NSM09_SB_02.fm Page 47 Wednesday, July 17, 2013 6:45 AM Algebra Recall 2 Prepare for this chapter by attempting the following questions. If you have difficulty with a question, go to Pearson Places and download the Recall Worksheet. 2 1 Simplify the following expressions. (a) 3x + 4x + 7x (b) 5y + 7y − 4y (c) 3k2 + 5k2 − 7k2 2 Simplify the following expressions. (a) 5y + 6h – 2h – 3y (b) 6x2 + 5y3 – 2x + 3y3 3 Simplify the following expressions. (a) 3x × 4y (b) 5k × 4k (c) -5g × 6g × 2g 4 Simplify, writing your answer in index form. (c) (52)2 5 Expand: 2 3 4 (a) ⎛ -----⎞ (b) (24 × 52)4 ⎝ 5 3⎠ (d) (8.6325 × 5.15)0 s (b) 68 ÷ 62 pa ge (a) 24 × 29 7 ×3 (c) ⎛ -----------------⎞ 4 2 2 ⎝ 45 ⎠ 6 Find (i) the perimeter and (ii) the area of each of the following. (a) 5 cm (b) 2m (c) 5 mm 5 mm 4 mm pl e 9m 6 mm Sa m 7 Expand the following expressions. (a) 4(x + 2) (b) 6(x − 3) (c) -2(x + 4) 8 Factorise the following expressions. (a) 6x − 12 (b) 10x − 25 (c) -3x + 15 Syllabus references Number and Algebra: Indices ✓ Extend and apply the index laws to variables, using positive-integer indices and the zero index (ACMNA212) (Stage 5.1) ✓ Simplify algebraic products and quotients using index laws (ACMNA231) (Stage 5.1) ✓ Apply index laws to numerical expressions with integer indices (ACMNA209) (Stage 5.1) ✓ Apply index laws to algebraic expressions involving integer indices (ACMNA209) (Stage 5.2) Number and Algebra: Algebraic Techniques ✓ Apply the distributive law to the expansion of algebraic expressions, including binomials, and collect like ter ms where appropriate (ACMNA213) (Stage 5.2) ✓ Factorise algebraic expressions by taking out a common algebraic factor (ACMNA230) (Stage 5.2) ✓ Expand binomial products and factorise monic quadratic expressions using a variety of strategies (ACMNA233) (Stage 5.2) ✓ Expand binomial products using a variety of strategies (ACMNA233) (Stage 5.3§) ✓ Factorise monic and non-monic quadratic expressions (ACMNA269) (Stage 5.3§) Number and Algebra: Equations ✓ Rearrange literal equations (Stage 5.3§) Measurement and Geometry: Numbers of any Magnitude ✓ Investigate very small and very large time scales and intervals (ACMMG219) (Stage 5.1) ✓ Express numbers in scientific notation (ACMNA210) (Stage 5.1) 2 • Algebra 47 NSM09_SB_02.fm Page 48 Wednesday, July 17, 2013 6:45 AM 2.1 Stage 5.1 3.1 Introducing index laws using variables Need to Know Index form 81 = 34 number index 81 = 3×3×3×3 34 Index form pa ge Expanded form = s base Index laws for integers: When multiplying numbers in index form that have the same base, keep the base and add the indices. 36 ----- = 36 − 4 34 = 32 When dividing numbers in index form that have the same base, keep the base and subtract the indices. (42)5 = 42 × 5 = 410 When raising a number in index form to a power, keep the base and multiply the indices. (2 × 5)3 = 23 × 53 When the product of factors in brackets is raised to a power, each factor in the brackets is raised to the same power. Sa m 4 ⎛ 7---⎞ = 7----4⎝ 3⎠ 34 pl e 24 × 2 5 = 2 4 + 5 = 29 20 = 1, 49750 = 1 0 5 6.3470 = 1, ⎛ ---⎞ = 1 ⎝ 7⎠ When the quotient of factors is raised to a power, each factor in the brackets is raised to the same power. Any number, except 0, raised to the zero index is equal to 1 (00 is undefined). Multiplication with indices To calculate a2 × a3, write each power in expanded form and count the number of factors in the multiplication string. a2 = a × a There are two factors of base a in the multiplication string. a3 = a × a × a There are three factors of base a in the multiplication string. a ×a =a×a×a×a×a = a5 There are five factors of base a in the multiplication string. 