How to learn Business Math

Step-by-Step Business Math and Statistics | download free pdf
SimonHaion Profile Pic
SimonHaion,Malaysia,Teacher
Published Date:09-07-2017
Your Website URL(Optional)
Comment
Step-by-Step Business Math and Statistics By Jin W. Choi Included in this preview: • Copyright Page • Table of Contents • Excerpt of Chapter 1 For additional information on adopting this book for your class, please contact us at 800.200.3908 x501 or via e-mail at infocognella.comMath. Chapter 1. Algebra Review Part 1. Business Mathematics There are 4 chapters in this part of business mathematics: Algebra review, calculus review, optimization techniques, and economic applications of algebra and calculus. Chapter 1. Algebra Review A. The Number System The number system is comprised of real numbers and imaginary numbers. Real numbers are, in turn, grouped into natural numbers, integers, rational numbers, and irrational numbers. 1. Real Numbers = numbers that we encounter everyday during a normal course of life Æ the numbers that are real to us. i. Natural numbers = the numbers that we often use to count items Æ counting trees, apples, bananas, etc.: 1, 2, 3, 4, … a. odd numbers: 1, 3, 5, … b. even numbers: 2, 4, 6, … ii. Integers = whole numbers without a decimal point: 0, +1, +2, +3, +4, …. a. positive integers: 1, 2, 3, 4, … b. negative integers: –1, –2, –3, –4, … iii. Rational numbers = numbers that can be expressed as a fraction of integers such as a/b (= a÷b) where both a and b are integers a. finite decimal fractions: 1/2, 2/5, etc. b. (recurring or periodic) infinite decimal fractions: 1/3, 2/9, etc. iv. Irrational Numbers = numbers that can NOT be expressed as a fraction of integers = nonrecurring infinite decimal fractions: 3 3 a. n-th roots such as 2, 5, 7 , etc. b. special values such as ʌ (=pi), or e (=exponential), etc. Chapter 1: Algebra Review 1Math. Chapter 1. Algebra Review v. Undefined fractions: a. any number that is divided by a zero such as k/0 where k is any number b. a zero divided by a zero = 0/0 f c. an infinity divided by an infinity = f 0 d. a zero divided by an infinity = f vi. Defined fractions: a. a one divided by a very small number Æ 1 1 10 10 10,000,000,000 a very large  10 0.0000000001 10 number such as a number that can approach ’ b. a one divided by a very large number Æ 1 1/(a large number) = a small number Æ 0 f c. a scientific notion Æ the use of exponent 2 2.345E+2 = 2.345 x 10 = 234.5 6 2.345E+6 = 2.345 x 10 = 2,345,000 1 1 -2 2.345E–2 = 2.345 x 10 = 2.345˜ 2.345˜ 0.02345 2 10 100 -6 2.345E–6 = 2.345 x 10 = 1 1 2.345˜ 2.345˜ 0.000002345 6 10 1,000,000 Similarly, a caret () can be used as a sign for an exponent: n 10 X = Xn Æ X = X10 Note: For example, E+6 means move the decimal point 6 digits to the right of the original decimal point whereas E-6 means move the decimal point 6 digits to the left of the original decimal point. 