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7Number and algebra

EquationsandlogarithmsHistorically, algebra dates back to ancient Egypt andBabylon where linear and quadratic equations were solved.In ancient Babylon, quadratic equations were solved by verysimilar methods to those still relevant and taught today.Logarithms were developed in the seventeenth century andare still in use, most recognised in the pH, decibel andRichter scales.

n Chapter outlineProficiency strands

7-01 Equations with algebraicfractions

U F R

7-02 Quadratic equationsx2 þ bx þ c ¼ 0

U F R C

7-03 Simple cubic equationsax3 ¼ c*

U F R C

7-04 Equation problems U F PS R C7-05 Equations and formulas U F PS R C7-06 Changing the subject of

a formula*U F R C

7-07 Graphing inequalities ona number line

U F C

7-08 Solving inequalities U F R7-09 Logarithms* U F R C7-10 Logarithm laws* U F R C7-11 Exponential and

logarithmic equations*U F R C

*STAGE 5.3

nWordbankcubic equation An equation involving a variable cubed(power of 3), such as 4x3 ¼ 500exponential equation An equation where the variable isa power, such as 3 x ¼ 243logarithm The power of a number, to a given base. Forexample, log10 1000 ¼ 3, meaning that the logarithm of1000 to base 10 is 3, because 1000 ¼ 103

inequality A mathematical statement that two quantitiesare not equal, involving algebraic expressions and aninequality sign (>, �,

n In this chapter you will:• solve linear equations involving simple algebraic fractions• solve simple quadratic equations using a range of strategies• substitute values into formulas to determine an unknown• solve linear inequalities and graph their solutions on a number line• (STAGE 5.3) use the definition of a logarithm to establish and apply the laws of logarithms• (STAGE 5.3) solve simple exponential equations• solve linear equations and problems involving equations• (STAGE 5.3) solve linear equations involving complex algebraic fractions• (STAGE 5.3) solve simple cubic equations of the form ax3 ¼ c• (STAGE 5.3) change the subject of a formula• (STAGE 5.3) solve simple logarithmic equations

SkillCheck

1 Solve each equation.a 4a þ 5 ¼ 2a � 19 b 3xþ 2

5¼ 4 c 4(2 � x) ¼ �24

2 Factorise each expression.a k2 þ 5k þ 4 b y2 � 10y þ 16 c m2 � m � 56d u2 þ 8u � 65 e w2 � 10w þ 21 f x2 � 2x � 24

7-01 Equations with algebraic equations

Example 1

Solve each equation.

a 2m3� m

2¼ 2 b 2aþ 4

5¼ 2

3

Solutiona 2m

3� m

2¼ 2

Multiply both sides by a common multiple of the denominators to remove the fractions.The lowest common multiple (LCM) of 3 and 2 is 6, so multiply both sides by 6.

62m3� m

2

� �¼ 6 3 2

6 2 32m3 1� 6 3 3 m

2 1¼ 12

4m� 3m ¼ 12m ¼ 12

Worksheet

StartUp assignment 6

MAT10NAWK10040

Video tutorial

Equations withalgebraic fractions

MAT10NAVT10026

Puzzle sheet

Equations code puzzle

MAT10NAPS10041

Puzzle sheet

Equations order activity

MAT10NAPS10042

Puzzle sheet

Solving linearequations 1

MAT10NAPS00035

Puzzle sheet

Solving linearequations 2

MAT10NAPS00036Check by substituting that thissolution is correct.

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Equations and logarithms

b 2aþ 45¼ 2

3Multiply both sides by 15, the LCM of 5 and 3.

2aþ 45 1

3 15 3 ¼ 23 1

3 15 5

3ð2aþ 4Þ ¼ 106aþ 12 ¼ 10

6a ¼ �2

a ¼ �26

¼ � 13

Example 2

Solve 2nþ 13� 3n� 2

2¼ �5

Solution2nþ 1

3� 3n� 2

2¼ �5

6 22nþ 1

3 1

� �� 6 3 3n� 2

2 1

� �¼ 6 3 ð�5Þ

2ð2nþ 1Þ � 3ð3n� 2Þ ¼ �304nþ 2� 9nþ 6 ¼ �30

�5nþ 8 ¼ �30�5n ¼ �38

n ¼ �38�5¼ 7 3

5

The LCM of 3 and 2 is 6.

Exercise 7-01 Equations with algebraic fractions1 Solve each equation.

a 3y5¼ 9 b 2a

9¼ 2 c mþ 5

2¼ 6 d k � 2

5¼ 11

e nþ 53¼ �10 f y� 1

4¼ �2 g xþ 1

4þ 2 ¼ 10 h y� 1

5� 6 ¼ 3

i mþ 25� 1 ¼ 3 j x� 6

5þ 7 ¼ 0 k 2ðxþ 1Þ

5¼ 10 l 3ðm� 2Þ

4¼ 6

m 8ðnþ 1Þ3

þ 2 ¼ 4 n 5ð1� nÞ2

� 1 ¼ 3 o 4ð1þ dÞ3

þ 1 ¼ 7 13

Stage 5.3

Video tutorial

Equations withalgebraic fractions

MAT10NAVT10026

See Example 1

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NEW CENTURY MATHS ADVANCEDfor the A u s t r a l i a n C u r r i c u l u m10þ10A

2 Solve each equation.

a 2k3¼ 5

4b 3w

10¼ 2

5c 5x

2¼ � 10

3d x� 1

2¼ xþ 1

4

e yþ 25¼ y� 1

2f aþ 5

3¼ a� 1

8g pþ 2

5¼ p� 5

2h 2y� 1

5¼ yþ 1

4

i 3yþ 23¼ 2yþ 1

4j w

5þ w

2¼ 7 k w

2� w

5¼ 15 l 2w

3� w

4¼ 4

m 3a2þ a

3¼ 1 n 2y

5� y

3¼ 4 o a

3þ 3a

4¼ 2

3 Solve each equation. Select the correct answer A, B, C or D.

a 4m5� m

3¼ 2

A m ¼ 10 B m ¼ 12 C m ¼ 307

D m ¼ 43

b mþ 12¼ 3þ 2m

5

A m ¼ 1 B m ¼ 5 C m ¼ 53

D m ¼ 23


a x� 14þ 2x

7¼ 0 b pþ 2

3þ pþ 1

4¼ 10 c mþ 2

3þ mþ 1

4¼ 12

d x� 35þ x� 2

2¼ 6 e 3x� 10

3þ x� 2

2¼ 11 f 3yþ 1

4� yþ 2

3¼ 4

g 7þ 2a5� a� 1

2¼ 6 h 6a� 1

4� aþ 2

3¼ 8 i wþ 3

6� w� 1

5¼ 1

3

j a� 105� 5� 2a

4¼ 1

2

7-02 Quadratic equations x2 þ bx þ c ¼ 0

An equation in which the highest power of the variable is 2 is called a quadratic equation;for example, x2 ¼ 5, 3m2 þ 7 ¼ 10, d2 � d � 6 ¼ 0 and 4y2 � 3y ¼ 8.

Stage 5.3

See Example 2

iSto

ckph

oto/

Lag

ui

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Solving ax2 ¼ cSummary

The quadratic equation x2 ¼ c (where c is a positive number) has two solutions,

x ¼ �ffiffifficpðwhich means x ¼

ffiffifficp

and x ¼ �ffiffifficpÞ

Example 3

Solve each quadratic equation.

a m2 ¼ 16 b 3x2 ¼ 75 c 3m2 � 12 ¼ 0

Solutiona m 2 ¼ 16

m ¼ �ffiffiffiffiffi16p

¼ �4Finding the square root of both sides.

b 3x 2 ¼ 75

x 2 ¼ 753

x 2 ¼ 25x ¼ �

ffiffiffiffiffi25p

¼ �5

c 3m 2 � 12 ¼ 03m 2 � 12þ 12 ¼ 0þ 12

3m 2 ¼ 12

m 2 ¼ 123

m 2 ¼ 4m ¼ �

ffiffiffi4p

¼ �2

Example 4

Solve 5x2

9¼ 25, writing the solution in exact (surd) form.