2 3 Notice that 2 + 3 = 5, so a2 × a3 = a2 + 3 = a5 m a =a×a×…×a×a There are m factors of base a in the multiplication string. an = a × a × … × a × a m There are n factors of base a in the multiplication string. n To multiply a by a , am × an = am + n 48 PEARSON There are m + n factors of base a in the multiplication string. mathematics New South Wales 9 Stages 5.1, 5.2 and 5.3 NSM09_SB_02.fm Page 49 Wednesday, July 17, 2013 6:45 AM 2.1 Division with indices As for multiplication, a pattern for the division of numbers in index form can be identified. To calculate a5 ÷ a3, write the division as a fraction, write each power in expanded form and cancel the common factor of a3. 5 a a×a×a×a×a ----- = --------------------------------------3 a×a×a a 2 =a Notice that 5 − 3 = 2 so a5 ÷ a3 = a5 − 3 = a2 To divide am by an, am ÷ an = am − n There are m − n factors of base a in the multiplication string. Power of a power s To calculate (a5)4, rewrite as a product and then use the index law of multiplication. = a4 × 5 or a5 × 4 = a20 To find the power of a power, (am)n = am × n Index laws for variables pa ge (a5)4 = a5 × a5 × a5 × a5 = a5 + 5 + 5 + 5 When multiplying variables in index form that have the same base, keep the base and add the indices. am am ÷ an = -----an = am − n When dividing variables in index form that have the same base, keep the base and subtract the indices. (am)n = am × n When raising a variable in index form to a power, keep the base and multiply the indices. This rule is known as the ‘power of a power’ law. Sa m pl e am × a n = a m + n Remember that a = a1. - Worked Example 1 1 Simplify each of the following, leaving your answers in index form. (a) 73 × 78 (a) 73 × 78 = 73 + 8 = 711 (b) c3 × c6 (b) c3 × c 6 = c3 + 6 = c9 (c) 4h5 × 3h9 (c) 4h5 × 3h9 = 4 × 3 × h5 × h 9 = 12h5 + 9 = 12h14 2 • Algebra 49 NSM09_SB_02.fm Page 50 Wednesday, July 17, 2013 6:45 AM 2.1 Worked Example 2 2 Simplify each of the following, leaving your answers in index form. 11 9 11 5 d7 d3 (a) -------- (a) (b) ----- 11 9 -----11 5 = 119 − 5 = 114 (b) d7 -----d3 = d7 – 3 = d4 (c) 20c 8e7 ÷ 4c 3e 2 8e7 20c -----------------4c 3 e 2 (c) 5 20 c 8 – 3 e 7 – 2 = ----------------------------------14 Worked Example 3 pa ge s = 5c8 – 3e7 – 2 = 5c5e5 3 Simplify each of the following, leaving your answers in index form. (b) (23)2 × (24)3 (c) (53)4 ÷ (55)2 (d) (r 3)5 (e) (p3)4 × (p5)2 (f) (m5)3 ÷ (m2)6 (63)4 = 63 × 4 = 6 12 Sa m (a) pl e (a) (63)4 (d) (r 3)5 = r3×5 = r 15 (b) (23)2 × (24)3 = 23 × 2 × 2 4 × 3 = 26 × 212 = 26 + 12 = 218 (e) (p3)4 × (p5)2 = p3 × 4 × p 5 × 2 = p12 × p10 = p12 + 10 = p22 (c) (f) (53)4 ÷ (55)2 = 53 × 4 ÷ 5 5 × 2 = 512 ÷ 510 = 512 − 10 = 52 ( m5 )3 --------------( m2 )6 m5 × 3 = -------------m2 × 6 m 15 = -------m 12 = m15 − 12 = m3 50 PEARSON mathematics New South Wales 9 Stages 5.1, 5.2 and 5.3 NSM09_SB_02.fm Page 51 Wednesday, July 17, 2013 6:45 AM 2.1 . 