1. science
2. mathematics
3. algebra

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
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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
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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
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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.