Solution5x 2

9¼ 25

5x 2 ¼ 25 3 9¼ 225

x 2 ¼ 2255

¼ 45x ¼ �

ffiffiffiffiffi45p

¼ �ffiffiffi9p ffiffiffi

5p

¼ �3ffiffiffi5p

As a surd

In simplest surd form

Worksheet

Equations review

MAT10NAWK10043

Video tutorial

Simple quadraticequations

MAT10NAVT10028

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Example 5

Solve 7x2 � 88 ¼ 0, writing the solution correct to one decimal place.

Solution7x 2 � 88 ¼ 0

7x 2 ¼ 88

x 2 ¼ 887

x ¼ �ffiffiffiffiffi887

r

x ¼ �3:54562 . . .� �3:5

Solving x2 þ bx þ c ¼ 0 by factorisingTo solve quadratic equations of the form x2 þ bx þ c ¼ 0, we need to factorise the quadraticexpression on the LHS, which we learnt in Chapter 5, Products and Factors.

Example 6

Solve x2 þ 5x þ 6 ¼ 0.

Solutionx2 þ 5x þ 6 ¼ 0(x þ 2)(x þ 3) ¼ 0

The LHS has been factorised into two factors, (x þ 2) and (x þ 3), whose product is 0.If two numbers have a product of 0, then one of the numbers must be 0.

) xþ 2 ¼ 0 or xþ 3 ¼ 0) x ¼ �2 or x ¼ �3

[ The solution to x2 þ 5x þ 6 ¼ 0 is x ¼ �2 or x ¼ �3.Check:

When x ¼ �2,LHS ¼ (�2)2 þ 5 3 (�2) þ 6 ¼ 0RHS ¼ 0Therefore LHS ¼ RHS.When x ¼ �3,LHS ¼ (�3)2 þ 5 3 (�3) þ 6 ¼ 0RHS ¼ 0Therefore LHS ¼ RHS.

Video tutorial

Quadratic equationsby factorising

MAT10NAVT10029

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Summary

When solving quadratic equations by factorising, the following property is used.If pq ¼ 0, then p ¼ 0 or q ¼ 0.

Example 7

Solve each quadratic equation.

a x2 � x � 2 ¼ 0 b u2 þ 3u � 28 ¼ 0c a2 � 2a ¼ 0 d p2 ¼ 5p þ 24

Solutiona x2 � x � 2 ¼ 0

(x � 2)(x þ 1) ¼ 0) x� 2 ¼ 0 or xþ 1 ¼ 0) x ¼ 2 or x ¼ �1

[ The solution to x2 � x � 2 ¼ 0 is x ¼ 2 or x ¼ �1.b u2 þ 3u � 28 ¼ 0

(u þ 7)(u � 4) ¼ 0) uþ 7 ¼ 0 or u� 4 ¼ 0) u ¼ �7 or u ¼ 4

[ The solution to u2 þ 3u � 28 ¼ 0 is u ¼ �7 or u ¼ 4.c a2 � 2a ¼ 0

This requires a simpler factorisation as thereare only two terms, both involving a.a(a � 2) ¼ 0) a ¼ 0 or a� 2 ¼ 0) a ¼ 0 or a ¼ 2[ The solution to a2 � 2a ¼ 0 is a ¼ 0 or a ¼ 2.

d p2 ¼ 5p þ 24p2 � 5p � 24 ¼ 0 Moving all terms to the LHS

and making the RHS ¼ 0(p � 8)(p þ 3) ¼ 0) p� 8 ¼ 0 or pþ 3 ¼ 0) p ¼ 8 or p ¼ �3

[ The solution to p2 ¼ 5p þ 24 is p ¼ 8 or p ¼ �3.

Note: Quadratic equations of the form ax2 þ bx þ c ¼ 0 will be met in Chapter 11, Quadraticequations and the parabola.

Video tutorial

Simple quadraticequations

MAT10NAVT10028

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Exercise 7-02 Quadratic equations x2 þ bx þ c ¼ 01 Solve each quadratic equation.

a m2 ¼ 144 b x2 ¼ 400 c y2 ¼ 225 d k2 � 169 ¼ 0e y2 � 1 ¼ 0 f w2 � 16 ¼ 0 g x2 þ 10 ¼ 14 h t2 � 9 ¼ 7

i a2

2¼ 8 j 5k2 ¼ 180 k 3w2 ¼ 300 l d2 þ 60 ¼ 204

m k2

2¼ 0:5 n w

2

10¼ 2:5 o 5y2 ¼ 5 p m

2

4¼ 9

q 4x2 ¼ 1 r 2p2 þ 3 ¼ 21 s 3k2

4þ 5 ¼ 8 t y

2

5� 2 ¼ 18

2 Solve each quadratic equation, writing the solution in exact (surd) form where necessary.

a 5m2 � 20 ¼ 0 b 4a2

9¼ 36 c m2 ¼ 28

d 9k2 þ 10 ¼ 13 e k2

16¼ 6 f 3k

2

10¼ 27

3 Solve each quadratic equation, writing the solution correct to two decimal places where necessary.

a 9m2 � 2 ¼ 32 b 2x2

5¼ 23 c 6y2 ¼ 0.726

d 2w2

5¼ 20 e 3a2 þ 11 ¼ 267 f 2y2 � 14 ¼ 63

4 Solve each quadratic equation.a x2 þ 3x þ 2 ¼ 0 b y2 þ 5y þ 4 ¼ 0 c y2 þ 16y þ 48 ¼ 0d x2 þ x � 12 ¼ 0 e x2 þ 2x � 3 ¼ 0 f x2 þ 3x � 40 ¼ 0

5 Solve each quadratic equation.a x2 � x � 30 ¼ 0 b x2 � 8x þ16 ¼ 0 c x2 � 5x � 66 ¼ 0d d2 � 2d ¼ 0 e x2 � 3x � 10 ¼ 0 f n2 þ 4n ¼ 0g k2 � 7k ¼ 0 h y2 ¼ 5y i v2 ¼ 12vj m2 ¼ 3m k a2 þ 24a ¼ �80 l n2 ¼ �10nm u2 þ 2u ¼ 8 n x2 ¼ x þ 42 o p(p þ 9) ¼ �20

6 Explain why the quadratic equation x2 ¼ �25 has no solutions.7 State which of these quadratic equations have no solutions. Give reasons.

a x2 ¼ �9 b 2k2 þ 5 ¼ 9 c 3m2 þ 8 ¼ 4

d 9w2

2� 1 ¼ 1 e 4þ d

2

3¼ 8 f 5a

2

2þ 3 ¼ 2

7-03 Simple cubic equations ax3 ¼ cAn equation in which the highest power of the variable is 3 is called a cubic equation, for example,x3 ¼ 12, 2m3 þ 1 ¼ 25, d3 � 14 ¼ 4 and x3 � 3x2 þ 5x þ 4 ¼ 0.

