College Algebra (Abridged)

College Algebra Abridged Study Guide 1st Edition 5/14/2020

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Copyright © 2020 by Achieve All rights reserved. This book or any portion thereof may not be reproduced or used in any manner whatsoever without the express written permission of the publisher except for the use of brief quotations in a book review. Printed in the United States of America First Printing, 2015 Achieve PO Box 10188 #29831 Newark, NJ 07101-3188 Tel: 888.900.8380 Visit the Achieve website for more information.

Table of Contents Chapter 1: Introduction to Algebra and Functions ................................................................................4 1.1 Set of Real Numbers .............................................................................................................................................4 1.2 Inequalities.............................................................................................................................................................6 1.3 Mathematical Translations of Equalities fromWords ...............................................................................8 1.4 Mathematical Translations of Inequalities fromWords............................................................................9 1.5 Solving Equations .................................................................................................................................................9 1.6 Solving Inequalities........................................................................................................................................... 10 1.7 Solving Inequalities with Modulus Symbol................................................................................................. 11 1.8 Systems of Linear Equations........................................................................................................................... 12 1.9 Polynomials......................................................................................................................................................... 17 1.10 Solving Quadratic Equations........................................................................................................................ 18 1.11 Functions ........................................................................................................................................................... 19 1.12 Exponential Functions ................................................................................................................................... 21 1.13 Logarithm Functions ...................................................................................................................................... 22 1.14 Chapter One Review ....................................................................................................................................... 24 1.15 Chapter One Review Answers...................................................................................................................... 27

Chapter 1: Introduction to Algebra and Functions Overview Mathematics is a common tool that is used in everyday life. Ranging from simple counting of inventory items to solving complex equations in computer and engineering work, every day involves the use of mathematics. There are a large number of incentives for an individual to study throughout this course to equip him/herself with the knowledge of mathematics in order to excel in their career. Mathematics provides us with a better understanding of the world around us. It helps to hone the individual’s skills in problem solving, logical reasoning and flexible thinking, which is of utmost importance in the modern business world. It is pervasive and useful in almost all the arenas of life, such as business management, predicting stock market prices, safeguarding credit transactions on the internet, science, engineering, and even managing day-to-day financial activities. In this chapter, the introduction to algebra and functions will be presented. The main focus will be on concept building and its application. The building blocks of this chapter will include solving equations, linear inequalities, systems of linear equations by analytical and graphical methods, functions, and linear and exponential growth. The aim of this chapter is to equip you with the knowledge of general algebra that helps in building the foundation of mathematics. Objectives By the end of this chapter, you should be able to recognize, understand, and solve the following: • Algebraic operations • Equations and inequalities • Functions and their properties • Number systems and operations 1.1 Set of Real Numbers Algebra is a tool that is used to solve real life problems in the domains of science, business, architecture, management, space travel, and many other fields. We begin this chapter by understanding the basic notations and symbols used to build and solve algebraic expressions. Sets represent the collection of similar elements, and are denoted by the enclosed brackets {}. The unique characteristic of all elements in a set is that they have some similarity in appearance of purpose. For instance, {2, 4, 6, 8} represents the set of all one-digit even numbers, and {a, e, i, o, u} represents the set of all vowels in English. Sets are denoted by a letter. The set of all positive numbers less than 10 is denoted by S = {1, 2, 3, 4, 5, 6, 7, 8, 9} and the set of all odd numbers less than 10 is denoted by T= {1, 3, 5, 7, 9}. In this case, T is known as the sub-set of S, since all elements in set T are present in set S. Another way to describe a set is by using a set-builder notation. For instance, the set {1, 3, 5} in a set-builder notation is written as {x l x is an odd number between 0 and 6}. Set-builder notation is read as follows:

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Real numbers are used extensively in mathematics and are represented on a continuous number line. The number line includes the set of all the whole numbers {0, 1, 2, 3, 4…}, natural numbers{1, 2, 3, 4…}, integers {…-3, -2, -1, 0, 1, 2, 3, …}, fractions (numbers written in the form of , such that the ratio is not equal to zero) and irrational numbers (like square root of 2, given by √ 2 ). On the number line, negative values are on the left side of zero while positive values are on the right side of zero, as shown below: The properties of real numbers deal with four major operations: addition (+), subtraction ( − ), multiplication ( ×, ∙ ) and division ( ÷,/ ). Addition, subtraction, and multiplication of real numbers will result in a real number. Division of real numbers is also possible, provided that the denominator of the fraction is not equal to zero. If the denominator is equal to zero, then the result is undefined. The Properties of Real Number are listed below (assume , , and are real numbers): 1. Cumulative l w states that when we add or multiply two real numbers, their order does not matter. That is, + = + and × = × 2. Associative law states that while adding or multiplying more than two real numbers then the effect of parentheses does not matter. That is, ( + ) + = + ( + ) and ( × ) × = × ( × ) 3. Distributive law applies to the cases where one makes use of both addition and multiplication operations. × ( + ) = × + × and ( + ) × = × + × 4. The additive identity is known as 0 and 1 is a multiplicative identity . That is, + 0 = and × 1 = 5. Additive inverse is denoted by a negative sign. That is, the additive inverse of would be ( − ). Addition of a real number with its additive inverse would be 0. So, + ( − ) = 0 6. Multiplicative inverse of would be −1 -1 or 1 . The multiplication of a real number with its multiplicative inverse would be equal to 1, Therefore, × −1 = 1 7. Cancellation law for addition: if + = + , then = . 8. Cancellation law for multiplication: if × = × , then = , such that ≠ 0 .

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9. Cancellation law for division: = , provided ≠ 0 and ≠ 0 . Absolute values , or the magnitude of an integer, are another vital concept to know in algebra. It is defined as the distance between the integer and the zero value on the number line. Absolute values are denoted by the modulus symbol | | where can be any integer, negative or positive. If we have to calculate the absolute value of -5, then it will be denoted by |-5|and is equal to 5, showing that the distance from -5 to 0 is 5 units. Some of the properties of absolute values are: | | = � ≥ 0 ( − ) < 0 | | ≥ 0 | − | = | | | ∗ | = | | ∗ | | Scientific Notation Sometimes we make use of some power of 10 that makes it convenient to write a very large number; this means we are making use of scientific notation. For instance, suppose we have to write the number 687657.788. We can write it as 6.87657788 × 10 5 . 1.2 Inequalities The value of different numbers can be compared by their relative position on the number line. For instance, in the given number line below, is less than and is denoted by < , which means lies to the left of on the number line. We can also say > , that is is greater than and lies on the right side of the number line, relative to the position of . The mathematical representation for different expressions is given in the following table: Expressio s Interpretations = is equal to < is less than

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> ≥ ≤ ≠ ≈

is greater than is greater than or equal to is less than or equal to is not equal to

is approximately equal to The symbols <, >, ≥, ≤ and ≠ denote the signs of inequalities and the expressions using these signs of inequalities, < , > , ≥ , ≤ and ≠ denotes inequalities. We can make use of set- builder notation for these inequalities as well. For instance, = { | ≥ 2} means set represents all the values of such that is greater than or equal to 2. This can be shown graphically on the number line as follows: It must be noted that in the above number line, there is a closed bracket [ at 2, which means that 2 is also included in the expression. In contrast, the expression = { | > 2} is represented on the number line as follows: When a parenthesis ( is used on the number line, it means that all numbers greater than 2, excluding 2, are to be considered here. In other words, while using inequalities, we may make use of different types of brackets that have different meanings. For instance, ( , ) means the set contains all the numbers between and , excluding and . On the contrary, [ , ] means all numbers between and , including and . Similarly, [ , ) means all numbers between and , inclusive of and exclusive of . This is known an interval notation . Let us understand this using a few examples: Set-Builder Notation Graphical Representation Interval Notation { | > 4} (4, ∞ )

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{ | < 2} { | ≤ 1} { | ≥ − 3}

( −∞ , 2) ( −∞ , 1] [ − 3, ∞ )

1.3 Mathematical Translations of Equalities fromWords An equation is used to show that two mathematical expressions are equal to each other. For instance, 2 + 5 = 9 . This is as an equation since the expressions on both sides of the sentence are equal and denoted by the equal to (=) sign. Some of the examples of equations are:

3 + 5 = 8 5 – 10 = 10 2 ∗ 3 = 8– 2 70 = 280

In these examples, the expressions on either side of the equal sign are equal to each other. Written sentences can be converted into mathematical expressions so that they can be solved to arrive at meaningful solutions. For instance, “a number is multiplied by 10 and then added to 8 to get 48”. Assume the number is and the mathematical equation becomes: 10 + 8 = 48 It should be noted that the two sides of the equation should balance each other using the equal to (=) sign. Let us solve a few more examples to get a better understanding. Word Problem Algebraic Translation Kathy asks her mother how old she is. Her mother replies, “If you double my age and add 10 to it, then divide by 2, you get 35.” (2 + 10) 2 = 35