2 Step by Step Business Math and Statistics Math. Chapter 1. Algebra Review 2. Imaginary Numbers = numbers that are not easily encountered and recognized on a normal course of life and thus, not real enough (or imaginary) to an individual. Æ Often exists as a mathematical conception. i  1  2 2i i 2 2  4 2i (5i) = –25 B. Rules of Algebra 1.  a b b a Æ 2 + 3 = 3 + 2 Æ 5 2. ab ba Æ 2 x 3 = 3 x 2 Æ 6  1 -1 0 3. aa 1 for az 0 Æ 2 x 2 = 2 = 1 4. a(b c) ab ac Æ 2 x (3 + 4) = 2 x 3 + 2 x 4 Æ 14 5.  a ( a) a ( a) 0Æ 2 + (–2) = 2 – (+2) = 2 – 2 = 0 6.  ( a)b a( b)  ab Æ (–2) x 3 = 2 x (–3) Æ –6 7.  ( a)( b) ab Æ (–2) x (–3) = 2 x 3 Æ 6 2 2 2 2 2 2 8.  (a b ) a 2ab b Æ (2 + 3) = 2 + 2(2)(3) + 3 Æ 25 2 2 2 2 2 2 9.  (a b ) a 2ab bÆ (2 – 3) = 2 – 2(2)(3) + 3 Æ 1 2 2 2 2 10.  (a b)( a b) a b Æ (2 + 3)(2 – 3) = 2 – 3 Æ –5  a 2 2 11.  ( a) /( b) a / b Æ  ( 2) /( 3) 2 / 3 Æ  b 3 3  a a a a 2 2 2 12. ˜ 1  Æ (2) /( 3) Æ  b b b b 3 3 3 b ac b 3 2˜ 4 3 11 13. a Æ 2 Æ c c 4 4 4 a c ad bc 2 4˜ 2 5 3˜ 4 22 14.  Æ  Æ b d bd 3 5 3˜ 5 15 Chapter 1: Algebra Review 3Math. Chapter 1. Algebra Review b b ab 3 3 2˜ 3 6 15. u a a˜ Æ u 2 2˜ Æ c c c 4 4 4 4 a 2 a c a d ad 2 4 2 5 2˜ 5 10 b 3 16. y u Æ y u Æ c 4 b d b c bc 3 5 3 4 3˜ 4 12 d 5 1/ 2 0.5 1/ 2 0.5 17. a a a where a• 0 Æ 2 2 2 Æ 1.4142 1/n 1/3 n 3 18. a = a where a 0 •Æ 2 = 2 Æ 1.2599 19. ab = a b Æ 2˜ 3 = 2 3 6 Æ 2.4495 a a 2 2 20. Æ Æ 0.8165 b 3 b 3 21.  a bz a b Æ  2 3z 2 3 a a 2 2 zz 22. Æ b b 3 3 C. Properties of Exponents Æ Pay attention to equivalent notations It is very important that we know the following properties of exponents: 0 0 1. X 1 Æ Note that 0 = undefined 1 1  b 10 2. X = X ( b) Æ X = X ( 10) b 10 X X a b a b a b a b 3. X X X˜ X X X XÆ X (a b) 2 3 2 3 2 3 2 3 5 Æ X X˜ X X X X X X 3 4 3 4 7 Æ 2˜ 2 2 2 = 128 a b ab a˜ b ab 4. (X ) X X X Æ X ab 2 3 23 2˜ 3 6 Æ (X ) X X X 4 Step by Step Business Math and Statistics Math. Chapter 1. Algebra Review 3 4 3˜ 4 12 Æ (2 ) 2 2 = 4096 a X a b a b 5. X˜ X X b X 2 X 1 2 3 2 3 1 Æ ˜ X X X X 3 X X a a a a a a a 6. (XY ) X Y X˜ Y X Y 2 2 2 2 2 2 2 Æ (XY ) X Y X˜ Y X Y 1 n 1/ n n 7. X X X 1 1/ 2 0.5 2 0.5 2˜ 0.5 1 2 Æ 4 4 4 4 (2 ) 2 2 2 q p / q 1/ q p p 1/ q p 8. X (X ) (X ) X 10/5 2 1/5 10 10 1/5 5 10 2 Æ 22 (2) (2) 2 = 2 = 4 2 / 3 3 2 / 3 2 3 2 3 Æ 8 (2 ) 2 8 64 4 D. Linear and Nonlinear Functions 1. Linear Functions Linear Functions have the general form of: Y = a + b X where Y and X are variables and a and b are constants. More specifically, a is called an intercept and b, a slope coefficient. The most visually distinguishable character of a linear function is that it is a straight line. Note that +b means a positive slope and –b means a negative slope. 