Summary

The cubic equation x3 ¼ c has one solution: x ¼ffiffiffic3p

See Example 3

See Example 4

See Example 5

See Example 6

See Example 7

Stage 5.3

NSW

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Example 8

Solve each cubic equation.

a y3 ¼ 64 b p3 ¼ 50 c �2x3 ¼ 2000

Solutiona y 3 ¼ 64

y ¼ffiffiffiffiffi643p

¼ 4Finding the cube root of both sides.

b p 3 ¼ 50p ¼

ffiffiffiffiffi503p

50 is not a cube number so leave the answer as a surd.

c �2x 3 ¼ 2000

x 3 ¼ 2000�2¼ �1000

x ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffi�10003p

¼ �10

Dividing both sides by (�2).

Example 9

Solve each cubic equation, writing the solution correct to one decimal place.

a 11x3 � 102 ¼ 0 b 2y3

7¼ �11

Solutiona 11x 3 � 102 ¼ 0

11x 3 ¼ 102

x 3 ¼ 10211

¼ 9:272 . . .x ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi9:272:::3p

¼ 2:1008 . . .� 2:1

b 2y 3

7¼ �11

2y 3 ¼ �11 3 7¼ �77y 3

¼ �772

¼ �38:5y ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffi�38:53p

¼ �3:3766 . . .� �3:4

Stage 5.3

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Exercise 7-03 Simple cubic equations ax3 ¼ c1 Solve each cubic equation, writing the solution in exact form where necessary.

a x3 ¼ 1 b m3 ¼ 125 c a3 ¼ 1331d u3 ¼ �8 e y3 ¼ �729 f n3 ¼ 20g h3 ¼ 11 h k3 ¼ �48 i 5m3 ¼ �75j 7m3 ¼ 448 k �4x3 ¼ 81 l 12x3 ¼ �480

2 Solve each cubic equation, writing the solution correct to one decimal place.a w3 � 16 ¼ 0 b m3 þ 6 ¼ 22 c 5m3 � 1080 ¼ 0

d 3t3 � 10 ¼ 87 e x3

3¼ 9 f 5x

3

7¼ �120

g 3x3

4¼ �10 h 2x

3

5¼ 0:2048 i 7a

3

9� 10 ¼ 121

j a3 � 0.064 ¼ 0 k � 7x3

9¼ 10 l 5t3 þ 46 ¼ �370

3 a Does a cubic equation of the form ax3 ¼ c always have a solution?b When is the solution to x3 ¼ c positive?c When is the solution to x3 ¼ c negative?d Can x3 ¼ c have two solutions?

7-04 Equation problems

Example 10

At a concert, an adult’s ticket costs $5 more than twice the cost of a child’s ticket. The totalcost for 3 adults and 7 children is $327. Find the cost of a child’s ticket and an adult’s ticket.

SolutionLet the cost of a child’s ticket be $c. Using a variable to represent an

[ Cost of an adult’s ticket ¼ $(2c þ 5) unknown quantity.

3ð2cþ 5Þ þ 7c ¼ 3276cþ 15þ 7c ¼ 327

13cþ 15 ¼ 32713c ¼ 312

c ¼ 24

Forming an equation.

Solving the equation.

A child’s ticket costs $24.) Cost of an adult’s ticket ¼ 2 3 $24þ 5

¼ $53[ A child’s ticket costs $24 and an adult’s ticket costs $53.Check: 3 3 $53 þ 7 3 $24 ¼ $327.

Stage 5.3

See Example 8

See Example 9

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Example 11

Jennifer is 7 years older than Amy. Ten years from now, the sum of their ages will be 43.How old are they now?

SolutionLet x ¼ Amy’s age now.[ Jennifer’s age now ¼ x þ 7.

Now In 10 years’ timeAmy x x þ 10Jennifer x þ 7 x þ 7 þ 10 ¼ x þ 17In 10 years’ time:

ðxþ 10Þ þ ðxþ 17Þ ¼ 432xþ 27 ¼ 43

2x ¼ 16x ¼ 8 Amy is 8 now.

Jennifer’s age now ¼ 8þ 7¼ 15

Amy is 8 years old now and Jennifer is 15 years old now.[ Check: In 10 years’ time, the sum of their ages will be 18 þ 25 ¼ 43.

Exercise 7-04 Equation problemsFor each question, write an equation and solve it to answer the problem.1 A rectangle is four times as long as it is wide. The perimeter of the rectangle is 250 cm. Find

the dimensions of the rectangle.

2 The equal sides of an isosceles triangle are twice as long as the other side. The perimeter of thetriangle is 90 mm. Find the lengths of the sides of the triangle.

3 At the football match, an adult’s ticket costs $6 more than twice the cost of a child’s ticket. Thetotal cost for 3 adults and 5 children is $249. Find the cost of a child’s ticket and an adult’s ticket.

4 The sum of three consecutive numbers is 186. Find the numbers.

5 The sum of three consecutive even numbers is 288. Find the numbers.

6 Sanjay is nine times the age of his son, Anand. In 5 years he will be four times the age ofAnand. How old are they now?

7 When 15 is subtracted from three times a certain number, the answer is 63. What is the number?

8 The product of 2 and a number is the same as 12 subtract the number. Find the number.

9 The sum of the present ages of Vatha and Chris is 36. In 4 years time, the sum of their ageswill equal twice Vatha’s present age. How old are they now?

10 Four consecutive numbers have a sum of 858. Find the numbers.

11 Find x.(2x + 45)°

5(x – 12)°

See Example 10

See Example 11

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12 Manori’s bag has 10-cent and 20-cent coins. She has 202 coins with a total value of $31.90.How many 20-cent coins does Manori have?

13 If 17 more than a number is 5 more than three times the number, what is the number?

14 If the perimeter of this parallelogram is 130, find x. 3(x + 2)

x – 3

15 The sum of Scott’s age and his mother’s age is 45. In 5 years’ time, three times Scott’s age less9 will be the same as his mother’s age. Find the present ages of Scott and his mother.

16 One angle in a triangle is double the smallest angle, and the third angle in the triangle is5 more than four times the smallest angle. Find the size of each angle.

17 A large container of water is 78

full. After 15 L has been taken out, the container is 23

full.

When full, how many litres does the container hold?

18 The total cost of a school camp for Year 10 students was $21 280. Each teacher paid $185 toattend and each student paid $165. There was one teacher for every 15 students. Find thenumbers of teachers and students that attended the camp.

Mental skills 7 Maths without calculators

Multiplying decimals

1 Study each example.

a 3 × 8 = 24, so 3 × 0.8 = 2.4

0 dp + 1 dp = 1 dp (dp = decimal places)

The number of decimal places in the answer is equal to the total number of decimal placesin the question. Also, the answer sounds reasonable because, by estimation:3 3 0.8 � 3 3 1 ¼ 3 (2.4 � 3)

b 6 × 5 = 30, so 0.6 × 0.5 = 0.30 = 0.3

1 dp + 1 dp = 2 dp

By estimation, 0:6 3 0:5 � 0:5 3 0:5 ¼ 12

3 12¼ 1

4¼ 0:25 (0.3 � 0.25)

c 7 × 3 = 21, so 0.07 × 0.3 = 0.021

2 dp + 1 dp = 3 dp

By estimation, 0:07 3 0:3 � 0:07 3 13� 0:02 (0.021 � 0.02)

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7-05 Equations and formulasA formula is an equation that describes a relationship between variables. For example, the formulafor the perimeter of a rectangle is P ¼ 2(l þ w), where P is the subject of the formula and appearson the LHS of the ‘¼’ sign.

Example 12

The cost, $C, of a taxi trip is C ¼ 5 þ 2.4d, where d is the distance travelled in kilometres.a Find the cost of a taxi trip if the distance travelled is 15 km.b Find the distance travelled by the taxi if the cost of the trip was $78.20.

