Contemporary Business Mathematics With Canadian Applications Chapter 2

Copyright © 2012 Pearson Canada Inc. 2-1

Chapter 2

Review of Basic Algebra

Contemporary Business Mathematics with Canadian Applications

Ninth Edition S. A. Hummelbrunner/K. Suzanne Coombs

PowerPoint: D. Johnston, Revised by P. Au


ObjectivesAfter 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)


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.


Formula SimplificationYou 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


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


Formula EvaluationI = 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


Exponents

• Power an

• Base a

• Exponent n

• The factor a is multiplied by itself n times.

• POWER = BASE TO THE EXPONENT


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)


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

a0 = 1 if a ≠ 0

a0 is undefined if a = 0


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


Fractional Exponents


Fractional Exponents


Examples of Fractional Exponents


Exponents and Logarithms

Exponential form Logarithmic form

N = b y y = log b N

8 = 2 3 3 = log 2 8

100 = 10 2 2 = log 10 100


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

Ln(ak) = k ln a ln (1.03)6 = 6 ln (1.03)


Equations

• An equation is an expression of equality between two algebraic expressions.

3x = 36

2x + 4 = 60

5x - .4 = 2.5


Solving Equations Using Addition

• X - 3 = 9

Add 3 to both sides.

X - 3 + 3 = 9 + 3

X = 12


Solving Equations Using Subtraction

• X + 3 = 8

Subtract 3 from both sides.

X + 3 - 3 = 8 - 3

X = 5


Solving Equations Using Multiplication

X = 6 3Multiply both sides by 3.

3 (X) = 6 (3) 3

X = 18


Solving Equations Using Division

4 X = 24

Divide both sides by 4.

4X = 24 4 4

X = 6


Using Two or More Operations to Solve an Equation

3 X - 5 = 2 X + 6

3X – 2X = 6 + 5

X = 11

You can substitute your result back into theoriginal equation to check your answer.


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 .


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.


Summary

Exponential expressions, logarithms, andalgebraic equations are important tools in thesolution of problems found in businessmathematics and finance.

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Contemporary Business Mathematics With Canadian Applications Chapter 2