2. Nonlinear Functions There are many different types of nonlinear functions such as polynomial, exponential, logarithmic, trigonometric functions, etc. Only polynomial, exponential and logarithmic functions will be briefly explained below. i) The n-th degree polynomial functions have the following general form: Chapter 1: Algebra Review 5Math. Chapter 1. Algebra Review 2 3 n 1 n  Y a bX cX dX ...... pX qX Or alternatively expressed as: nn132 Yq X pX ......dX cX bX a where a, b, c, d, …, p and q are all constant numbers called coefficients and n is the largest exponent value. Note that the n-th degree polynomial function is named after the highest value of n. For example, when n = 2, it is most often called a quadratic function, instead of a second-degree polynomial function, and has the following form: 2 Y a bX cX When n = 3, it is called a third-degree polynomial function or a cubic function and has the following form: 2 3 Y a bX cX dX ii) Finding the Roots of a Polynomial Function Often, it is important and necessary to find roots of a polynomial function, which can be a challenging task. An n-th degree polynomial function will have n roots. Thus, a third degree polynomial function will have 3 roots and a quadratic function, two roots. These roots need not be always different and in fact, can have the same value. Even though finding roots to higher-degree polynomial functions is difficult, the task of finding the roots of a quadratic equation is manageable if one relies on either the factoring method or the quadratic formula. If we are to find the roots to a quadratic function of: 2 aX bX c 0 we can find their two roots by using the following quadratic formula: 2  br b 4ac X , X 1 2 2a iii) Examples: Find the roots, X and X , of the following quadratic equations: 1 2 2 (a) X 3X 2 0 6 Step by Step Business Math and Statistics Math. Chapter 1. Algebra Review 1 Factoring Method : 2  X 3X 2 ( X˜ 1) (X 2) 0 Therefore, we find two roots as: X = 1 and X = 2. 1 2 2 Quadratic Formula : Note: a 1, b  3, and c 2 2 2  (r3) ( 3) 4(1)(2)  br b 4ac X , X 1 2 2a 2˜ 1 3r 9 8 3r 1 = 1, 2 2 2 2 (b) 4X 24X 36 0 Factoring Method: 222 4XX 24 3 6(26X˜ )(26 X )  (26X ) 4(3X ) 0 Therefore, we find two identical roots (or double roots) as: X X  3 1 2 Quadratic Formula: Note: a 4 , b 24 , and c 36 2 2  (24r) (24) 4(4)(36)  br b 4ac X , X 1 2 2a 2˜ 4 1 The factoring method often seems more convenient for people with great experience with algebra. That is, the easiness comes with experience. Those who lack algebraic skill may be better off using the quadratic formula. 