Solutiona When d ¼ 15:

C ¼ 5þ 2:4 3 15¼ 41

The cost was $41.

b When C ¼ 78.20:78:20 ¼ 5þ 2:4d73:20 ¼ 2:4d

d ¼ 73:202:4

¼ 30:5The distance travelled was 30.5 km.

2 Now evaluate each product.a 0.7 3 5 b 12 3 0.2 c 0.4 3 0.3 d (0.6)2

e 8 3 0.1 f 0.03 3 0.9 g 4 3 0.05 h 1.1 3 8i 0.3 3 0.8 j 0.2 3 0.06 k 9 3 0.2 l 0.07 3 0.4

3 Study each example.Given that 15 3 23 ¼ 345, evaluate each product.a 1.5 × 2.3 = 3.45

(Estimate: 1.5 × 2.3 ≈ 2 × 2 = 4)1 dp + 1 dp = 2 dp

b 150 × 0.23 = 15 × 10 × 0.23 = 15 × 0.23 × 10 = 3.45 × 10 = 34.5

15

(Estimate: 150 × 0.23 ≈ 150 × 0.2 = 150 × = 30)0 dp + 2 dp = 2 dp

c 0.15 × 2300 = 0.15 × 23 × 100 = 3.45 × 100 = 345

15(Estimate: 0.15 × 2300 ≈ 0.2 × 2300 = × 2300 = 460)

2 dp + 0 dp = 2 dp

4 Now given that 39 3 17 ¼ 663, evaluate each product.a 3.9 3 17 b 39 3 170 c 39 3 0.17 d 0.39 3 1.7e 3.9 3 1.7 f 390 3 1.7 g 3.9 3 0.17 h 3.9 3 170i 3900 3 1.7 j 39 3 1.7 k 39 3 0.017 l 0.39 3 0.17

Puzzle sheet

Getting the rightformula

MAT10NAPS10044

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Example 13

The surface area of a sphere is SA ¼ 4pr2, where r is the radius. Find, correct to one decimalplace, the radius of a sphere with surface area 40 m2.

SolutionWhen r ¼ 40:

40 ¼ 4pr 2

10 ¼ pr 2

r 2 ¼ 10p

¼ 3:183 . . .r ¼

ffiffiffiffiffiffiffiffiffiffiffi3:183p

¼ 1:784 . . .� 1:8 m

The radius of the sphere is 1.8 m.

r is positive

Exercise 7-05 Equations and formulas1 The formula for the circumference of a circle is C ¼ 2pr, where r is the radius. Find, correct to

one decimal place:a the circumference of a circle with radius 2.4 mb the radius of a circle whose circumference is 200 cm

2 The perimeter of a rectangle is P ¼ 2(l þ w). Find the width of a rectangle whose perimeter is58 m and length is 12 m.

3 The formula for converting speed expressed in m/s to a speed expressed as km/h is k ¼ 3.6M,where M is the speed in m/s. Calculate in m/s the speed of a car travelling at 110 km/h.

4 Use the formula from question 3 to convert each speed to km/h.a 10 m/s b 24 m/s c 50 m/s

5 The average of m and n is A ¼ mþ n2

. If two numbers have an average of 28 and one of themis 13, find the other number.

6 The formula for converting a temperature recorded in �F to a temperature in �C isC ¼ 5

9ðF � 32Þ. Convert each temperature to �C, correct to the nearest degree.

a 80�F b 32�F c 212�F d 102�F7 The body mass index (BMI) of an adult is B ¼ M

h 2, where M is the mass in kilograms and h is

the height in metres. Find, correct to one decimal place:a the BMI of Dean who is 1.85 m tall and has a mass of 72 kgb the mass of a person with a BMI of 24, who is 2.1 m tall.

8 The volume of a sphere is V ¼ 43

pr 3, where r is the radius. Find, correct to one decimal place,

the radius of a sphere with a volume of 500 m3.

See Example 12

See Example 13

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9 The average speed in km/h of a car is given by the formula S ¼ DT

, where D is the distancecovered in kilometres and T is the time taken in hours. Find, correct to the nearest wholenumber:a the distance travelled, if a car maintains a speed of 87.2 km/h for 5 hoursb the time taken, if a distance of 650 km is covered at a speed of 91 km/h

10 The cost, $C, of hiring a car is C ¼ 45 þ 0.15d, where d is the number of kilometres travelled.Calculate:a the cost of hiring a car to travel 350 kmb the distance travelled, if the cost is $138.

11 The surface area of a cylinder is given by the formula SA ¼ 2pr2 þ 2prh. Calculate, correct toone decimal place, the height of a cylinder with surface area 1255.38 cm2 and radius 9 cm.

7-06 Changing the subject of a formula

Example 14

Change the subject of the formula:

a A ¼ 12

bh to h b v2 ¼ u2 þ 2as to s c aþ 2aþ 10 ¼ k to a

Solutiona A ¼ 1

2bh

12

bh ¼ A

bh ¼ 2A

h ¼ 2Ab

Swapping sides so that h appears on the LHS.

Multiplying both sides by 2.

Dividing both sides by b.

b v 2 ¼ u 2 þ 2asu 2 þ 2as ¼ v 2

2as ¼ v 2 � u 2

s ¼ v2 � u 2

2a

Swapping sides so that s appears on the LHS.

Subtracting u2 from both sides.

Dividing both sides by 2a.

c aþ 2aþ 10 ¼ k

aþ 2 ¼ kðaþ 10Þ¼ ak þ 10k

a� ak ¼ 10k � 2að1� kÞ ¼ 10k � 2

a ¼ 10k � 21� k

Multiplying both sides by a þ 10.ExpandingMoving the a-terms to the LHS, the 2 to the RHS.

Factorising a from the LHS.

Dividing both sides by 1 � k.

Stage 5.3

NSW

Video tutorial

Changing the subjectof a formula

MAT10NAVT10005

Worksheet

Changing the subjectof a formula

MAT10NAWK10211

2619780170194662


Exercise 7-06 Changing the subject of a formula1 Make y the subject of each formula.

a x þ 2y ¼ 5 b m þ py ¼ k c P � ky ¼ 8

d m3¼ y

5e D ¼ K � My f 5þ 8y

d¼ 4

g ay� k2¼ c h yþ 3

5¼ 4m

3i xy2 þ 5 ¼ w

j x ¼ffiffiffiyk

rk n ¼ d

5� y l T ¼ffiffiffiffiffiffiffiffiffiffiffiyþ k

c

r

2 Change the subject of each formula to the variable indicated in brackets.

a a2 þ b2 ¼ c2 [b] b s ¼ ut þ 12

at 2 [a] c v ¼ u þ at [a]

d V ¼ 43pr 3 [r] e A ¼ p(R2 � r2) [R] f A ¼ prl þ pr2 [l]

g S ¼ 180(n � 2) [n] h 1xþ 1

r¼ 1

s[r] i x ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib 2 � 4acp

[b]

j x þ y ¼ 5 � 3x [x] k m ¼ 5A2Aþ n [A] l S ¼

aðp� 1Þp

[p]

m X(a þ b) ¼ Y(a � b) [a] n 5þ x3xþ a ¼ 2 [x] o y ¼

uþ bxuþ ab [b]

Stage 5.3

Investigation: Restricting values of variables

1 Consider the formula x2 þ y2 ¼ 4.a Explain why the least value that x can take is �2 and the largest value that x can take is 2.b Does the same restriction apply to the values that the variable y can take? Explain why.c By making y the subject, show that y ¼ �

ffiffiffiffiffiffiffiffiffiffiffiffiffi4� x2p

.d Are the values that x and y can take in y ¼ �

ffiffiffiffiffiffiffiffiffiffiffiffiffi4� x2p

different from the values thatthey can take in x2 þ y2 ¼ 4? Give reasons.