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Marlin gets his monthly pocket money. He saves half of it, and then makes two purchases worth $6 and $5. Finally when he reaches home, he is left with $3. 2 – 6 – 5 = 3 A bus starts leaves the station filled to capacity. At the first stop, half the passengers get off and 10 get on board. At the next station, 12 get off while 17 get on board. Now, the number of passengers in the bus is 28. − 2 + 10 − 12 + 15 = 28 1.4 Mathematical Translations of Inequalities fromWords Inequalities can also be translated from words into mathematical equations. For instance, we need to translate the sentence, “a number minus 6 is less than 2” into an inequality. “A number” states the presence of an unknown variable. Let us denote that by . After subtracting 6 from , this expression should be less than (denoted by inequality sign <) 2. Now the expression becomes: – 6 < 2 Similarly, mathematical expressions can be formed using different inequality notations like more than (>), less than (<), at most (≤), at least (≥), etc., as shown below. The sum of 5 more than the number is no more than 15 + 5 < 15 Five boxes of cereal costs at most 25 dollars 5 ≤ 25 The difference between a number and 5 is at least 17 − 5 ≥ 17 Jenny’s height is greater than 65 inches > 65 1.5 Solving Equations In order to solve the equations, it is important to move all variables to one side using inverse operations (addition  subtraction & multiplication  division). It should be noted that applying inverse operations on both sides of an equation keeps the balance. For instance, the equation – 3 = 7 can be simplified by adding 3 on each side, − 3 + 3 = 7 + 3 . Thus; the final answer is = 10 . More examples are shown below. Equation Steps

(2 + 10) 2 = 35 2 + 10 = 70 Step 1: Multiply both sides by 2 2 + 10 2 × 2 = 35 × 2 2 = 60 Step 2: Subtract 10 from both sides

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2 + 10 − 10 = 70 − 10 Step 3: Divide both sides by 2 2 2 = 60 2 2 – 6– 5 = 3 Step 1: Simplify the equation 2 – 11 = 3 2 = 14 Step 2: Add 11 to both sides 2 – 11 + 11 = 3 + 11 = Step 3: Multiply both sides by 2 2 × 2 = 14 × 2 Step 1: Simplify the equation 2 × 2 2 × 1 − 2 + 13 = 28 3 2 = 15 Step 2: Subtract 13 from both sides 3 2 + 13– 13 = 28– 13 = Step 3: Multiply both sides by 2 3 3 2 × 2 3 = 15 × 2 3 =

2 – 2 + 10 − 12 + 15 = 28 3 2 + 13 = 28

1.6 Solving Inequalities While solving inequalities, we need to apply the same rules of operations (addition, subtraction, multiplication, and division). As with equations, whatever is done to one side must be done to the other side. However, there is one additional rule: If multiplying or dividing by a negative number, reverse the sign of inequality. For instance: − 3 ≤ 10 , multiplying both sides by − 3 will become, − 3 × − 3 ≥ 10 × − 3

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It should be noted that in this case, the sign of inequality has been reversed from ‘less than or equal to’ to ‘greater than or equal to’. Now, let us practice some examples to solve the inequalities. Inequalities Solution + ( + 2) < 16 2 + 2 < 16 Step 1: Simplify the inequality + + 2 < 16 2 < 14 Step 2: Subtract 2 from both sides 2 + 2– 2 < 16– 2 < 7 Step 3: Divide both sides by 2 2 2 < 14 2 5 ≤ 25 ≤ 5 Step 1: Divide both sides by 5 5 5 ≤ 25 5 – 5 ≥ 17 ≥ 22 Step 1: Add 5 to both sides – 5 + 5 ≥ 17 + 5 − 2 + 10 < 64 − 2 < 54 Step 1: Subtract 10 from both sides − 2 + 10– 10 < 64– 10 > − 27 Step 2: Divide both sides by ( − 2 ) & reverse the inequality sign − 2 − 2 > 54 − 2 1.7 Solving Inequalities with Modulus Symbol Solving inequalities involving absolute values needs a little more attention. For instance, 3 > | | means that is 3 units from 0, which means could be 3 or − 3 . The range of values would be > − 3 and < 3 . This can also be written as − 3 < < 3 . This is what is shown on the number line below: Let us look at another example. Solve: |2 − 7| ≥ 5 . 1. Set up a compound inequality: − 5 ≥ 2 − 7 ≥ 5 2. Add seven to both sides: 2 ≥ 2 ≥ 12 3. Divide both sides by 2: 1 ≥ ≥ 6