2 a, b, In order to use the quadratic formula successfully, one must match up the values for and c correctly. Chapter 1: Algebra Review 7Math. Chapter 1. Algebra Review  24r 576 576 24r 0 24 =   3 8 8 8 2 2 (c) 4X 9Y 0 Factoring Method: 2 2  4X 9Y (2 X 3Y˜ ) (2X 3Y) 0 Therefore, we find two roots as: 3Y 3Y X 1.5Y and X  1.5Y 1 2 2 2 3 Quadratic Formula : 2 Note: a 4 , b 0 , and cY  9 22 2 r (0) (0) 4( 4)( 9Y ) rbb 4ac XX , 12 22 a˜4 2 0r 0 144Yr 12Y 3 = r Y 1.5Y, 1.5Y 8 8 2 E. Exponential and Logarithmic Functions 1. Exponential Functions X An exponential function has the form of Y a˜ b where a and b are X constant numbers. The simplest form of an exponential function is Y b where b is called the base and X is called an exponent or a growth factor. A unique case of an exponential function is observed when the base of e is X used. That is, Y e where e 2.718281828 . Because this value of e is 4 often identified with natural phenomena, it is called the “natural” base . 3 One must be very cognizant of the construct of this quadratic equation. Because we are to find the roots 2 associated with X, –9Y should be considered as a constant term, like c in the quadratic equation. n 1 §· 4 Technically, the expression 1 approaches e as n increases. That is, as n approaches f , ¨¸ n ©¹ e 2.718281828 . 8 Step by Step Business Math and Statistics Math. Chapter 1. Algebra Review Examples In order to be familiar with how exponential functions work, please verify the following equalities by using a calculator. 2 4 2 4 6 a. ˜ 5e e 5e 5e 5˜ 403.4287935 2017.143967 3 4 7 b. (5e˜ ) (3e ) 15e 15˜ 1096.633158 16449.49738 10 5 5 3 4 3 4 1 y 10e 2e˜ e 5e 1.839397206 c. 2 e 2.718281828 2. Logarithmic Functions The logarithm of Y with base b is denoted as “ log Y ” and is defined as: b X log Y X if and only if b Y b provided that b and Y are positive numbers with bz 1. The logarithm X X enables one to find the value of X given 2 4 or 5 25. In both of these cases, we can easily find X=2 due to the simple squaring process X involved. However, finding X in 2 5 is not easy. This is when knowing a logarithm comes in handy. Examples Convert the following logarithmic functions into exponential functions: X log 8 X Æ 2 8 Æ X = 3 2 0 log 1 0 Æ 5 1 5 1 log 4 1 Æ 4 4 4  2 1 2 §·  1 2 2 log 4  2 Æ 2 2 2 4 ¨¸ 1/ 2 2 ©¹ a. Special Logarithms: A common logarithm and a natural logarithm. i) A Common Logarithm = a logarithm with base 10 and often denoted without the base value. That is, log X log XÆ read as "a (common) logarithm of X." 10 ii) A Natural Logarithm = a logarithm with base e and often denoted as ‘ln”. Chapter 1: Algebra Review 9Math. Chapter 1. Algebra Review log X ln X That is, Æ read as "a natural