2 a If Z ¼ ax2, what range of values can Z, a and x take?

b If x ¼ffiffiffiZa

r, what range of values can Z, a and x take?

3 a In the formula A ¼ pr2, explain why there are no restrictions on r but A � 0.b Make r the subject of the formula given that the formula is for the area of a circle.

Have the restrictions on the variables r and A changed?4 a What are the restrictions on the variables x and y in the formula y ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi16� x2p

?b Change the subject of the formula to x. Are the restrictions on the variables the same

as for part a? Explain.5 Consider the formula y ¼ 1

x� 3.a What are the restrictions on the variables x and y? Give reasons.b Make x the subject of the formula. Are the restrictions on the variables the same as in

part a or different? Explain.c Compare your answers to the above questions with those of other students in

your class.

See Example 14

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7-07 Graphing inequalities on a number lineAn inequality looks like an equation except that the equals sign (¼) is replaced by an inequalitysymbol >, �, < or �.2x � 7 ¼ 15 is an equation. There is only one value of x that makes it true.2x � 7 � 15 is an inequality. There is a range of values of x that make it true.

Summary

> means ‘is greater than’ � means ‘is greater than or equal to’< means ‘is less than’ � means ‘is less than or equal to’

The inequality x � 3 is read ‘x is greater than or equal to 3’ and includes 3 and all the numbersabove 3, such as 3.01, 4, 10, 20 000, etc.The inequality x > 3 is read ‘x is greater than 3’ and means all the numbers above 3, but not 3.

Inequality In words Meaningx > 3 x is greater than 3 Values above 3x < 3 x is less than 3 Values below 3x � 3 x is greater than or equal to 3 Values above and including 3x � 3 x is less than or equal to 3 Values below and including 3

For convenience, we can represent all the values in an inequality using a number line.

Example 15

Graph each inequality on a number line.

a x � 1 b x < 5 c x > �3

Solutiona x � 1 means that x can be any number greater than 1 or equal to 1.

–3 –2 –1 0 1 2 3 4 5 6x

b x < 5 means that x can be any number less than 5, but not including 5.

–3 –2 –1 0 1 2 3 4 5 6x

c x > �3 means that x can be any number greater than �3, but not including �3.

–3 –2 –1 0 1 2 3 4 5 6x

Worksheet

Graphing inequalities

MAT10NAWK10045

The filled circle at 1means we include 1.

The open circle on 5 meansthat 5 is not included.

2639780170194662


Exercise 7-07 Graphing inequalities on a number line1 Graph each inequality on a separate number line.

a x � 2 b x < �3 c x � 1 d x > 7e x � 4 f x > 0 g x � �2 h x < 10

2 Write the inequality illustrated by each number line.

–2

a b

c d

0 2 4 6x

0 1 2 3 4 5 6x

–8–10 –6 –4 –2 0x

–8–10 –6 –4 –2 0 2x

3 Which inequality is graphed below? Select the correct answer A, B, C or D.

–7 –6 –5 –4 –3 –2 –1 0 1 2x

A x > �2.5 B x < �2.5 C x < �3.5 D x > �3.54 Write the inequality represented on each number line.

a –3 –2 –1 0 1 2 3 x b –1 0 1 2 3 4 5 x

c 0 2 4 6 8 10 12 x d –3 –2 –1 0 1 2 3 x

e –9 –6 –3 0 3 6 9 x f –3 –2 –1 0 1 2 3 x

g –10 –8 –6 –4 –2 0 2 x h –5 0 5 10 15 20 25 x

i –1 0 1 2 3 4 5 x

Investigation: The language of inequalities

Work in pairs to complete this activity.Use inequality symbols to write each statement algebraically.a The minimum height (H) for rides at an amusement park is 1.3 m.b The speed limit in a school zone is 40 km/h.c To be eligible to vote, you must be at least 18 years old (A ¼ age).d The overseas tour is only for people whose age (A) is from 18 to 35.e The cost (A) of a tennis racquet will be at least $95 but no more than $360.f A new flute (F) costs at least $475.g The price of units (U) in a new block start at $240 000.

See Example 15

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Investigation: Solving inequalities

We have solved equations by doing the same thing to both sides (keeping the equation‘balanced’). Will this method work with inequalities, such as x þ 4 > 10 or 6x < 13?1 Start with an inequality that is true, such as 7 > 4.2 Add 5 (or any number you choose) to both sides of the inequality; for example 7 > 4

becomes 12 > 9. Is the new inequality true or false?3 Subtract 9 (or any number you choose) from each side of the original inequality; for

example 7 > 4 becomes �2 > �5. Is the new inequality true or false?4 Multiply both sides of the original inequality by 4 (or any positive number you choose);

for example 7 > 4 becomes 28 > 16. Is the new inequality true or false?5 Divide both sides of the original inequality by 2 (or any positive number you choose);

for example 7 > 4 becomes 312> 2. Is the new inequality true or false?

6 Multiply both sides of the original inequality by �3 (or any negative number you choose);for example 7 > 4 becomes �21 > �12. Is the new inequality true or false?

7 Divide both sides of the original inequality by �4 (or any negative number you choose),for example 7 > 4 becomes �13

4> �1. Is the new inequality true or false?

8 Which of the six operations used in questions 2 to 7 can be used on inequalities to givea true result?

9 Which of the six operations used in questions 2 to 7 cannot be used with inequalitiesbecause they give a false result?

10 Copy and complete the following inequality statements.a 6 < 8

6 3 3 < 8 3 ___ (multiplying both sides by 3)[ 18 __ 24

b 10 > �410 4 2 __ �4 4 __ (dividing both sides by 2)[ __________

Does the inequality sign (< or >) stay the same when multiplying or dividing by apositive number?

11 a Is it true that 5 < 8?b Multiply both sides by �2. Is it true that �10 < �16?c What needs to be reversed to change �10 < �16 into a true inequality statement?d Copy and complete the following to make a true inequality statement: �10 ______ �16.

12 a Is it true that 18 > �6?b Divide both sides by �3. Is it true that �6 > 2?c What needs to be reversed to change �6 > 2 into a true inequality statement?d Copy and complete the following to make a true inequality statement: �6 ____ 2.

13 Copy and complete:When multiplying or d__________ both sides of an inequality by a n__________number, the inequality sign must be r__________.

2659780170194662


7-08 Solving inequalities

Example 16

Solve each inequality and graph its solution on a number line.

a 2x � 10 � 16 b 2(y � 1) � 12 c wþ 32

> �1

Solutiona 2x� 10 � 16

2x� 10þ 10 � 16þ 102x � 262x2� 26

2x � 13

10 11 12 13 14 15x

b 2ðy� 1Þ � 122y� 2 � 12

2y� 2þ 2 � 12þ 22y � 142y2� 14

2y � 7

0 1 2 3 4 5 6 7 8 10 119y

c wþ 32

> �1wþ 3

23 2 > �1 3 2

wþ 3 > �2wþ 3� 3 > �2�3

w > �5

–6 –5 –4 –3 –2 –1 0 1w

Summary

Inequalities can be solved algebraically in the same way as equations, by using inverseoperations. However, when multiplying or dividing both sides of an inequality by a negativenumber, you must reverse the inequality sign.