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It means that constitutes all values less than and equal to 1 and greater than and equal to 6. This can be represented on the number line as follows:

1.8 Systems of Linear Equations In this section, wewill learn how to solvemore than one linear equationwithmore than one unknown variable. This can be done by solving both the equations simultaneously to find the break-even, or equilibrium, point. The linear equations are generally in the form of: + = + = It should be noted that in the above stated equations: , , , , , and are constants and the values of both and b or and cannot be zero. Further, the system of linear equations could have any of three solutions: 1. The lines (graphical representation) formed by these equations will intersect at one point. Therefore, there will be one solution to the system. 2. The lines formed by these equations will be identical. Therefore, there will be an infinite number of solutions. 3. The lines formed by these equations will be parallel and distinct from each other. Therefore, there will be no solutions. The graphical representation of these cases is given as follows:

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Let us solve a few sets of linear equations to gain a better understanding of the concept. Case I: System of Equations with Only One Solution 2 – = 1 ………………….……………………………(1) 3 + 2 = 12 …………………………………………….(2) Solving the first equation for , we get: = 2 – 1 ……………………………………………….(3) Substituting the value of into equation 2, we get: 3 + 2 × (2 − 1) = 12 Solving the equation 3 + 4 – 2 = 12 7 = 14 = 2 Substituting the value of into equation 3, we get: = 2 × 2– 1 = 3 This set of these linear equations has one solution. In other words, these two lines cross at only one point: (2, 3). The graphical representation of this set of linear equations is as follows:

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Case II: System of Equations with Indefinitely Many Solutions 2 – = 1 …………………………………………………..(1) 8 – 4 = 4 ………………………………………………….(2) Solving equation (1) for the value of = 2 – 1 Substituting the value of in equation 2 8 – 4(2 − 1) = 4 Solving the equation for the value of 8 – 8 + 4 = 4 0 = 0 In this case, the unique value(s) of cannot be obtained since the values of variables, as well as constants, cancel out. This is primarily because both the given equations are actually the same. If we multiply equation (1) by 4, we get equation (2). The solution for this set of linear equations is any set of ordered pairs ( , ) which satisfies the equation 2 − = 1 . In order to find the solution to such problems, let us consider = 0 , so we get the values of and as (0, − 1) which becomes a possible solution. Now, consider the value of = 1 . The values of ordered pair ( , ) would be (1,1) . This is another possible solution.

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Plotting these points on the graph, gives us we get the infinitely many solutions to this set of equations. The graph is given as follows:

Case III: System of Equations with No Solution 2 – = 1 …………………………………………………..(1) 8 – 4 = 12 ………………………………………………….(2) Solving equation (1) for the value of ,

= 2 – 1 Substituting the value of in equation 2, 8 – 4(2 − 1) = 12 Solving the equation for the value of , 8 – 8 + 4 = 12 4 = 12 This is clearly impossible, hence we can say that the given system of equations has no solution. Plotting both these equations on the graph (using the same basis as in case II), we get the following representation:

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It can be seen that the two equations form parallel lines with each other and thus have no points of intersection. 1.9 Polynomials Up to this point, we have simplified and solved the linear equations whose variables have the power of 1. We shall now discuss how to factor, simplify, and solve linear equations using polynomials. Polynomials are defined as those mathematical expressions that take the following form: ( ) = + −1 −1 + −2 −2 + ⋯ + 1 + 0 For example: ( ) = 4 3 − 6 2 + 9 + 12 In this case, 3 = 4, 2 = − 6, 1 = 9 and 0 = 12 The values of , −1 , −2 , …, 1 and 0 are known as the coefficients of polynomials and the power of x determines the degree of ( ) , also known as the leading term. The equation with a maximum of two degrees of the unknown variable is known as a quadratic equation. Addition and subtraction of polynomials are possible only between like terms --those terms which have the same variable(s) raised to the same exponent. For instance, we are given these two polynomials: ( ) = 4 3 − 6 2 + 9 + 12