logarithm of X." e b. Properties of Logarithms i) Product Property: log mn log m log n b b b m ii) Quotient Property: log log m log n b b b n n iii) Power Property: log m n˜ log m b b Example 1 Using the above 3 properties of logarithm, verify the following equality or inequality by using a calculator. i) ln 30 ln(˜ 5 6) ln5 ln 6 Æ 3.401197 20 ii) ln ln 20 ln 40 ln 0.5 Æ –0.693147 40 ln 20 20 iii) z ln ln 40 40 3 iv) ln10 3˜ ln10 ln1000 Æ 6.907755 X Example 2 Find X in 2 5. (This solution method is a bit advanced.) In order to find X, (1) we can take a natural (or common) logarithm of both sides as: X ln 2 ln5 (2) rewrite the above as: X˜ ln 2 ln5 by using the Power Property ln 5 (3) solve for X as: X ln 2 (4) use the calculator to find the value of X as: ln5 1.6094379 X 2.321928095 ln 2 0.6931471 10 Step by Step Business Math and Statistics Math. Chapter 1. Algebra Review Additional topics of exponential and logarithmic functions are complicated and require many additional hours of study. Because it is beyond our realm, no 5 additional attempt to explore this topic is made herein . F. Useful Mathematical Operators n n 1. Summation Operator = Sigma = Ȉ Æ XX X ¦¦ ¦ ii i i 1 i 1 n X X X ..... X X = Sum X ’s where i goes from 1 to n. i ¦ i 1 2 n 1 n i 1 Examples: Given the following X data, verify the summation operation. i = 1 2 3 4 5 X = 25 19 6 27 23 i 3 a. X X X X 25 19 6 50 ¦ i 1 2 3 i 1 5 b .  X X X X X X 25 19 6 27 23 100 ¦ i 1 2 3 4 5 i 1 5 c. X X X X 6 27 23 56 ¦ i 3 4 5 i 3 3 5  X X (X X X ) (X X X ) d. ¦¦ i i 1 2 3 3 4 5 ii 1 3  ( 25 19 6) (6 27 23 ) 50 56 106 3 5 e.  X X (X X X ) (X X X ) ¦¦ i i 1 2 3 3 4 5 ii 1 3  (25 19 6 ) (6 27 23) 50 56  6 n 2. Multiplication Operator = pi = Ȇ Æ X X – i– i i 1 n X X˜ X˜ ....˜ X˜ X = Multiply X ’s where i goes from 1 to n. i – i 1 2 n 1 n i 1 5 For detailed discussions and examples on this topic, please consult high school algebra books such as Algebra 2, by Larson, Boswell, Kanold, and Stiff. ISBN=13:978-0-618-59541-9. Chapter 1: Algebra Review 11Math. Chapter 1. Algebra Review Examples: Given the following X data, verify the multiplication operation. i = 1 2 3 4 5 X = 3 5 6 2 4 i a. 3 X˜ X˜ X X ˜ 3 5˜ 6 90 – i 1 2 3 i 1 5 b. X˜ X˜ X˜ X˜ X X˜ 3˜ 5˜ 6 2˜ 4 720 – i 1 2 3 4 5 i 1 5 c. X˜ X˜ X X ˜ 6 2˜ 4 48 – i 3 4 5 i 3 3 5 d.  X X (˜X X˜ X ) (X˜ X˜ X ) – i– i 1 2 3 3 4 5 i 1 i 3 ˜˜ (35 6)˜ (6˜ 2 4) 90 48 138 3 5 e.  X X (X˜ X˜ X ) (X˜ X ) ––i i 1 2 3 4 5 ii 1 4 ˜˜ (3 5 6)˜ (2 4) 90 8 72 2 5  X X (X X ) (X˜ X˜ X ) f. ¦ i– i 1 2 3 4 5 i 1 3  (3 5)˜ (6˜ 2 4) 8 48 40 G. Multiple-Choice Problems for Exponents, Logarithms, and Mathematical Operators: Identify all equivalent mathematical expressions as correct answers. 