2 1 Introducing index laws using variables Navigator Answers page 424 Stage 5.1 Q1 Column 1, Q2 Column 1, Q3 Column 1, Q4, Q5, Q6 Columns 1 & 2, Q7, Q8, Q9, Q10, Q11, Q14, Q18 Q1 Column 2, Q2 Column 2, Q3 Column 2, Q4, Q6 Columns 2 & 3, Q7, Q8, Q9, Q10, Q11, Q12, Q15, Q17, Q18 Q1 Column 3, Q2 Column 3, Q3 Column 3, Q4, Q6 Columns 2 & 3, Q7, Q8, Q9, Q10, Q11, Q12, Q13, Q15, Q16, Q18 Fluency and understanding 1 Simplify each of the following, leaving your answers in index form. 1 (b) 37 × 34 (c) 511 × 57 (d) a3 × a4 (e) b6 × b5 (f) 5c3 × c2 (g) 3d3 × 2d2 (h) 4b4 × 10b4 (i) 7b2 × 4b2 (j) 4a5 × 3a (k) 7a3 × 2a (l) 20a3 × 10a4 (m) 6a4b2 × 4a5 (n) 9a3b3 × 3ab2 (p) -5a4b6 × 7b2 (q) -9e 5f 5 × -4ef pa ge s (a) 24 × 25 (o) 5b4c 5 × 6b3c 2 (r) 3a5b8 × -8ab2 2 Simplify each of the following, leaving your answers in index form. (a) 38 ÷ 35 (b) 47 ÷ 42 (d) m5 ÷ m2 (e) 4b5 ÷ b3 n6 n4 14b 7 7b 5 (j) 6a 5 -------3a 4 (h) ----------- (i) 6 22c ----------11c 2 (k) ----------- 12a 7 8a 3 (l) 4f 5 14e ---------------10ef 2 (n) -24a5b3 ÷ 21a2b (o) 144a6b8 ÷ -12a5b7 Sa m (m) 10a7b7 ÷ 5a4b3 2 (f) 5g7 ÷ g3 pl e (g) ----- (c) 95 ÷ 93 3 Simplify each of the following, leaving your answers in index form. 3 (a) (23)4 (b) (32)5 (c) (54)2 (d) (32)5 × (34)2 (e) (43)2 × (45)3 (f) (72)5 × (74)2 (g) (52)8 ÷ (53)2 (h) (85)8 ÷ (82)3 (i) (114)8 ÷ (113)9 (j) (b7)3 (k) (k8)4 (l) (a6)2 (m) y 5 × (y 3)4 (n) (k 3)8 × k 3 (o) (m6)3 × m2 (p) (g 2)5 × (g 4)2 (q) (y 2)4 × (y 5)3 (r) (t 3)3 × (t 2)6 (s) (p3)4 ÷ (p5)2 (t) (n7)2 ÷ (n2)4 (u) (b8)3 ÷ (b3)7 4 (a) -4r3 × -r2 can be simplified to give: A 4r1 B -4r5 C 4r5 D 4r6 C 10abc D 10a2bc (b) The product of -abc, -5a and -2a is: A -10a3bc B -7a2bc 2 • Algebra 51 NSM09_SB_02.fm Page 52 Wednesday, July 17, 2013 6:45 AM 2.1 5 Copy and complete each of the following by writing the correct index number in the box. 6y 4 × y 6 y2 × y5 x7 × x9 x 12 (a) ---------------- (b) -------------------- x = ------12 x (Add indices) 6y = ------y (Add indices) =x (Subtract indices) = 6y (Subtract indices) 6 Simplify each of the following by completing any multiplication of variables first. x8 × x7 x6 (b) ------------------ j4 × j8 j2 × j5 (e) ----------------------- (d) -------------- c6 × c9 c5 × c3 t7 t 2 × 3t 4 (c) --------------- 4a 7 × 3a 4 a3 (f) f8 × f6 f 10 7 (a) ---------------- simplifies to: f4 B f6 15x 9 × 3x 6 9x 10 × x 4 (b) -------------------------- simplifies to: A 5x9 B 9x C f 48 D f 12 C 5x D 9x9 C -4kn D 45k12n6 C 3xy5z 3 D 3x9y9z9 pa ge A 8q × 3q 6 --------------------12q 4 s (a) ---------------- (c) -36k 4n3 ÷ 9n2k 3 can be simplified to: 4k4n 42x 5 y 7 z 6 14x 4 y 2 z 3 B -4k7n6 pl e A (d) ---------------------- simplifies to: A 3xy5z 2 B 28xy5z 3 Sa m 8 (a) How many groups of 3a2 will divide into 15a4? (b) Use your answer to (a) to rewrite this as a multiplication that has 15a4 as the product. a8 × b5 a2 × a6 9 (a) ----------------- simplifies to: A a16b5 B a10b11 C b5 D b11 d4 ----2 C 4d4 c--------24 D d 4----24 3c 7 × 8d 4 12c 3 × 4c 4 (b) ------------------------- simplifies to: A 2d 4 B 10 Simplify each of the following. (a) (m5)2 ÷ m7 (b) (p6)4 ÷ p10 (k 4 ) 3 (c) ------------ k 9 (n 8 ) 2 (d) ------------ n 6 11 (a) (k2)3 × (k5)2 simplifies to: A k16 B k12 C k60 D k15 b C b5 D b12 (b) (b5)4 ÷ (b3)5 simplifies to: A 52 PEARSON b4 B mathematics New South Wales 9 Stages 5.1, 5.2 and 5.3 NSM09_SB_02.fm Page 53 