529
4.603 4.613 4.624 4.634
4.710 4.721 4.732 4.742
4.819 4.831 4.842 4.853
4.932 4.943 4.955 4.966
5.047 5.058 5.070 5.082
5.164 5.176 5.188 5.200
5.284 5.297 5.309 5.321
5.408 5.420 5.433 5.445
5.534 5.546 5.559 5.572
5.662 5.675 5.689 5.702
5.794 5.808 5.821 5.834
5.929 5.943 5.957 5.970
6.067 6.081 6.095 6.109
6.209 6.223 6.237 6.252
6.353 6.368 6.383 6.397
6.501 6.516 6.531 6.546
6.653 6.668 6.683 6.699
6.808 6.823 6.839 6.855
6.966 6.982 6.998 7.015
7.129 7.145 7.161 7.178
7.295 7.311 7.328 7.345
7.464 7.482 7.499 7.516
7.638 7.656 7.674 7.691
7.816 7.834 7.852 7.870
7.998 8.017 8.035 8.054
8.185 8.204 8.222 8.241
8.375 8.395 8.414 8.433
8.570 8.590 8.610 8.630
8.770 8.790 8.810 8.831
8.974 8.995 9.016 9.036
9.183 9.204 9.226 9.247
9.397 9.419 9.441 9.462
9.616 9.638 9.661 9.683
9.840 9.863 9.886 9.908
7
3.289
3.365
3.443
3.524
3.606
3.690
3.776
3.864
3.954
4.046
4.140
4.236
4.335
4.436
4.539
4.645
4.753
4.864
4.977
5.093
5.212
5.333
5.458
5.585
5.715
5.848
5.984
6.124
6.266
6.412
6.561
6.714
6.871
7.031
7.194
7.362
7.534
7.709
7.889
8.072
8.260
8.453
8.650
8.851
9.057
9.268
9.484
9.705
9.931
8
3.296
3.373
3.451
3.532
3.614
3.698
3.784
3.873
3.963
4.055
4.150
4.246
4.345
4.446
4.550
4.656
4.764
4.875
4.989
5.105
5.224
5.346
5.470
5.598
5.728
5.861
5.998
6.138
6.281
6.427
6.577
6.730
6.887
7.047
7.211
7.379
7.551
7.727
7.907
8.091
8.279
8.472
8.670
8.872
9.078
9.290
9.506
9.727
9.954
9
3.304
3.381
3.459
3.540
3.622
3.707
3.793
3.882
3.972
4.064
4.159
4.256
4.355
4.457
4.560
4.667
4.775
4.887
5.000
5.117
5.236
5.358
5.483
5.610
5.741
5.875
6.012
6.152
6.295
6.442
6.592
6.745
6.902
7.063
7.228
7.396
7.568
7.745
7.925
8.110
8.299
8.492
8.690
8.892
9.099
9.311
9.528
9.750
9.977
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2001, Lawrence Morales; MAT107 Chapter 6 – Page 54
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
4
4
4
4
4
4
4
4
4
4
4
5
Thousandth Parts
3 4 5 6 7
2 3 4 5 5
2 3 4 5 5
2 3 4 5 6
2 3 4 5 6
2 3 4 5 6
3 3 4 5 6
3 3 4 5 6
3 4 4 5 6
3 4 5 5 6
3 4 5 6 7
3 4 5 6 7
3 4 5 6 7
3 4 5 6 7
3 4 5 6 7
3 4 5 6 7
3 4 5 6 8
3 4 5 7 8
3 4 6 7 8
3 5 6 7 8
4 5 6 7 8
4 5 6 7 8
4 5 6 7 9
4 5 6 8 9
4 5 6 8 9
4 5 7 8 9
4 5 7 8 9
4 6 7 8 10
4 6 7 8 10
4 6 7 9 10
4 6 7 9 10
5 6 8 9 11
5 6 8 9 11
5 6 8 10 11
5 6 8 10 11
5 7 8 10 12
5 7 9 10 12
5 7 9 10 12
5 7 9 11 12
5 7 9 11 13
6 7 9 11 13
6 8 10 11 13
6 8 10 12 14
6 8 10 12 14
6 8 10 12 14
6 8 10 13 15
6 9 11 13 15
7 9 11 13 15
7 9 11 13 16
7 9 11 14 16
8
6
6
6
7
7
7
7
7
7
7
8
8
8
8
8
9
9
9
9
9
10
10
10
10
11
11
11
11
12
12
12
12
13
13
13
14
14
14
15
15
15
16
16
16
17
17
18
18
18
9
7
7
7
7
7
8
8
8
8
8
9
9
9
9
9
10
10
10
10
11
11
11
11
12
12
12
12
13
13
13
14
14
14
15
15
15
16
16
16
17
17
18
18
18
19
19
20
20
21
1648
1649
1650
PART 8: Appendix C – Napier’s Rods
Index
Rod
1651
1652
1653
2001, Lawrence Morales; MAT107 Chapter 6 – Page 55
1654
2001, Lawrence Morales; MAT107 Chapter 6 – Page 56
1655
1656
1657
1658
1659
1660
1661
1662
1663
1664
1665
1666
1667
1668
1669
1670
1671
1672
1673
1674
1675
1676
1677
1678
1679
1680
1681
1682
1683
1684
1685
1686
1687
1688
1689
1690
1691
1692
1693
1694
1695
1696
1697
1698
1699
PART 9: Homework
Multiplication with the Gelosia Grid System
Multiply the following using the gelosia grid system. There is a sheet of blank gelosia grids at the
end of the homework section if you need them.