2 1. (X + Y) = 2 2 2 2 a. X + 2XY + Y b. X – 2XY + Y 2 2 2 2 c. X + XY + Y d. X + 2XY + 2Y e. none of the above 12 Step by Step Business Math and Statistics Math. Chapter 1. Algebra Review 2 2. (X – Y) = 2 2 a. X + 2XY + Y b. (X – Y) (X – Y) 2 2 2 2 c. X – 2XY + Y d. X – XY + Y e. only (b) and (c) of the above 2 3. (2X + 3Y) = 2 2 2 2 a. 4X + 6YX + 9Y b. 4X + 12XY + 9Y 2 2 2 2 c. 2X + 6XY + 3Y d. 4X + 9Y e. none of the above 2 4. (2X – 3Y) = 2 2 2 2 a. 4X – 9Y b. 2X + 6XY + 3Y 2 2 2 2 c. 4X – 12XY + 9Y d. 4X + 9Y e. none of the above 3 10 5. (2X )(6X ) = 3+10 30 a. 2X b. 12X 3/10 13 c. 48X d. 12X e. none of the above 6 2 3 2 3 4 6. (12X Y )(2Y X )(3X Y ) = 11 9 12 8 a. 72X Y b. 72X Y 10 10 8 12 c. 17X Y d. 72Y X e. only (b) and (d) of the above 2 2 7. X (X + Y) = Chapter 1: Algebra Review 13Math. Chapter 1. Algebra Review 2 2 2 2+2 1+2 2 2 a. X (X + 2XY + Y ) b. X + 2X Y + X Y 4 3 2 2 c. X + 2X Y + X Y d. all of the above e. none of the above 3 X 6 8. ˜ = 2 2 X 5 a. 3X b. 3X  1 c. 3X d. 12X e. none of the above 3 10 9. (2X )/(6X ) = 1  7 a. 0.33333333X b. 7 3X 1  7 c. X d. only (a) and (c) of the above 3 e. all of the above 10 5 3 10. ˜ X Y = 9 5 X Y 10  4 2 a. 10X Y b. 4 2 X Y 9 5 5 3 c. 10X Y X Y d. only (a) and (b) of the above e. all of the above 0.5 1.5 1.5 0.5 11. 24X Yy 12X Y = 2Y 2X Y a. b. c. X Y 2X Y X d. e. X Y 14 Step by Step Business Math and Statistics Math. Chapter 1. Algebra Review 1 1 4 2.5 2 3 3 12. (64X ) (8Y )y 8X Y = 2 2 2 a. b. c. 2 2 X Y XY Y X 2Y d. e. none of the above 2 X 2 3 13. (2X ) = 6 6 5 a. 2X b. 8X c. 8X 6 3 2 d. 16X e. (2X ) 4 3 2 2 14. (3X Y ) = 8 7 16 12 16 10 a. 9X Y b. 9X Y c. 81X Y 16 12 8 7 d. 81X Y e. 81X Y 4 3 2 2 2 4 15. (4X Y ) /(2X Y ) = -3 2 2 a. Y b. Y c. 1/Y 2 2 d. X e. 1/X Using the following data, answer Problems 16 – 20. i = 1 2 3 4 5 6 7 X = 30 52 67 22 16 42 34 i 3 16. X ¦ i i 1 a. 6 b. 140 c. 149 d. 104520 e. none of the above Chapter 1: Algebra Review 15Math. Chapter 1. Algebra Review 6 17. X – i i 4 a. 6 b. 80 c. 120 d. 14784 e. none of the above 2 5 18. X X ¦ i– i i 1 3 a. – 23 b. 0 c. 2352 d. 2350 e. none of the above 5 7 19. X X ¦ i– i i 5 3 a. 92 b. 23584 c. 23676 d. 46432 e. none of the above 7 4 6 20. X X X –¦– i i i i 6 i 3 5 a. 0 b. 1428 c. 89 d. 672 e. 84 X 21. Find the value of X in 3 59049 . a. 20 b. 15 c. 10 d. 5 e. none of the above 22. Identify the correct relationship(s) shown below: X X X a. X log20 log X b. 15z 3˜ 5 20 2 ln X c. 5ln ln 2 d. ln X lnY 5 lnY e. none of the above is correct. Answers to Exercise Problems for Exponents and Mathematical Operators 2 1. (X + Y) = 2 2 a. X + 2XY + Y because 2 2 2 2 (X + Y) (X + Y) = X + XY + YX + Y = X + 2XY + Y 16 Step by Step Business Math and Statistics Math. Chapter 1. Algebra