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( ) = 8 4 + 3 2 − 17 + 10 If we wish to add or subtract these two expressions, then we combine the coefficients of the like terms. The addition for these expressions is given as follows: ( ) = 4 3 – 6 2 + 9 + 12 + 8 4 + 3 2 − 17 + 10 ( ) = 8 4 + 4 3 + (– 6 2 + 3 2 ) + (9 − 17 ) + (12 + 10) ( ) = 8 4 + 4 3 − 3 2 − 8 + 22 Similarly, subtraction of these polynomials will be done as follows: ( ) = 4 3 – 6 2 + 9 + 12– (8 4 + 3 2 − 17 + 10) ( ) = − 8 4 + 4 3 + (– 6 2 − 3 2 ) + (9 + 17 ) + (12 − 10) ( ) = − 8 4 + 4 3 − 9 2 + 26 + 2 The rules for multiplication and division of polynomials are different. The coefficients are multiplied or divided while the exponents of the variables are added (in the case of multiplication) or subtracted (in the case of division). For instance, if we wish to multiply the above equations ( ) and ( ) , we will be multiplying each and every term of both the equations. ( ) = (4 3 – 6 2 + 9 + 12) ∙ (8 4 + 3 2 − 17 + 10) ( ) = 8 4 (4 3 – 6 2 + 9 + 12) + 3 2 (4 3 – 6 2 + 9 + 12) − 17 (4 3 – 6 2 + 9 + 12) + 10(4 3 – 6 2 + 9 + 12) ( ) = (32 7 – 48 6 + 72 5 + 96 4 ) + (12 5 – 18 4 + 27 3 + 36 2 ) +( − 68 4 + 102 3 − 153 2 – 204 ) + (40 3 – 60 2 + 90 + 120) Simplifying the terms as per the rule of addition and subtraction ( ) = 32 7 − 48 6 + (72 5 + 12 5 ) + (96 4 – 18 4 − 68 4 ) +(27 3 + 102 3 + 40 3 ) + (36 2 − 153 2 – 60 2 ) +( − 204 + 90 ) + 120 ( ) = 32 7 – 48 6 + 84 5 + 10 4 + 169 3 – 177 2 − 114 + 120 1.10 Solving Quadratic Equations The system of equations with a maximum leading power of two is termed as a quadratic equation, while the system of equations with a maximum leading power of three are known as cubic equations. The graphs for quadratic equations are generally a U-shaped curve. The general form of a quadratic equation is:

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2 + + = 0

Where & are coefficients and is a constant. In order to find the values of unknown variables in a quadratic equation, one may guess and check, factor, or use the quadratic formula : = − ± √ 2 − 4 2 Let us solve an example: 5 2 + − 1 = 0 Plugging the values of = 5 , = 6 and = 1 in the above equation, we get: = − (6) ± � 6 2 − 4(5)(1) 2(5) = − 6 ± √ 16 10 = − 1, − 0.2 These solutions represent the -intercepts of the quadratic equation, as shown below.

1.11 Functions A function determines a relationship between different constants and/or variables. In other words, it maps a number or variable to another unique number. For instance, a function can be derived whereby a number added to 4 gives another number. In this case, if we apply this function to 3, we get the output of 7. This function can be mathematically written as: ( ) = + 4

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The input for the function is known as the independent variable while the output derived from this function is known as dependent variable. We can plot a function on a graph. For instance, take the example: ( ) = 3 2 – 4 In this function, is the independent variable while the expression 3 2 – 4 represents a dependent variable. In order to plot this function, find the value of this output for different values of . − ( ) ( , ) − 2 3( − 2) 2 – 4 = 8 8 ( − 2,8) − 1 3( − 1) 2 – 4 = − 1 − 1 ( − 1, − 1) 0 3(0) 2 – 4 = − 4 − 4 (0, − 4) 1 3(1) 2 – 4 = − 1 − 1 (1, − 1) 2 3(2) 2 – 4 = 8 8 (2,8) Plotting these points on the graph and joining them, we get the following graph:

Similarly, we can graph any equation with number of leading terms. However, there are some translation rules for graphing functions: ( ) + is ( ) shifted upward units ( )– is ( ) shifted downward units ( + ) is ( ) shifted left units ( – ) is ( ) shifted right units – ( ) is ( ) reflected over the -axis f(–x) is the mirror of ( )