Worksheet

Inequalities review

MAT10NAWK10046

266 9780170194662

Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16


Example 17

Solve each inequality.

a 1 � 2x � �11 b 4 � r < 7 c aþ 5�3 > 4

Solutiona 1� 2x � �11

1� 2x�1 � �11�1�2x � �12�2x�2 �

�12�2

x � 6

Dividing both sides by a negative numberreverses the inequality sign.

b 4� r < 74� r � 4 < 7� 4

�r < 3�r�1 >

3�1

r > �3

Dividing both sides by a negative numberreverses the inequality sign.

c aþ 5�3 > 4

aþ 5�3 3 �3ð Þ < 4 3 �3ð Þ

aþ 5 < �12aþ 5�5 < �12�5

a < �17

Multiplying both sides by a negativenumber reverses the inequality sign.

Exercise 7-08 Solving inequalities1 Solve each inequality and graph its solution on a number line.

a x � 1 > 6 b 3y � 12 c m þ 4 � 2d x

5� �20 e 12x < 60 f 5y > �20

g 4a � 2 h 3w � �30 i 8a þ 5 � 45j 3a þ 1 � 10 k 6a þ 4 � �2 l 3w � 3 < �12

2 Solve each inequality.a 3(x þ 2) � 9 b 5(m � 4) � 10 c 2(y þ 5) � �6

d x� 12� 2 e w� 2

5> �1 f 2aþ 1

3< 3

g 2ðmþ 1Þ3

� 3 h 4ðm� 2Þ3

� �6 i 3þ x5< 10

j 3 þ 2x < 9 þ x k 11 � 5y � 9 � 6y l 2(3 þ 5a) � 5(4 þ a)

See Example 16

2679780170194662


3 What is the solution to 3ðx� 2Þ5

� �1 ? Select the correct answer A, B, C or D.

A x � 0 B x � �1 C x � 13

D x � � 113

4 Solve each inequality and graph its solution on a number line.a 5 � x � 2 b 15 > 7 � y c 1 � k < 12d 7 � m � 7 e 2 � p > 8 f �t þ 6 � 10

5 Solve each inequality.a �2x < 6 b k�3 � 4 c �5t >12

d �x3� �4 e 4 � 3w > 7 f �4y þ 3 � 11

g 8 � 5a < 3 h �2d � 3 > 8 i 5þ w�3 > 2

j �pþ 2�3 < �2 k 1 � 3m < 9 � 5m l 3(3x þ 4) � 6(1 � 2x)

Just for the record Film and game classificationIn Australia, films, publications and computer games are rated by the Classification Board.Films and videos are rated G, PG, M, MA15þ or R18þ, with each category containing a list ofguidelines related to the film’s use of violence, coarse language, adult themes, sex and nudity.

General (G) means suitable for all ages. Children can watch filmsclassified G without adult supervision.

Parental guidance (PG) means that parental guidance is recommendedfor persons under 15 years of age. These films contain material that maybe confusing or upsetting to children, but not harmful or disturbing.Parents should watch the film with their children or preview it to checkelements such as language used or inappropriate themes.

Mature (M) means recommended for mature audiences, 15 years andover. The film or computer game may contain material that is harmfulor disturbing to children, but the impact is not so strong as to requirerestriction.

Mature accompanied (MA15þ) means legal restrictions apply to personsunder the age of 15. Children are not allowed to see MA15þ films unlessaccompanied by a parent or guardian, because they contain materialthat is likely to be harmful or disturbing to them.

Restricted (R18þ) means legally restricted to adults, 18 years and over.It applies to films that deal with issues and scenes that require anadult perspective, and so are unsuitable for persons under 18 yearsof age. A person will be asked for proof of age before buying, hiringor viewing films or computer games in this category.

1 Each of the classifications is represented by a logo (as shown) with the letter insidea particular shape. What shape is each logo?

2 Write each classification category as an inequality.

See Example 17

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7-09 LogarithmsThe logarithm of a number is the power of the number, to a given positive base.For example, the logarithm of 256 to the base 2 is 8, written log2 256 ¼ 8, because 28 ¼ 256.

Example 18

Evaluate each expression.

a log3 81 b log4 16 c log10 10 000

Solutiona log3 81 means 3? ¼ 81 b log4 16 means 4? ¼ 16

Since 34 ¼ 81then log3 81 ¼ 4.

Since 42 ¼ 16then log4 16 ¼ 2.

c log10 10 000 means 10? ¼ 10 000Since 104 ¼ 10 000then log10 10 000 ¼ 4.

Summary

If y ¼ ax, then loga y ¼ xwhere a is the base, a > 0, x is the power, and y > 0.

Since a > 0, ax > 0 and y > 0.Logarithms are only meaningful for positive numbers, y.

Investigation: Power tables

1 Copy and complete this table of powers of 2 from 0 to 20.

x 0 1 2 … 202x

2 Use the table to calculate 32 3 128. Explain the method you used.3 Use the table to calculate:

a 16 3 1024 b 128 3 2048 c 256 3 64 d 4096 3 324 Use the table to calculate 262 114 4 8192. Explain the method you used.5 Use the table to calculate:

a 16 384 4 512 b 128 4 8 c 8192 4 1024 d 1 048 576 4 65 536When powers are used this way in calculations, they may be called logarithms.

Stage 5.3

Puzzle sheet

Logarithms 1

MAT10NAPS00059

Puzzle sheet

Logarithms 2

MAT10NAPS00060

‘3 to the power of what equals 81?’ ‘4 to the power of what equals 16?’

2699780170194662


Example 19

Write each expression as a logarithm.

a 243 ¼ 35 b 0.01 ¼ 10�2 c 2 ¼ 813 d p ¼ qr

Solutiona 243 ¼ 35

[ log3 243 ¼ 5b 0.01 ¼ 10�2

[ log10 0.01 ¼ �2

c 2 ¼ 813

) log8 2 ¼13

d p ¼ qr[ logq p ¼ r

Example 20

Rewrite logn m ¼ x in index form.

Solutionlogn m ¼ x[ m ¼ nx n is the base, x is the power.

Exercise 7-09 Logarithms1 Evaluate each expression.

a log5 25 b log2 8 c log7 49 d log2 16e log3 243 f log10 1000 g log5 125 h log6 36i log2 64 j log3 6561 k log10 1000 000 l log8 512

2 Write each expression in logarithmic form.a 52 ¼ 25 b 43 ¼ 64 c 10 000 ¼ 104 d 2512 ¼ 5

e 116¼ 2�4 f 3�2 ¼ 1

9g 823 ¼ 4 h 0.01 ¼ 10�2

iffiffiffi2p¼ 414 j 1612 ¼ 4 k 932 ¼ 27 l 1ffiffiffi

6p ¼ 6�12

3 Write each expression in index form.

a log5 125 ¼ 3 b log10 10 ¼ 1 c log ffiffi3p 27 ¼ 6 d log2 8 ffiffiffi2p ¼ 3:5e log2 64 ¼ 6 f log3

181¼ �4 g log5

1125¼ �3 h log8

ffiffiffi2p¼ 1

6i log100 10 ¼

12

j log5 5ffiffiffi5p¼ 3

2k log8 2 ¼

13

l log1001

100¼ �1

4 Why can’t you find the logarithm of a negative number or zero?

Stage 5.3

See Example 18

See Example 19

See Example 20

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7-10 Logarithm lawsThe index laws from Chapter 5, Products and factors, are related to the logarithm laws.