1)
356×706
2) 45,632×673
3)
56.983×2.98
4) 0.03652×0.00045
Show with gelosia grids how to do the following computations. (Recall how the Babylonians did
division.)
5)
784.32÷4
6) 8943.2÷3
7)
873÷20
8) 838.45÷8
Checking Solutions of Cubic Equations
Determine whether or not the given values of x are solutions to the associated equation
9)
Is x = 3 a solution of x3 + 2 x 2 − 5 x + 3 = 33 ?
10) Is x = 2 a solution of x3 + 2 x 2 − 5 x + 3 = 33 ?
11) Is x = 4 a solution of 2 x3 − 4 x 2 + 3x + 1 = 77 ?
12) Is x = 5 a solution of 2 x 3 − 4 x 2 + 3x + 1 = 156 ?
Using Cardano’s Formula on Depressed Cubics
Use Cardano’s formula to solve the following equations. Show all steps so that it is clear how
your solution is found. Give your answer in two forms: (1) Exact from with all radicals and
simplified fractions preserved and (2) in decimal form rounded to four decimal places, where
appropriate.
13)
x3 + 4 x = 39
14) x3 + 9 x = 54
15)
x3 − 8 x + 32 = 0
16) x3 + 24 x = 16
17)
x3 = 9 x + 12
18) x3 + 5 x = 5
2001, Lawrence Morales; MAT107 Chapter 6 – Page 57
1700
1701
1702
1703
1704
1705
1706
1707
1708
1709
1710
1711
1712
1713
1714
1715
1716
1717
1718
1719
1720
1721
1722
1723
1724
1725
1726
1727
1728
1729
1730
1731
1732
1733
1734
1735
1736
1737
1738
1739
1740
1741
1742
1743
Using Cardano’s Formula on Non-Depressed Cubics
Solve the following cubic using Cardano’s method of first producing a depressed cubic and then
using Cardano’s equation. Show all algebraic steps. Give your answer in two forms: (1) Exact
from with all radicals and fractions preserved and (2) in decimal form rounded to four decimal
places, where appropriate.
19)
x3 − 3x 2 + 4 x − 70 = 0
20)
x3 + 12 x 2 + 51x − 64 = 0
21)
x3 − 15 x 2 + 69 x − 135 = 0
22)
x3 − 12 x 2 + 56 x − 1176 = 0
23)
2 x3 + 24 x 2 + 86 x + 64 = 0
Stevin’s Notation
24) Write in Stevin notation: 678.298
25) Write in Stevin notation: 559.0391
26) Write in Stevin notation: 0.00397
27) What calculation is being demonstrated in the picture shown from
Simon Stevin? Clearly explain your reasoning and conclusions.
28) Use the figure referred to in Problem (27) to show how Stevin might add the following
numbers. Include a final result, written in Stevin’s notation, as part of your work.
34.98 + 238.924 + 183.85
29) Use the figure referred to in Problem (27) to show how Stevin might add the following
numbers. Include a final result, written in Stevin’s notation, as part of your work.
8935.34 + 98.03765 + 3.498
30) Use the figure referred to in Problem (27) to show how Stevin might add the following
numbers. Include a final result, written in Stevin’s notation, as part of your work.
34.982038 + 23.998371 + 33.4981272
2001, Lawrence Morales; MAT107 Chapter 6 – Page 58
1744
1745
1746
1747
1748
1749
1750
1751
Basic Logarithm Rules
Write each of the following as statements with logarithms.