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1.12 Exponential Functions An exponential function is a function that is written in the form of ( ) = , such that > 0 For example, when = 1 , then ( ) = 1 . If > 1 , then the pattern of -values for different values of will look as follows: Suppose = 2 2 ( ) ( , ) Plotting these values on a graph, we get the following representation:

− 2 2 −2 = 1 4 = 0.25 0.25 ( − 2,0.25) − 1 2 −1 = 1 2 = 0.5 0.5 ( − 1,0.5) 0 2 0 = 1 1 (0,1) 1 2 1 = 2 2 (1,2) 2 2 2 = 4 4 (2,4) 3 2 3 = 8 8 (3,8)

It should be noted that for all values of a, (0) = 1 , since 0 is always equal to one for any value of . Moreover, the value of ( ) will always be positive and greater than 0, since is always greater than 0, and hence, is always greater than zero. There are some rules of exponents that help simplify the exponential expressions: 1. = + 2. ( ) = 3. = − 4. ( ) = 5. � � = , b ≠ 0 6. − = 1 , a ≠ 0 7. 1 = √ 8. = ( √ )

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Let us solve an example using these rules of exponents. Simplify the following expression: (5 −1 6 −2 ) −3 (10 3 −3 ) 3 = 5 −1 ( −3 ) 6 ( −3 ) −2 ( −3 ) 10 1 ( 3 ) 3 ( 3 ) −3 ( 3 ) = 5 3 −18 6 10 3 9 −9 = 125 −18−9 6− ( −9 ) 1000 = −27 15 8 = 15 8 27 1.13 Logarithm Functions A logarithm function is the function that is written in the form of: ( ) = log , such that > 0 and ≠ 1 If = 2 , then ( ) = log 2 2 ( ) = Assuming different positive values of (it is necessary that 2 ( ) > 0) , log 2 ( ) ( , ) When we plot these values in the graph, we get the following graphical representation: 1 4 log 2 � 1 4 � = − 2 − 2 � 1 4 , − 2 � 1 2 log 2 � 1 2 � = − 1 − 1 � 1 2 , − 1 � 1 log 2 (1) = 0 0 (1,0) 2 log 2 (2) = 1 1 (2,1) 4 log 2 (4) = 2 2 (4,2) It should be noted that for all values of , (1) = 0 . Moreover, the value of ( ) will always be positive and greater than 0. The shape of the above graph becomes inverse (positive side only) whenever the values of a lies between 0 and 1.

The properties of logarithms are as follows: 1. = + 2. ln � � = ln − 3. = . 4. = 5. = 1 6. 1 = 0 Relationship between Logarithm and Exponential Functions The relation between log and exponential functions can be explained by looking at their graphs:

The log function is given by ( ) = ln and the exponential function is given by ( ) = . It should be noted that the logarithm function is the reflection of the exponential function over the line = . Hence, logarithm function is the inverse of exponential function.

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1.14 Chapter One Review Question 1: A firm that manufactures leather belts evaluates its costs and revenues. The results show that the firm incurs an average fixed cost of $2400 per week while the variable costs amount to $4.50 per belt. The average revenue earned by the firm is $7.50 per belt. Calculate the number of belts the firm should manufacture in order to break even. a. 800 b. 300 c. 200 d. 1000 Question 2: The functions ( ) and ( ) are defined as: ( ) = + 5 ( ) = | − 7| What is the value of ( ( − 2)) ? a. 10 b. 0 c. 14 d. − 4 Question 3: Which of the following describes the possible value of for this quadratic equation? 2 − 3 − 10 = 0

a. 2, − 5 b. 5, − 2 c. − 2, − 5 d. 2, 5

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Question 4: Solve for :

− 2 − 1 = + 4 (2 + 2)

a. − 2, 1 b. 0, 2 c. 5, − 1 d. 5, 0 Question 5: During practice, a softball pitcher throws a ball whose height can be modeled by the equation ℎ = 2 2 − 9 + 11 , where ℎ = height in feet and = time in seconds. How long does it take for the ball to reach a height of 7 feet? a. 1 2 seconds in going up, 4 seconds in coming down b. 1 4 seconds in going up, 2 seconds in coming down c. 2 seconds in going up, 2 seconds in coming down d. 4 seconds in going up, 2 seconds in coming down Question 6: What is the -intercept of the function, ( ) = ( − 5) 2 ? a. (0, 5) b. (0, − 5) c. (0, − 25) d. (0, 25) Question 7: For the exponential function ( ) = given that, 0 < < 1 , what is the -intercept? a. (0, 1)

b. (0, 2) c. (0, 0) d. (0, 4)