Summary

The logarithm of a product is equal to the sum of the logarithm of each factor.

loga (xy) ¼ loga x þ loga y

For example, log2 (8 3 4) ¼ log2 8 þ log2 4.This law corresponds to the index law am 3 an ¼ amn.Proof:Let m ¼ loga x and n ¼ loga y.[ x ¼ am and y ¼ an[ xy ¼ am 3 an ¼ am þ n

) logaðxyÞ ¼ mþ n¼ loga xþ loga y

Summary

The logarithm of a quotient is equal to the difference between the logarithm of each term.

logax

y

� �¼ loga x� loga y

For example, log324327

� �¼ log3 243� log3 27

This law corresponds to the index law am 4 an ¼ am�n.Proof:Let m ¼ loga x and n ¼ loga y.[ x ¼ am and y ¼ an

) xy¼ a

m

an¼ am�n

) logax

y

� �¼ m� n

¼ loga x� loga y

Summary

The logarithm of a term raised to a power is equal to the power multiplied by the logarithmof the term.

loga xn ¼ n loga x

Stage 5.3

2719780170194662


For example, log4 82 ¼ 2 log4 8

This law corresponds to the index law (am)n ¼ amn.Proof:Let m ¼ loga x[ x ¼ am

) xn ¼ ðamÞn

¼ amn

) loga xn ¼ mn¼ loga x 3 n¼ n loga x

Summary

Properties of logarithmsloga a

x ¼ x loga 1 ¼ 0, because a0 ¼ 1loga a ¼ 1, because a1 ¼ a loga

1x

� �¼ � loga x

Proof:

loga1x

� �¼ log x�1

¼ � loga x Using the law loga xn ¼ n logax

Example 21

Evaluate each expression.

a log5 0.04 b log2 5 � log2 10 c 2 log3 6 � log3 4 d log5 10 þ log5 2 � log5 4

Solutiona log5 0:04 ¼ log5

4100

� �

¼ log5125

� �

¼ log51

5 2

� �

¼ log5 5�2

¼ �2

b log2 5� log2 10 ¼ log2510

� �

¼ log212

� �

¼ log2 2�1

¼ �1

c 2 log3 6� log3 4 ¼ log3 6 2 � log3 4¼ log3 36� log3 4

¼ log3364

� �

¼ log3 9¼ 2

d log5 10þ log5 2� log5 4 ¼ log5ð10 3 2Þ � log5 4¼ log5 20� log5 4

¼ log5204

� �

¼ log5 5¼ 1

Stage 5.3

Video tutorial

Logarithm laws

MAT10NAVT10001

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Example 22

Simplify each expression.

a 6 loga a þ loga a4 � loga a9 b log2 x þ log2 w � 2 log2 y clog3 a

3

5 log3 a

Solutiona 6 loga aþ loga a 4 � loga a 9 ¼ 6 3 1þ 4� 9

¼ 1b log2 xþ log2 w� 2 log2 y ¼ log2ðxwÞ � log2 y 2

¼ log2xw

y 2

� �

c log3 a 3

5 log3 a¼ 3 log3 a

5 log3 a

¼ 35

Example 23

Given log10 7 � 0.8451, find the value of each expression.

a log10 49 b log10 700 c log10 (0.07)

Solutiona log10 49 ¼ log10 7 2

¼ 2 log10 7� 2 3 0:8451¼ 1:6902

b log10 700 ¼ log10 ð7 3 100Þ¼ log10 7þ log10 100� 0:8451þ 2¼ 2:8451

c log10ð0:07Þ ¼ log107

100

� �

¼ log10 7� log10 100� 0:8451� 2¼ �1:1549

Exercise 7-10 Logarithm laws1 Evaluate each expression.

a log2 128 b log10 1000 c log8 64 d log515

e log2ffiffiffi2p

f log319

g log10 0.0001 h log2116

i log8 2 þ log8 4 j log4 32 � log4 2 k log10 4 þ log10 25l log5 200 � log5 8 m log2 18 � 2 log2 3 n 3 log10 2 þ log10 12.5

Stage 5.3

See Example 21

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2 Simplify each expression.a logx 5 þ logx 6 b logx 10 � logx 2 c 3 logx 2

d 2 logx 4 � logx 8 e logx 10 þ logx 4 f 12 logx 100

g �logx 4 h logx 8 � (logx 10 þ logx 4) i 12 ðlogx 8þ logx 18Þ

3 If log10 4 ¼ 0.6021, find the value of each expression.a log10 16 b log10 400 c log10 4000 d log10 2e log10 0.4 f log10 160 g log10 2.5 h log10

ffiffiffiffiffi40p

4 Evaluate each expression.a log3 4 þ log3 15 � log3 20 b log3 270 � (log3 2 þ log3 5)

c log4 20 þ (log4 32 � log4 10) d 2 log10 25 � log10 6.25

e 2 log10 2 � (log10 5 þ log10 8) f log100 50 � log100 5

g 2 log5 10 þ (log5 50 � log5 40) h 5 log8 2þ12

log8 4

i 12

log4 25� 2 log4ffiffiffiffiffi20p

j 13

log2 125� 3 log2ffiffiffiffiffi803p

5 Simplify each expression.a loga a2 þ 3 loga a b loga a 3ð Þ c 5 loga a � loga a4

d loga x7ð Þ

loga xe loga y3 � 3 loga y f loga

ffiffiffixp � loga

1x

7-11 Exponential and logarithmic equationsExponential equations are equations like 3 x ¼ 243, where the variable is a power.Logarithms can be used to solve exponential equations rather than using a ‘guess-and-check’ method.The log key on your calculator can be used to evaluate log10 x, that is, logarithms to the base 10.

Example 24

Solve each exponential equation.

a 3 x ¼ 243 b 4mþ1 ¼ 18ffiffiffi2p

Solutiona 3 x ¼ 243

log10 3x ¼ log10 243

x log10 3 ¼ log10 243

x ¼ log10 243log10 3

¼ 5

Taking log10 of both sides.

Enter on a calculator:log 243 ÷ log 3 =

Note: The log key means log10, and forconvenience we will write log to mean log10.

Stage 5.3

See Example 22

See Example 23

Worksheet

Logarithms review

MAT10NAWK10212

Puzzle sheet

Exponential equations

MAT10NAPS00040

274 9780170194662

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b 4mþ1 ¼ 18ffiffiffi2p

log 4mþ1� �

¼ log 18ffiffiffi2p

� �

ðmþ 1Þ log 4 ¼ log 18ffiffiffi2p

� �

ðmþ 1Þ log 4 ¼ � log 8ffiffiffi2p� �

mþ 1 ¼ � log 8ffiffiffi2p

log 4

¼ �1:75

m ¼ �2:75

Taking log10 of both sides.

Example 25

Solve 5 x ¼ 17, writing the solution correct to three decimal places.

Solution5x ¼ 17

log 5x ¼ log 17x log 5 ¼ log 17

x ¼ log 17log 5

¼ 1:7603 . . .� 1:760

Logarithmic equations are equations like log5 x ¼ �3, which can be solved by rewriting theequation in index form.