Wednesday, July 17, 2013 6:45 AM 2.1 Reasoning, communicating and open-ended 12 Solve for x in the following equations. (a) 5x + 2 = 52x − 1 (b) 3x + 1 = 3 (c) 22x + 4 = 1 (d) 4(7x + 7x) = 56 13 When Josh started his part-time job, he decided to save some of his wages on a weekly basis. He was paid $50.00 a week, so he decided to start by saving 5c the first week, 10c the next week, 20c the third week, continuing to double the amount he saved each week. He was shocked to find that, by the 11th week, he could not afford to keep up his plan. (a) Copy the following table and complete the second column to show how much Josh Hint If the bases on both sides of the equation are the same, the expression for the powers must be equal. saved each week for the first 5 weeks. Week Amount saved each week (c) Amount saved each week (c) in index form 1 2 s 3 4 pa ge 5 (b) Complete the third column in this table by writing these amounts in index form. (c) Write a rule in index form for finding the amount Josh saved in the nth week. (d) Use your rule to calculate how much Josh was supposed to save in the 11th week. (e) How much would Josh need to earn each week if he was to keep up his saving plan for 6 months (26 weeks)? statement: ■ × ▲ = -3a2b3 pl e 14 Find three different expressions for ■ and ▲ in terms of a and b to make this a true Hint Look for a pattern. What values do you get if 1 is subtracted from a power of 2? 15 Write three possible divisions involving indices that will give an answer of 5a2b3. Sa m 16 Create three index number division expressions and simplify them using any of 24a2b3, 3ab, 48a5b6, 22a2b2 or 2 × 32a. Check your answers by giving the divisions to a friend to simplify. 17 Write three different sets of values for m and n so that (am)n simplifies to a24. 18 The index laws are sometimes used incorrectly. Below is a list of equations that ‘look’ reasonable, but are actually incorrect. In each case, explain the mistake that has been made and give the correct answer in its place. Explain why your answer is correct. (a) x 2x 4 = x8 x9 x3 (b) ----- = x 3 4 (c) (x 2)4 = x 2 = x16 2 • Algebra 53 NSM09_SB_02.fm Page 54 Wednesday, July 17, 2013 6:45 AM 2.2 Stage 5.1 2.2 More index laws Need to Know To expand an expression is to remove the brackets in the expression. Expanding products Expand (2 × a)3. (2 × a)3 = (2 × a) × (2 × a) × (2 × a) =2×2×2×a×a×a = 23 × a 3 s So (2 × a)3 = 23 × a3. Expand (kp)4. (kp)4 = kp × kp × kp × kp =k×k×k×k×p×p×p×p = k4 × p4 = k4 p 4 pa ge Each factor in the brackets has been raised to the power of 3. Each factor in the brackets has been raised to the power of 4. So (ab)m = ambm. m 5 Expand ⎛ ----⎞ . ⎝ n⎠ pl e Expanding quotients Sa m 5 ⎛m ----⎞ = m ---- × m ---- × m ---- × m ---- × m ---⎝ n⎠ n n n n n 5 m = ------5 n Each variable in the brackets has been raised to the power of 5. m a m a So ⎛ --⎞ = ------ . ⎝ b⎠ m b The zero index If 20 = 1, 45 6780 = 1 and 0.000 1230 = 1, is a0 = 1 also? m a To simplify ------ , subtract indices (second index law). m a m a ------ = am − m m a = a0 Alternatively, treating it as a fraction division, m a ------ = 1 m a Therefore a0 = 1. 