Download Chapter 2

Contemporary Business Mathematics
with Canadian Applications
Eighth Edition
S. A. Hummelbrunner/K. Suzanne Coombs
PowerPoint: D. Johnston
Chapter 2
Review of Basic Algebra
Copyright © 2008 Pearson Education Canada
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Objectives
After completing chapter two, the student will
be able to:
• Simplify algebraic expressions.
• Evaluate expressions with positive,
negative, and exponent zero.
• Use a calculator to evaluate expressions
with fractional exponents.
• Write exponential expressions in
logarithmic form.
(continued)
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Objectives
(continued)
• Use a calculator to determine the value of
natural logarithms.
• Solve algebraic equations using addition,
subtraction, multiplication, division and
formula rearrangement.
• Solve word problems by creating equations.
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Formula Simplification
You can combine like terms.
•
•
•
•
•
3x + 2x + 7x = 12x
6x - 4y -2 x +8y = 4x + 4y
7xy - 3xy - xy = 3xy
4c - 5d -3c +4d = c - d
3x2 + 2.5x2 = 5.5x2
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Formula Simplification
• If brackets are preceded by a + sign, do not
change the sign of the terms inside the
brackets.
• (7a - 2b) + (4a -5b) = 11a -7b
• If brackets are preceded by a - sign, change
the sign of each term inside the brackets.
• (4c - 5d) - (2c -3d) = 2c -2d
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Formula Evaluation
I = 60, R = 0.05, T = 3 P = 1000, A=1500
Formula
Evaluation
I
RT
60 = 400
.05x3
A
1+RT
1500
= 1304.35
1+.05x3
P(1+RT)
1000(1+.05x3) =
1150
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Exponents
•
•
•
•
•
Power
an
Base
a
Exponent n
The factor a is multiplied by itself n times.
POWER = BASE TO THE EXPONENT
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Using Exponents
•
•
•
•
63 = 6 x 6 x 6
(-2)4 = (-2)(-2)(-2)(-2)
(1+i)5 = (1+i)(1+i)(1+i)(1+i)(1+i)
(1/4)2 = (.25)(.25)
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Operations with Powers
am x an = a m+n
23 x 22 = 2 3+2 = 25
am an = a m-n
2523 = 2 5-3 =22
(am)n = a mn
(23) 2 = 2 (3x2) = 26
(ab)m = ambm
(4x5)2 = 42 x 52
(a/b)m = am/bm
(4/2)3 = 43/23
a = 1 if a  0
a0 is undefined if a = 0
0
Copyright © 2008 Pearson Education Canada
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Negative and Zero Exponents
•
•
•
•
•
•
a -n = 1/an
2 -4 = 1/24 = 1/16
(1+i) -3 = 1/(1+i)3
(1.05)0 = 1
(-4) -2 = 1/(-4)2 = 1/16
(¾) -3 = 1/(¾)3 = 2.37
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Fractional Exponents
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Fractional Exponents
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Examples of Fractional
Exponents
49
27
36
(1/2)
-(1/3)
(3/2)
= 7
= 1/3
= 216
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Exponents and Logarithms
Exponential form
N=b
8=2
y
y = log b N
3
100 = 10
Logarithmic form
3 = log 2 8
2
2 = log 10 100
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Logarithm
A logarithm is defined as the exponent to which a base must be
raised to produce a given number.
64 = 26

6 is the logarithm of 64 to the base 2,
written 6 = log264
** Logarithm with base 10 are called Common Logarithms.
Log 1000 = 3 or Log101000 = 3
** x = Loge y or x = ln y is called Natural Logarithm
where e = 2.718282 approximately.
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Properties of Logarithms
ln(ab) = ln a + ln b
ln[(3)(6)] = ln 3 + ln 6
ln(a/b) = ln a – ln b
ln (50/3) = ln 50 – ln 3
k
Ln(a ) = k ln a
6
ln (1.03) = 6 ln (1.03)
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Equations
• An equation is an expression of equality
between two algebraic expressions.
• 3x = 36
• 2x + 4 = 60
• 5x - .4 = 2.5
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Solving Equations Using
Addition
•
•
•
•
X-3=9
Add 3 to both sides .
X-3+3=9+3
X = 12
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Solving Equations Using
Subtraction
•
•
•
•
X+3=8
Subtract 3 from both sides.
X + 3 -3 = 8 - 3
X=5
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Solving Equations Using
Multiplication
X =6
3
Multiply both sides by 3.
3 (X) = 6 (3)
3
X = 18
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2-20
Solving Equations Using
Division
4 X = 24
Divide both sides by 4.
4X = 24
4
4
X=6
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Using Two or More Operations
to Solve an Equation
3X-5=2X+6
3X – 2X = 6 + 5
X = 11
You can substitute your result back into the
original equation to check your answer.
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Solving Word Problems
• Step 1 - Describe in words the unknown X.
• Step 2 - Translate information in the word
problem in terms of the unknown X.
• Step 3 - Set up an algebraic equation
matching the expression from step #2 to a
specific number.
• Step 4 - Solve equation, state conclusion,
and check result .
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Solving a Word Problem
A VCR was advertised for $110 which
represented a reduction of 30% off the
original price. Find the original price.
Step 1 Let X represent the original price.
Step 2 Sales price = X – 0.30X
Step 3 X – 0.30 X = 110
Step 4 0.70 X = 110 X = 110/0.70 = 157.14
Substitute 157.14 back into original statement.
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Solving word problems
Problem 1.
**Material Cost of product is $4 less than twice the cost of direct
labour.
**Overhead cost is 5/6 of direct labour.
**Total cost of product is $157
**What is the amount of each of the three elements of cost?
Problem 2.
**Company laid off one-sixth of workforce.
**Number of employees after layoff is 690.
**How many employees were laid off?
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Summary
Exponential expressions, logarithms, and
algebraic equations are important tools in the
solution of problems found in business
mathematics and finance.
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