Example 26

Solve each logarithmic equation.

a log5 x ¼ �3 b logx 18 ¼ 3

Solutiona log5 x ¼ �3) x ¼ 5�3

¼ 15 3

¼ 1125

b logx 18 ¼ 3[ 18 ¼ x3x ¼

ffiffiffiffiffi183p

¼ 2:6207 . . .� 2:62

Stage 5.3

NSW

2759780170194662


Stage 5.3 Exercise 7-11 Exponential and logarithmic equations1 Solve each exponential equation.

a 2 k ¼ 512 b 5m ¼ 78 125 c 3d ¼ 59 049

d 5x ¼ 25ffiffiffi5p

e 2y ¼ 116

ffiffiffi2p f 4a ¼ 128

g 3kþ2 ¼ 27ffiffiffi3p

h 6n�2 ¼ 1216

ffiffiffi6p i 91�d ¼ 1

27ffiffiffi3p

2 Solve each exponential equation, writing the solution correct to three decimal places.a 7 x ¼ 16 b 5x ¼ 36 c 11 x ¼ 420 d 2 x ¼ 0.52

e 3 x ¼1.6 f 4x ¼ 25

g 2 x�2 ¼ 47 h 3 xþ4 ¼ 72

i 6 xþ3 ¼ 29 j 85�x ¼ 4000 k 5 y ¼ 4.8 l 7 kþ5 ¼ 3003 Solve each exponential equation by expressing both sides to base 2.

a 2 xþ2 ¼ 16 b 8x ¼ 32 c 4x�1 ¼ffiffiffi2p

d 81�x ¼ 16ffiffiffi2p

e 42�x ¼ 18

f 8xþ1 ¼ 18ffiffiffi2p g 1

4

� �1�x¼

ffiffiffi2p

h 5 12

� �x¼ 20

4 Solve each logarithmic equation, expressing your answer correct to three decimal places wherenecessary.

a log2 x ¼ 3 b log10 x ¼ 3 c log5 x ¼ �2 d log4 x ¼ �3e log27 x ¼ 13 f log4 x ¼ 12 g log10 x ¼ �3 h log8 x ¼ 32i log10 x ¼ � 12 j log4 x ¼ � 32 k log4 x ¼ 3 12 l log ffiffi5p x ¼ �4m logx 4 ¼ 2 n logx 5 ¼ �1 o logx 14 ¼ 2 p logx 0.01 ¼ 2q logx 16 ¼ 1 r logx 8 ¼ 3 s logx 60 ¼ 3 t logx 4:8 ¼ 12

5 Use the compound interest formula A ¼ P(1 þ r)n to determine the number of years (to thenearest year) it will take an investment of $1000 to grow to $2000, if it earns compoundinterest at a rate of 6% p.a.

6 Penny invests $12 000 at 1% per month compound interest. How many whole months will ittake for Penny’s investment to grow to $15 000?

7 A radioactive substance with a mass of 150 grams decays according to the equationA ¼ 150 3 2�t20

� �, where A (grams) is the amount remaining after t days. Find, correct to the

nearest whole number:a the mass of substance remaining after 10 daysb the time taken for the substance to decay to half its original massc the time taken for the substance to decay to a mass of 20 g.

See Example 24

See Example 25

See Example 26

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Power plus


a 2� x3� x� 5

6¼ 5þ x

4þ x� 2

3b 1

x� 1þ2

xþ 1 ¼ 0

c x2(x � 2) � 2(x þ 1) ¼ x(x2 � 3) � 2(5 þ x2)

2 Graph each inequality on a number line.a 1 � x � 4 b �2 � x � 3 c �12 < 4x � 4

3 The number of diagonals, D, in a polygon with n sides is D ¼ 12 n n� 3ð Þ. Show that thereis no polygon that has exactly 100 diagonals.

4 The two solutions of x2 � 8x � 11 ¼ 0 are in the form x ¼ p� qffiffiffi3p

, where p and q areintegers. Find p and q.

5 Solve each logarithmic equation.a log a þ log 3 ¼ log 21 b log x � log 4 ¼ log 5c log 7 þ log m ¼ log (m þ 12) d log (h þ 7) � log 3 ¼ log (h � 1)

2779780170194662


Chapter 7 review

n Language of maths

check cubic equation exact expand

exponential equation factorise formula fraction

greater than inequality LHS less than

logarithm logarithmic equation lowest common multiple (LCM) number line

quadratic equation RHS solution solve

subject surd variable

1 What type of equation has 2 as the highest power of x? Write an example of this type ofequation.

2 Write log7 a ¼ �3 in index form.

3 What is the difference between an equation and an inequality?

4 Why is it possible for a quadratic equation to have more than one solution?

5 When checking the solution to an equation, you need to show that ‘LHS ¼ RHS’. What doesthat mean?

6 What does the inequality symbol ‘�’ mean?

n Topic overview

Copy and complete this mind map of the topic, adding detail to its branches and using pictures,symbols and colour where needed. Ask your teacher to check your work.

Equations with algebraic fractions

Quadratic and cubic equations

Equation problems

Equations andformulas

Graphing inequalities on a number line

Equations andlogarithms

Solving inequalitiesLogarithms

Exponential andlogarithmic equations

Puzzle sheet

Equations andinequalities crossword

MAT10NAPS10047

Quiz

Equations

MAT10NAQZ00011

9780170194662278


a 3wþ 25¼ 4 b y

5¼ 7

4c 2aþ 1

2¼ 3a� 1

4

d 3mþ 56¼ 10� m

3e 2s

3� s

6¼ 2 f x

10þ x

2¼ 1


a m� 14þ m

2¼ 6 b yþ 1

3þ y� 1

4¼ 1

2c 2m� 1

4� m� 4

3¼ 4

33 Solve each quadratic equation.

a y2 ¼ 4 b p2 � 100 ¼ 0 c 4x2 ¼ 36

d 3m2 � 3 ¼ 0 e 2w2

5¼ 10 f x2 þ 8x þ 7 ¼ 0

g h2 � 8h � 9 ¼ 0 h u2 þ 4u � 77 ¼ 0 i k2 þ 5k ¼ 0

4 Solve each cubic equation, correct to one decimal place.

a u3 � 7 ¼ 0 b 5m3 � 125 ¼ 0 c x3

2¼ 1:5

5 Grace is three years younger than her sister Jane. Twice the sum of their ages is 4 more thantheir father’s age. If their father is 54, find the ages of Grace and Jane.

6 The braking distance (in metres) of a bicycle travelling at a speed of v metres/second is

d ¼ vðvþ 1Þ2

: Calculate the braking distance when the speed of the bicycle is 15 m/s.

7 Make a the subject of each formula.a y ¼ ax þ b b P ¼

ffiffiffiffiam

qc M(1 þ a) ¼ 1 � a

8 Graph each inequality on a number line.a x � 0 b x < 3 c x � �2 d x > �5

9 Solve each inequality.a y � 6 � 10 b 2y � �15 c 3a þ 10 > �5

d 10 � 6x < 28 e aþ 2�4 >72

f 3� 5x2

� 9

10 Write each expression in index form.a log6 216 ¼ 3 b log2 116 ¼ �4 c log7 7

ffiffiffi7p¼ 32

11 Evaluate each expression.a log7 84 � log7 12 b log2 3þ log2 13 c 2 log3 9 þ log3 2 � log3 6

12 If log10 3 � 0.4771, find the value of:a log10 9 b log10 300 c log10 103 d log10

ffiffiffiffiffi90p

13 Solve each exponential equation, correct to three decimal places.a 5 x ¼ 11 b 2 x ¼ 0.52 c 3 xþ4 ¼ 105 d 162�x ¼ 5

See Exercise 7-01

Stage 5.3

See Exercise 7-01

See Exercise 7-02

Stage 5.3

See Exercise 7-03

See Exercise 7-04

See Exercise 7-05

See Exercise 7-06

See Exercise 7-07

See Exercise 7-08

Stage 5.3

See Exercise 7-09

See Exercise 7-10

See Exercise 7-10

See Exercise 7-11

9780170194662 279

Chapter 7 revision

Chapter 7: Equations andlogarithmsSkillCheck7-01 Equations with algebraic equations7-02 Quadratic equations x2 + bx + c = 07-03 Simple cubic equations ax3 = c*7-04 Equation problemsMental skills 7: Multiplyingdecimals7-05 Equations and formulas7-06 Changing the subjectof a formula*Investigation: Restrictingvalues of variables7-07 Graphing inequalities on a number lineInvestigation: The language of inequalitiesInvestigation: Solving inequalities7-08 Solving inequalitiesInvestigation: Power tables7-09 Logarithms*7-10 Logarithm laws*7-11 Exponential andlogarithmic equations*Power plusChapter 7 review

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