54 PEARSON mathematics New South Wales 9 Stages 5.1, 5.2 and 5.3 NSM09_SB_02.fm Page 55 Wednesday, July 17, 2013 6:45 AM 2.2 Negative indices Evaluating 2-6. Consider 26 = 64, 25 = 32, 24 = 16, 23 = 8, 22 = 4, 21 = 2 and 20 = 1. The index on the left-hand side of each equation is one less than the previous one and the number on the right-hand side of each equation is half the previous one. Continuing this pattern: 1 2-1 = --2 1 2-2 = ----2 2 1 2-3 = ----3 2 1 So, 2-6 = ----6 2 1 In general, 2-m = ------ . m 2 1 1 Note that any base number can be used, e.g. 3-4 = ----- , 81-6 = -------- . 4 6 3 81 s Index laws for powers with the same base Multiplication of powers am × a n = am + n Division of powers am am ÷ an = -----an = am − n Power of a power (am)n = amn Expanding products (ab)m = ambm Expanding quotients m ⎛ a--⎞ = a-----m⎝ b⎠ bm pa ge The following laws are true if a, b ≠ 0 (a and b cannot be zero). Only integer values for m and n are considered here. When multiplying numbers with the same base, keep the base and add the indices. When dividing numbers with the same base, keep the base and subtract the indices. pl e When raising a number in index form to a power, keep the base and multiply the indices. Sa m When removing brackets around a product raised to a power, raise each number to that power. The zero index a0 = 1 (a ≠ 0) Negative indices with a numerical base Example 1 2-m = -----2m When removing brackets around a quotient raised to a power, raise each number to that power. Any variable or product of variables raised to the power of 0 is equal to 1. Note that 00 is undefined. A number raised to a negative power is equal to 1 divided by the same number raised to the positive power. Worked Example 4 4 Use an index law to expand each of the following, leaving your answers in index form. (a) (3 × 5)4 (a) (3 × 5)4 = 34 × 54 (b) (ab)5 (b) (ab)5 = a5 × b5 = a 5b 5 (c) (5p4q)3 (c) (5p4q)3 = 53 × (p4)3 × q3 = 53 × p4 × 3 × q 3 = 53 × p12 × q3 = 53p12q3 2 • Algebra 55 NSM09_SB_02.fm Page 56 Wednesday, July 17, 2013 6:45 AM 2.2 Worked Example 5 5 Use an index law to expand each of the following, leaving your answers in index form. 5 p (b) ⎛ --⎞ 5 (b) 45 = -----35 ⎛ --p⎞ ⎝ q⎠ 2a⎞ 3 ⎛ ----⎝ c 4-⎠ 7 (c) (2a) 3 = -----------(c 4 ) 3 p7 = ----7q Worked Example 6 3 ⎝ c4 ⎠ s ⎛ --4-⎞ ⎝ 3⎠ 2a (c) ⎛ ------⎞ ⎝ q⎠ ⎝ 3⎠ (a) 7 pa ge 4 (a) ⎛ ---⎞ 6 Simplify each of the following, leaving your answers in index form. (a) 23 × 63 23 × 6 3 = (2 × 6)3 = 123 (b) 54 × h 4 = (5h)4 (c) m2 × n2 (c) pl e (a) (b) 54 × h4 m 2 × n2 = (mn)2 Sa m Worked Example 7 7 Simplify each of the following, leaving your answers in index form. 16 3 43 (a) -------- (a) 16 3 -------43 16 3 = ⎛ -----⎞ ⎝ 4⎠ t4 24 (b) ----- (b) t4 -----24 t 4 = ⎛ ---⎞ ⎝ 2⎠ = 43 56 PEARSON mathematics New South Wales 9 Stages 5.1, 5.2 and 5.3 c8 d8 (c) ----- (c) c8 -----d8 c 8 = ⎛ ---⎞ ⎝ d⎠ NSM09_SB_02.fm Page 57 Wednesday, July 17, 2013 6:45 AM 2.2 Worked Example 8 8 Simplify each of the following. 