Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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 2-1 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) Copyright © 2008 Pearson Education Canada 2-2 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. Copyright © 2008 Pearson Education Canada 2-3 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 Copyright © 2008 Pearson Education Canada 2-4 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 Copyright © 2008 Pearson Education Canada 2-5 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 Copyright © 2008 Pearson Education Canada 2-6 Exponents • • • • • Power an Base a Exponent n The factor a is multiplied by itself n times. POWER = BASE TO THE EXPONENT Copyright © 2008 Pearson Education Canada 2-7 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) Copyright © 2008 Pearson Education Canada 2-8 Operations with Powers am x an = a m+n 23 x 22 = 2 3+2 = 25 am an = a m-n 2523 = 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 2-9 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 Copyright © 2008 Pearson Education Canada 2-10 Fractional Exponents Copyright © 2008 Pearson Education Canada 2-11 Fractional Exponents Copyright © 2008 Pearson Education Canada 2-12 Examples of Fractional Exponents 49 27 36 (1/2) -(1/3) (3/2) = 7 = 1/3 = 216 Copyright © 2008 Pearson Education Canada 2-13 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 Copyright © 2008 Pearson Education Canada 2-14 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. Copyright © 2008 Pearson Education Canada 2-15 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) Copyright © 2008 Pearson Education Canada 2-16 Equations • An equation is an expression of equality between two algebraic expressions. • 3x = 36 • 2x + 4 = 60 • 5x - .4 = 2.5 Copyright © 2008 Pearson Education Canada 2-17 Solving Equations Using Addition • • • • X-3=9 Add 3 to both sides . X-3+3=9+3 X = 12 Copyright © 2008 Pearson Education Canada 2-18 Solving Equations Using Subtraction • • • • X+3=8 Subtract 3 from both sides. X + 3 -3 = 8 - 3 X=5 Copyright © 2008 Pearson Education Canada 2-19 Solving Equations Using Multiplication X =6 3 Multiply both sides by 3. 3 (X) = 6 (3) 3 X = 18 Copyright © 2008 Pearson Education Canada 2-20 Solving Equations Using Division 4 X = 24 Divide both sides by 4. 4X = 24 4 4 X=6 Copyright © 2008 Pearson Education Canada 2-21 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. Copyright © 2008 Pearson Education Canada 2-22 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 . Copyright © 2008 Pearson Education Canada 2-23 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. Copyright © 2008 Pearson Education Canada 2-24 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? Copyright © 2008 Pearson Education Canada 2-25 Summary Exponential expressions, logarithms, and algebraic equations are important tools in the solution of problems found in business mathematics and finance. Copyright © 2008 Pearson Education Canada 2-26