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Question 8: Find the value of in the following log function: log 2 = 4 a. 2 b. 4 c. 16 d. 8 Question 9: For the given inequalities, which values of and hold true? − > 1 and < 2 – 1

a. ( − 1, 2) b. (3, − 1)

c. (0, 0) d. (1, 2) Question 10: Which of the following inequalities is represented by the graph shown below?

a. ≤ 4 – 7 b. > 4 + 7 c. < 4 + 7 d. ≥ 4 – 7

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1.15 Chapter One Review Answers Question 1: A Solution: In this case, we have to first determine the mathematical equation from the wording of the question. The “break even point” for any firm is when total costs of the company equal the total revenues. Costs may be fixed (they do not change with the change of output) or variable (they change as per the change in output). • The cost function of the firm is: ( ) = 2400 + 4.5 , where = number of leather belts produced. • The revenue function of the firm is: ( ) = 7.5 • At the breakeven point, ( ) = ( ) 2400 + 4.5 = 7.5 Solve the equation by subtracting 4.5 2400 = 3 Divide both sides by 3 = 800 The graphical representation for the equation, = 3 – 2400 is given as follows:

Question 2: C

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Solution: First, determine the value of ( − 2) , then take this value and substitute it into ( ) . ( ) = | − 7| ( − 2) = | − 2 − 7| = | − 9| = 9 Now, ( ) = + 5 , so by substituting = 9 , we get: (9) = 9 + 5 = 14 Question 3: B Solution: Solving the above quadratic equation, use the formula: = − ± √ 2 − 4 2 Where, = 1 , = − 3 and = − 10 . Substituting the values in above formula, we get: = − ( − 3) ± � ( − 3) 2 − 4(1)( − 10) 2(1) = 3 ± √ 49 2 = 3 ± 7 2 3+7 2 = 10 2 = 5 and 3−7 2 = − 4 2 = − 2 Question 4: D Solution: Solve −2 −1 = +4 ( 2 +2 ) using cross-multiplication. ( − 2)(2 + 2) = ( − 1)( + 4) 2 2 − 2 − 4 = 2 + 3 − 4 2 − 5 = 0 ( − 5) = 0 Therefore, = 0 and = 5 Question 5 : A

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Solution: In this case, ℎ = 7 , substituting the value of ℎ in the equation, we get: 7 = 2 2 − 9 + 11 2 2 − 9 + 4 = 0 Using the formula, = − ± √ 2 − 4 2 Where, = 2 , = − 9 and = 4 , substituting the values we get: = − ( − 9) ± � ( − 9) 2 − 4(2)(4) 2(2) = 9 ± √ 49 4 = 9 ± 7 4 So the ball will be at a height of 7 feet when at ½ second and 4 seconds. Question 6: D Solution: Using FOIL, we get: ( ) = ( − 5)( − 5) = 2 − 5 − 5 + 25 = 2 − 10 + 25 To find the -intercept, substitute 0 in for : (0) = 0 2 − 10(0) + 25 = 25 Question 7: A Solution: To find the -intercept, substitute 0 in for : ( ) = (0) = 0 = 1 (any Real number raised to the 0 power equals 1) Question 8: C Solution: We know that log = 4 . Next, convert from logarithmic form to exponential form: 2 4 = . Therefore, 16 = . Question 9: B

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Solution: We can solve this question by substituting the given values of and into the inequalities. The option that satisfies both the inequalities is the correct answer. Substitute = − 1 and = 2 : − 1– 2 > 1 − 3 > 1 Not true Now substitute (3, − 1) : 3– ( − 1) > 1 = 4 > 1 True − 1 < 2 × 3– ( − 1) = − 1 < 7 True Hence, option b is the answer. One may check for other options as well for double checking. Question 10: D Solution: The -intercept of the line is − 7 , therefore options (b) and (c) can be eliminated. The shaded area includes all points that will satisfy the equation. Therefore, select any point in the shaded area, plug it into each equation, and if it proves to be a true statement, that is the answer. For example, if the point (0,0) is chosen: ≤ 4 − 7 0 ≤ 4(0) − 7 0 ≤ − 7 This is false, so the answer cannot be (a). The answer must be (d). To check: ≥ 4 − 7 0 ≥ 4(0) − 7 0 ≥ − 7

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