2y 4 × y 3 y5 × y2 (a) 4p0 m3n6 m3n (b) -------------------- (c) ------------- 4 2y × y 3 -------------------y5 × y2 (b) (c) 2y 4 + 3 = ---------------y5 + 2 2y 7 = -------y7 = 2y7 − 7 = 2y 0 =2×1 =2 Worked Example 9 m3n6 ------------m3n = m3 – 3n6 – 1 = m0n5 = 1 × n5 = n5 s 4p0 = 4 × p0 =4×1=4 pa ge (a) 9 Simplify the following, writing your answer with positive indices. (b) 5-3 1 (a) 3-2 = -----32 1 (b) 5-3 = -----53 Sa m . (c) 8-5 pl e (a) 3-2 1 (c) 8-5 = -----85 2 2 More index laws Answers page 424 Navigator Q1 Column 1, Q2 Column 1, Q3 Column 1, Q4 Column 1, Q5 Column 1, Q6, Q7, Q8, Q9, Q10 Columns 1 & 2, Q11 Columns 1 & 2, Q12, Q13, Q14, Q15, Q16, Q19, Q20, Q21 Q1 Column 2, Q2 Column 2, Q3 Column 2, Q4 Column 2, Q5 Column 2, Q6, Q7, Q8, Q9, Q10 Columns 2 & 3, Q11 Columns 2 & 3, Q12, Q13, Q14, Q15, Q16, Q17, Q19, Q20, Q21 Stage 5.1 Q1 Column 3, Q2 Column 3, Q3 Column 3, Q4 Column 3, Q5 Column 3, Q6, Q7, Q8, Q9, Q10 Columns 3 & 4, Q11 Columns 3 & 4, Q12, Q13, Q14, Q15, Q16, Q17, Q18, Q19, Q20, Q21 Fluency and understanding 1 Use an index law to expand each of the following, leaving your answers in index form. (a) (6 × 5)2 (b) (11 × 3)4 (c) (12 × 9)5 (d) (4x)2 (e) (p3q)2 (f) (a4f)3 (g) (2p)5 (h) (p3q5)7 (i) (c2d7)4 (j) (bc)3 (k) (6a4)3 (l) (9p5q4)2 (m) (7m3n6)2 (n) (2ab2)3 (o) (3ef 4)4 4 2 • Algebra 57 NSM09_SB_02.fm Page 58 Wednesday, July 17, 2013 6:45 AM 2.2 2 Use an index law to expand each of the following, leaving your answers in index form. 12 (a) ⎛ ------⎞ 2 ⎝ 5⎠ y (d) ⎛ --⎞ 3 4 2 5 4 ⎝ s⎠ 3 ⎞4 ⎛ ----⎝ w 6⎠ 8 (k) ⎛ -------⎞ 3 5 3 4 a (n) ⎛ -----⎞ ⎝ g 2⎠ ⎝ c5⎠ 4 6 3 c (q) ⎛ -----⎞ ⎝ b⎠ ⎝ 2x⎠ 2⎞3 ⎛ --⎝ r 4⎠ 4 3 t (o) ⎛ -----⎞ ⎝ v 3⎠ 6 3 f (r) ⎛ ------⎞ ⎝ 6g⎠ pa ge (e) 34 × p4 (f) 79 × r9 (g) a5 × m5 (h) k7 × t7 (i) p8 × n8 4 Simplify each of the following, leaving your answers in index form. 24 3 63 pl e (b) -------- 132 5 11 5 (c) ----------- n7 67 (f) g9 ----49 (h) ------- k 11 n 11 (i) g 10 ------h 10 (a) k0 (b) 7g0 (c) 7 + g0 (d) a0b2 (e) 3m0n0 (f) c3d0e0 (g) -3a4b0 (h) -8g11h0 (i) -(4pr)0 (j) y4 ÷ y4 (k) 3x7 ÷ x7 (l) 24m6 ÷ 8m6 x6 y6 Sa m (e) ----- 5 Simplify each of the following. 3x 4 z z (n) ----------- a5b4 a2b4 (q) -------------- 12e 3 f 2 16e 3 (t) (m) ----------- (p) ----------- (s) --------------- h5k8 h5 (o) ---------------- 8m 3 n 4 16m 3 6h 4 r 3 2h 4 r 3 (r) 2s9 15r ---------------27r 2 (u) ----------------- 4d3 9b -------------6b 4 d 3 25k 4 p 2 15k 4 6 Simplify the following, writing your answers with positive indices. (a) 4 2 (b) 2 3 (c) 5 4 (d) 3 3 (e) 5 2 (f) 4 3 - PEARSON (l) 2 (d) 48 × h8 (g) ----- 58 d 3⎞ ⎛ ---⎝ g⎠ (c) 84 × 74 r5 35 9 (i) 4 (b) 46 × 56 (d) ----- Remember that any number (except 0) divided by itself equals one. m 3⎞ ⎛ -----⎝ n⎠ (a) 32 × 42 15 4 34 Hint (f) 3 Simplify each of the following, leaving your answers in index form. (a) -------- 8 3 ⎝ m 2⎠ h (m) ⎛ -----⎞ 7 5 8 ⎝ 2⎠ r (h) ⎛ ----⎞ ⎝ z⎠ 6 d (e) ⎛ ---⎞ 9 (c) ⎛ ---⎞ ⎝ e⎠ w (g) ⎛ ------⎞ 5a (p) ⎛ ------⎞ 4 ⎝ 3⎠ ⎝ x⎠ (j) 7 (b) ⎛ ---⎞ s 5 - mathematics New South Wales 9 Stages 5.1, 5.2 and 5.3 - NSM09_SB_02.fm Page 59 Wednesday, July 17, 2013 6:45 AM 2.2 7 (a) (2k)6 simplifies to: 2k6 A B 12k C 64k6 D 12k6 B 9a12 C 9a7 D 27a7 B 81 -----g4 C 12 -----g D 81 -----g B 16 k-----49 C 16 k-----7 D 10 k-----49 (b) (3a4)3 simplifies to: 27a12 A 3 (c) ⎛ ---⎞ 4 ⎝ g⎠ simplifies to: 12 -----g4 A 8 2 k (d) ⎛ -----⎞ ⎝ 7⎠ simplifies to: k 10 ------7 A completing the multiplication, applying other index laws if necessary. (b) (gh)7 × (gh)2 (d) (4y)2 × (5xy)2 (e) (cd)3 × (c3d2)3 (c) (3x)3 × (4x)2 pa ge (a) (xy)2 × (xy)4 s 8 Simplify each of the following by first applying the index law for expanding products, then (f) (x2y3)2 × (xy)2 9 Simplify each of the following by first applying the index law for expanding quotients, completing the multiplication, then simplifying if possible. 