Review 2 2. (X – Y) = e. only (b) and (c) of the above because 2 2 2 2 (X – Y) (X – Y) = X – XY – YX + Y = X – 2XY + Y 2 3. (2X + 3Y) = 2 2 b. 4X + 12XY + 9Y because 2 2 (2X + 3Y) (2X + 3Y) = 4X + 6XY + 6YX + 9Y 2 2 = 4X + 12XY + 9Y 2 4. (2X – 3Y) = 2 2 c. 4X – 12XY + 9Y because 2 2 2 (2X – 3Y) = (2X – 3Y) (2X – 3Y) = 4X – 6XY – 6YX + 9Y 2 2 = 4X – 12XY + 9Y 3 10 5. (2X )(6X ) = 13 3+10 3+10 13 d. 12X because (2)(6)X = 12X = 12X 6 2 3 2 3 4 6. (12X Y )(2Y X )(3X Y ) = 11 9 6+2+3 2+3+4 11 9 a. 72X Y because (12)(2)(3)X Y = 72X Y 2 2 7. X (X + Y) = d. all of the above because 2 2 2 2+2 1+2 2 2 4 3 2 2 X (X + 2XY + Y ) = X + 2X Y + X Y = X + 2X Y + X Y 3 X 6 ˜ 8. = 2 2 X 3 6X 3 2 b. 3X because 3X 3X 2 2X 3 10 9. (2X )/(6X ) = 1 3-10 -7 7 e. all of the above because (2/6)X = (1/3)X = 0.3333X 7 3X 10 5 3 10. ˜ X Y = 9 5 X Y d. only (a) and (b) of the above because Chapter 1: Algebra Review 17Math. Chapter 1. Algebra Review 10  9 5 5 3 9 5 5 3 4 2 10X Y X Y 10X Y 10X Y 4 2 X Y 0.5 1.5 1.5 0.5 11. 24X Yy 12X Y = 0.5 1.5 24X Y 2Y 2Y 0.5 1.5 1.5 0.5 1 1 a. because 2X Y 2X Y 1.5 0.5 X 12X Y X 1 1 4 2.5 2 3 3 12. (64X ) (8Y )y 8X Y = 1 1 1 1 1 1 2 3 2 2 3 3 2 (64X ) (8Y ) (64) X (8) Y a. because 2 4 4 X Y 2.5 2.5 3 3 8X Y 8X Y 1 1 1 1 4 2 3  2.5 (8)X (2)Y 2  2 1 2 3 3 = X (2)Y 2X Y 4 2 X Y 2.5 3 8X Y 2 3 13. (2X ) = 6 2 2 2 3 2+2+2 3 2x3 6 b. 8X because (2X ) (2X ) (2X ) = (2) X = 2 X = 8X 4 3 2 2 14. (3X Y ) = 16 12 2 4x2 3x2 2 2x2 8x2 6x2 16 12 d. 81X Y because 3 X Y = 3 X Y = 81X Y 4 3 2 2 2 4 15. (4X Y ) /(2X Y ) = 2 c. 1/Y because 4 3 2 2 2 -4 2 2 8 6 -4 -8 -8 -2 2 (4X Y ) (2X Y ) = (2 ) X Y (2) X Y = Y = 1/Y 3 16. X ¦ i i 1 3 c. 149 because X X X X 30 52 67 149 ¦ i 1 2 3 i 1 6 17. X – i i 4 6 d. 14784 because X˜ X˜ X X 22˜ 16˜ 42 14784 – i 4 5 6 i 4 18 Step by Step Business Math and Statistics Math. Chapter 1. Algebra Review 2 5 18. XX ¦ ii– i 1 i 3 e. none of the above 5 2 becauseXX () X X˜(X˜XX) ¦ ii– 1 2 345 i 1 i 3  (30 52)˜ (67˜ 22 16 ) 82 23584 23502 7 5 XX 19. ¦– ii i 5 i 3 c. 23676 7 5 XX () X XX˜(˜XXX) because ¦– ii 56 7 3 4 5 i 5 i 3  (16 42 34)˜ (67˜ 22 16 ) 92 23584 23676 76 4 20 XX  X 20. ¦ ––ii i ii 65 i 3 e. 845 76 4 because XX  X () XX(X X˜)(XX) ––¦ ii i 67 3 4 5 6 ii 65 i 3 ˜ (42 34 ) (67 22)˜ (16 42) 1428 89 672 845 X 21. Find the value of X in 3 59049 . c. 10 In order to find X, (1) we can take a natural (or common) logarithm of both sides as: X ln3 ln59049 (2) rewrite the above as: X˜ ln 3 ln 59049 by using the Power Property ln59049 (3) solve for X as: X ln3 (4) use the calculator to find the value of X as: ln59049 10.9861 X 10 ln 3 1.09861 22. Identify the correct relationship(s) shown below: e. none of the above is correct. Note that Chapter 1: Algebra Review 19