31)
43 = 64
32) 811/ 4 = 3
33)
x5 = y
34) 90 = 1
1752
35)
4−2 =
1753
1754
1755
Write each of the following as statements with powers.
1756
37)
log 4 16 = 2
38) log 25 5 =
39)
log z 3 = 2
40) log100 1 = 0
41)
log m x = y
42) log 4 x = 10
1757
1758
1759
1760
1761
1762
1763
1764
1765
1766
1767
1768
1769
1770
1771
1772
1773
1774
1775
1
16
36) A1 / x = 3
1
2
Use the following rules of logarithms to write the following logarithms as sum, difference, or
product.
log( AB) = log A + log B
log( A / B ) = log A − log B
log b nc = c log b n
43)
log(184 × 332)
45)
log
47)
log
49)
log 3 129.3 × 17.04
83.5
19.1
44) log 72.88
46) log 9547
1776
1777
104.8
33.8
48) log ( 3.205 )
7
1778
1779
50) log
3
58
1780
2001, Lawrence Morales; MAT107 Chapter 6 – Page 59
1781
1782
1783
1784
1785
1786
1787
1788
1789
1790
1791
1792
1793
1794
1795
1796
1797
1798
1799
1800
1801
1802
1803
1804
1805
1806
1807
1808
1809
1810
1811
1812
1813
1814
1815
1816
1817
1818
1819
 A
51) Prove the logarithm rule log   = log A − log B . See the proof earlier in the chapter for
 B
log( AB) = logA + logB for an example on how to approach the proof.
Using the Log Tables
Use the log tables to practice finding the values of the following base-10 logarithms.
52) log 7.32
53) log 4.394
54) log 85.66
55) log 79.329
56) log 6345
57) log 9386.55
58) log 18,387.34
59) log 523,764.88
Using the Antilog Tables
Use the antilog tables to practice finding the values of N in the given expressions.
60) log N = 0.536
61) log N = 0.7256
62) log N = 2.349
63) log N = 4.6301
64) log N = 6.8833
65) log N = 7.45838
Using Log and Antilog Tables to Do Calculations
Using the rules of logarithms and the logarithm and antilog tables, compute the values of the
following problems. Show all computations (by hand), including logarithm manipulations.
Express your answer to as many decimal places as the tables yield. Do not use a calculator at
all.
66) 57.3×87.2
68)
9,838×3,427,200
67) 732.3×8750
69) 59×0.0328
70)
48.3
22.9
71)
5932
286.3
72)
3,865,800
23971
73)
2.783
0.273
1820
1821
1822
2001, Lawrence Morales; MAT107 Chapter 6 – Page 60
4.873
75) 87, 450,0002
1823
1824
1825
1826
74)
1827
78)
18
5.73
79) 2.456 × 34.8
80)
3.8710
456.12
81)
76)
77) 184.5721/10
372.6
1828
1829
1830
1831
1832
1833
1834
1835
1836
1837
1838
1839
1840
1841
1842
1843
1844
1845
1846
1847
1848
1849
1850
1851
1852
1853
1854
1855
1856
1857
1858
1859
1860
1861
1862
5490 × 3599
Applications of Logarithms
If interest is compounded n times a year, then when you invest P dollars for t years at an interest
rate of r%, the value of your investment after t years is given by the formula:
 r
A = P 1 + 
 n
nt
82) Suppose you invest $1000 for one year at a 6% interest rate. Interest is compounded once a
month (12 times a year).
a.
Use a calculator to find the value of the account, A, at the end of that year.
b. Without using a calculator, find the value of the account by using log rules and the log
tables. You can do basic computations by hand, but the exponent must be computed with
the log tables.
_____
83) Suppose you’re lived more than a hundred years ago and you make the interesting
discovery that the time t that it takes for a pendulum to swing from one side to the other
L
and back (called its period) is given by the formula t = 2π
. In this equation, L is the
32
length of the pendulum (feet), π can be approximated as 3.142, and the 32 inside the square
root is the gravity constant for Earth.
a.
Suppose you are working on a new, large clock as a gift to a political leader. You will need
an accurate measure of the period of the pendulum you will use for the clock. The
pendulum is 15 feet long. Use log rules, log tables, and pencil and paper to find the period
of the clock to four decimal places. Do not use a calculator for any calculations. Show all
steps clearly and neatly.
b. Use regular algebra to solve for L in terms of t and π. Do not approximate for a value of π.