2 4 2 a b (d) ⎛ -----⎞ × ⎛ --⎞ ⎝ b 2⎠ 6 p p (b) ⎛ --⎞ × ⎛ --⎞ ⎝ q⎠ 4 6 6 2 ⎝ q 4⎠ 2 2a a (e) ⎛ ------⎞ × ⎛ --⎞ ⎝ a⎠ ⎝ b⎠ 5 3 d d (c) ⎛ -----⎞ × ⎛ -----⎞ ⎝ q⎠ pl e 3 ⎛ f⎞ ⎛ f⎞ (a) ⎜ --⎟ × ⎜ --⎟ ⎝ g⎠ ⎝ g⎠ 3 3 ⎝ b⎠ ⎝ q 2⎠ (f) 3x 2⎞ 2x ⎛ ------- × ⎛ -----⎞ ⎝ y⎠ ⎝ y ⎠ 2 10 Simplify each of the following by applying the expanding products and the powers of Sa m powers index laws, then the index law for division. (w 4 z 3 ) 3 (w 5 z 2 ) 2 (a) ------------------- (c 7 e 5 ) 2 (c 4 e 2 ) 3 (pq) 12 (p 2 q 3 ) 3 (b) ----------------- (vw) 9 (v 3 w 2 ) 2 (c) ------------------ (d) ------------------- 11 Simplify each of the following. (a) (x3)0 (b) (d5)0 (c) (k0)4 (d) (p0)2 (e) (m3)4 × (m2)0 (f) (f 5)4 × (f 3)0 (g) (t0)3 × t 4 (h) ((q4)5 × (q5)4)0 3k 2 -------------k × 5k (k) ----------------- d6 × e2 3e × 2e (o) ---------------------- (i) x6 × x5 ---------------x 11 (j) 2b 6 × 5c 8 c 7 × 4c (n) ----------------- (m) ----------------------- 4 × n3 n ---------------n5 × n2 4a 3 a 2 × 7a (l) c4 × d 2c 2 × 5c 2 (p) ----------------------- p7 × q5 15p 2 × p 5 12 Use the negative index law to write the following with positive indices, then evaluate. (a) 4-2 (b) 3-3 (d) ------- 1 2 -1 (e) ------- (f) (g) 4 × 3-2 (h) 3 × 4-3 (i) 2 × 5-2 7 × 3 -2 3 (j) − ----------------- 1 4 -3 6 × 2 -4 2 -3 (k) − ----------------- (c) 5-4 1 ------3 -2 3 × 3 -4 3 -6 (l) − ----------------- 2 • Algebra 59 NSM09_SB_02.fm Page 60 Wednesday, July 17, 2013 6:45 AM 2.2 1 2 -4 13 (a) ------- is the same as: 16 A B 8 C 1----16 D -16 B 2-3 C 3-2 D 1 ------2 -3 1 8 (b) --- is the same as: 23 A Reasoning, communicating and open-ended 14 Kellie and Eva are given the Here is Kellie’s working: 2 ⎛ 4x --------⎞ ⎝ 2x ⎠ Both girls had the correct answer, but Eva finished much faster. ‘How did you do it so fast?’ Kellie asked Eva. Explain why Eva’s method took fewer steps. 43 × x2 × 3 = ------------------------23 × x3 64x 6 = ------------8x 3 2 3 ⎛ 4x --------⎞ ⎝ 2x ⎠ = (2x)3 = 8x3 pa ge 4x 2 3 Simplify ⎛ --------⎞ . ⎝ 2x ⎠ 3 Here is Eva’s working: s following question as part of their indices homework. = 8x3 15 By rewriting the expression if required and then using index laws, simplify each of the following using the smallest possible base. 9 2n 9n (b) 2 n × 4 n + 1 125 5n 5 2n 25 (c) --------- × -------- pl e (a) -------- 16 (a) Using index notation, write the numbers 125, 25, 5 and 1, using base 5. (b) Using index notation, write the next five numbers using the pattern in part (a). (c) Write the whole sequence of nine numbers in this pattern without using index Sa m notation. 17 A model for the number of planets, N, created in a section of a galaxy is given by the n2T3p 2 formula N = ⎛ ---------------⎞ . In the formula: ⎝ b ⎠ n is the number of suns T is the temperature of the surrounding space p is the number of planets already present b is the number of black holes. (a) Write the formula without brackets. (b) (i) How many new planets does the model predict if T = 1, n = 1000 and b = p? (ii) If the temperature of the surrounding space is a negative number, will we obtain a positive or a negative answer? Explain your reasoning. 60 PEARSON mathematics New South Wales 9 Stages 5.1, 5.2 and 5.3 NSM09_SB_02.fm Page 61 Wednesday, July 17, 2013 6:45 AM 2.2 18 The intensity of sound is measured in watts per square metre. The level of sound (loudness) is measured in decibels. The threshold of human hearing is 10-12 watts/m2 and is defined as a level of sound of 0 decibels (dB). All other decibel levels are calculated by comparing the sound intensity with the threshold of human hearing. To do this we follow a simple rule: 1 Divide the intensity measured in watts/m2 by the sound intensity of the threshold of human hearing. For example, normal conversation is measured at 10−6 watts/m2. 10 -6 ------------- = 10-6 − (-12) 10 -12 = 106 This means that normal conversation is 1 million times as loud as the threshold of human hearing. 2 Multiply the index by 10. This is the intensity level in decibels. s Intensity level = 10 × 6 = 60 dB pa ge To convert from decibels to find the sound intensity, we reverse the process. For example, the sound level for a quiet library = 40 dB. To find the sound intensity: 1 Divide the number of decibels by 10. 40 ------ = 4 10 2 Use this number as an index for 10 and multiply by the threshold for human hearing. pl e The sound intensity of a quiet library = 104 × 10-12 = 10-8 watts/m2 (a) Copy and complete the following table. Sa m Source Sound level (loudness; dB) Sound intensity (watts/m2) Jet aircraft, 50 m away 100 Threshold of pain 10 Chainsaw, 1 m distance 110 Nightclub, 1 m from speaker 0.01 Busy traffic 80 Vacuum cleaner, 1 m distance 70 Average home Rustling of leaves in the distance 0.000 000 1 10 (b) If a whisper is 20 dB, how many times as loud is normal conversation? (c) If the intensity of a rock concert is 100 watts/m2 and the level of intensity of the threshold of discomfort is 120 dB, what comment could you make comparing these two measures? (d) Which is louder, thunder at a sound level of 110 dB or a loud TV at a sound intensity of 10-4 watts/m2? 2 • Algebra 61 NSM09_SB_02.fm Page 62 Wednesday, July 17, 2013 6:45 AM 2.2 19 (a) Write at least two expressions involving indices and multiplication that, when simplified, equal 36m4n8. (b) Write at least two expressions involving indices and division that can be simplified 4m 4 to ---------- . 5n 6 20 Write three different terms that 27x3y7 could be divided by so that no x appears in the answer. 3x 2 y3 21 Write a question that could be simplified to -------- by using: (a) any two index laws with an expression that has only positive indices (b) any three index laws with an expression that has only positive indices. Square cubes pa ge Problem Problem oblem solving s Show, step by step, how your question can be simplified to obtain the final answer. Equipment required: 1 brain, calculator ‘I have been studying the cube numbers. You know, some of them are also square numbers!’ It was smart Alec, bragging again. pl e ‘You’re kidding!’ said Sue. ‘A number can’t be a square and a cube.’ ‘Yes it can,’ said Alec. ‘Like, 64 is both 8 squared and 4 cubed.’ Sa m If Alec is right, there may be more. 62 PEARSON mathematics New South Wales 9 Stages 5.1, 5.2 and 5.3 1 Can you find any other two-digit numbers that are a square and a cube? 2 With the help of your calculator, how many can you find under 1000? Strategy options • Test all possible combinations. • Look for a pattern. • Break the problem into manageable parts.

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