Explain what this new equation would tell you.
c.
Suppose you are building a clock that is to have a period of approximately t = 2.896
seconds (that’s the total time to go both back and forth). Use log rules, log tables, and
pencil and paper to find the length of the pendulum to four decimal places. Do not use a
calculator for any calculations. Show all steps clearly and neatly.
2001, Lawrence Morales; MAT107 Chapter 6 – Page 61
1863
1864
1865
1866
1867
1868
1869
1870
1871
1872
1873
1874
1875
1876
1877
1878
1879
1880
1881
1882
1883
1884
1885
1886
1887
1888
1889
1890
1891
1892
1893
1894
Writing
Write a short essay on the given topic. It should not be more than one page and if you can type it
(double−spaced), I would appreciate it. If you cannot type it, your writing must be legible.
Attention to grammar is important, although it does not have to be perfect grammatically…I just
want to be able to understand it.
84) Was Cardano justified in publishing the solution to the cubic equation? Did he violate his
oath? Explain your position.
85) Research Cardano’s life and give a brief biography of the man that extends the information
given in this chapter or in class.
86) Research the live of Evariste Galois and give a brief biography of the man that extends the
information given in this chapter or in class.
87) Research the history of the slide rule and give a brief account of its invention, evolution,
use, etc.
88) Research the role of geometry in the Renaissance and give a brief description of what you
find.
89) Research the Golden Ratio (Golden Mean) in nature and art and give a description of what
you find.
90) Research the Golden Ratio (Golden Mean) and its current connection to “Sacred
Geometry” and give a description of what you find.
91) Research the Fibonacci numbers; describe them, where they are found, and how they are
used. If possible, try to make a connection between the Fibonacci numbers and the Golden
Ratio (Golden Mean)
2001, Lawrence Morales; MAT107 Chapter 6 – Page 62
PART 10: Endnotes
1895
1
Solution to Check Point A
Your grid should look like this. The answer is 30,004,188
2
Swetz, Learning Activities in the History of Math, page 181
Katz, page 260
4
Picture from http://www-groups.dcs.st-andrews.ac.uk/~history/PictDisplay/Tartaglia.html
5
Picture from http://www-groups.dcs.st-andrews.ac.uk/~history/PictDisplay/Cardan.html
6
Dunham, page 140.
7
Dunham, page 141.
8
Dunham, page 142.
9
Dunham, pp. 142-145.
10
Solution to Check Point B
Answer: You should get x = 3.
11
Solution to Check Point C
You should get x ≈ 1.8113.
12
Solution to Check Point D
You should use x = y – 4 to get a depressed cubic of y3 – 44y = 333. This gives y = 9 as a
solution which in turn gives x = 5 as the final solution.
13
Katz, page 365
14
Calinger, page 477
15
http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Stevin.html
16
Solution to Check Point E
3,456.28547 = 3,456b2c8d5e4f7g
17
Calinger, page 482
18
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Viete.html
19
Calinger, page 479
20
http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Napier.html
21
Five Fingers to Infinity, page 428
22
Calinger, page 491
23
Solution to Check Point F
log 4 16 = y . The value of y is 2 since 42 = 16.
24
Solution to Check Point G
3
yw = z
25
Sojlution to Check Point H
log(1298) + log(3892)
26
Solution to Check Point I
log(z) – log3.5
27
Solution to Check Point J
2001, Lawrence Morales; MAT107 Chapter 6 – Page 63
2
log 3 x 2 = log x 2 / 3 = log x
3
28
Solution to Check Point K
29
Solution to Check Point M
30
Solution to Check Point N
log10 8.55 = 0.9320
31
Did you get 4.5518?
You should get 1.525. N = 100.1834
Solution to Check Point O
419,800
Solution to Check Point P
You should get a total of 21.2153 from the log tables. The antilog tables gives you
1.642×1021 as a final answer.
33
Solution to Check Point Q
You should get a total of 1.1362 from the log tables. The antilog tables gives you 13.689
as a final answer.
34
Solution to Check Point R
The log tables should give you 24.4446. The antilog tables gives a final answer of 2.784×1024
which is off by quite a bit.
35
Solution to Check Point S
The log tables should give you 2.0426. The antilog tables give a final answer of 110.4
36
http://www.cs.nyu.edu/courses/spring00/V22.0004-002/history/napier2.html
37
CRC Handbook, page 12
32
2001, Lawrence Morales; MAT107 Chapter 6 – Page 64