cameronsmaths.files.wordpress.com · 8 bindi is 18 years old and investing $20000 in a ﬁxed term...

8Bindi is 18 years old and investing $20 000 in a fixed term deposit paying 6% p.a. compound interest. When Bindi has $30 000 she intends to put a deposit on a home. How long will it take for Bindi’s $20 000 to grow to $30 000?

The investment that Bindi has made is an example of an exponential function. In this chapter you will learn how to use an exponential function to model the growth of such an investment and other similar problems.

Exponentialfunctions

MQ10 VIC ch 08 Page 257 Tuesday, November 20, 2001 11:23 AM

258 M a t h s Q u e s t 1 0 f o r V i c t o r i a

Index lawsIn Maths Quest 9 we looked at indices and the index laws. These laws are the basis forexponential functions, which we will cover later in the chapter. Let’s first revise thiswork. A number in index form has two parts, the base and the index and is written as:

The first two index laws relate to multiplication and division of index expressions.

First Index Law: When numbers with the same base are multiplied, the indices areadded.

am × an = am + n

Second Index Law: When numbers with the same base are divided, the indices aresubtracted.

am ÷ an = am − n

ax indexbase

Simplify each of the following.

a m4n3p × m2n5p3 b 2a2b3 × 3ab4 c

THINK WRITE

a Write the expression. a m4n3p × m2n5p3

Multiply the numbers with the same base by adding the indices. Note that p = p1.

= m4 + 2 n3 + 5 p1 + 3

= m6b8p4

b Write the expression. b 2a2b3 × 3ab4

Simplify by multiplying the coefficients, then multiply the numbers with the same base by adding the indices.

= 2 × 3 × a2 + 1 × b3 + 4

= 6a3b7

c Write the expression. c

Simplify by dividing both coefficients by the same factor, then divide numbers with the same base by subtracting the indices.

=

=

2x5 y4

10x2 y3------------------

1

2

1

2

12x5y4

10x2y3-----------------

21x5 2– y4 3–

5---------------------------

x3y5

--------

1WORKEDExample

MQ10 VIC ch 08 58 Tuesday, November 20, 2001 11:23 AM

C h a p t e r 8 E x p o n e n t i a l f u n c t i o n s 259The Third Index Law is used in calculations when a zero index is involved.

Third Index Law: Any term (excluding 0) with an index of 0, is equal to 1.

a0 = 1

The Fourth, Fifth and Sixth Index Laws involve removing brackets from an indexexpression.

Fourth Index Law: To remove brackets, multiply the indices inside the brackets by theindex outside the brackets. Where no index is shown, assume that it is 1.

(am)n = amn

Fifth Index Law: To remove brackets containing a product, raise every part of theproduct to the index outside the brackets.

(ab)m = ambm

Sixth Index Law: To remove brackets containing a fraction, multiply the indices ofboth numerator and denominator by the index outside the brackets.


a (2b3)0 b −4(a2b5)0

THINK WRITEa Write the expression. a (2b3)0

Apply the Third Index Law which states that any term (excluding 0) with an index of 0, is equal to 1.

= 1

b Write the expression. b −4(a2b5)0

The term inside the brackets has an index of zero so the bracket is equal to 1.

= −4 × 1

Simplify. = −4

1

2

1

2

3

2WORKEDExample

ab---

m am

bm------=


a (2n4)3 b (3a2b7)3 c

Continued over page

THINK WRITEa Write the expression. a (2n4)3

Apply the Fourth Index Law by multiplying the indices inside the brackets by the index outside the brackets (2 = 21).

= 21 × 3n4 × 3

= 23n12

= 8n12

2x3

y4---------

4

12

3WORKEDExample



Index laws

1 Simplify each of the following.

a a3 × a4 b a2 × a3 × a c b × b5 × b2

d ab2 × a3b5 e m2n6 × m3n7 f a2b5c × a3b2c2

g mnp × m5n3p4 h 2a × 3ab i 4a2b3 × 5a2b × b5

j 3m3 × 2mn2 × 6m4n5 k 4x2 × xy3 × 6x3y3 l 2x3y2 × 4x × x4y4

THINK WRITE

b Write the expression. b (3a2b7)3

Apply the Fifth Index Law by multiplying the indices inside the brackets by the index outside the brackets. Where no index is shown, assume that it is 1.

= 31 × 3 × a2 × 3 × b7 × 3

= 33a6b21

Simplify. = 27a6b21


Apply the Sixth Index Law by multiplying the indices of both numerator and denominator by the index outside the brackets.

=

Simplify. =

1

2

3

12x3

y4--------

4

221 4× x3 4××

y4 4×------------------------------

316x12

y16-------------

rememberTo simplify expressions with numbers and/or pronumerals in index form, the following index laws are used.1. am × an = am + n

2. am ÷ an = am − n

3. a0 = 14. (am)n = amn

5. (ab)m = ambm

6.ab---

m am

bm------=

remember

8A

EXCEL

Spreadsheet

Index laws

WWORKEDORKEDEExample

1a, b

12---

12--- 1

2---

Mathca

d

Index laws


C h a p t e r 8 E x p o n e n t i a l f u n c t i o n s 2612 Simplify each of the following.

a a4 ÷ a3 b a7 ÷ a2 c b6 ÷ b3

d e f

g h i 6x7y ÷ 8x4

j 7ab5c4 ÷ ab2c4 k l


a a0 b (2b)0 c (3m2)0

d 3x0 e 4b0 f −3 × (2n)0

g 4a0 − h 5y0 − 12 i 5x0 − (5xy2)0


a (a2)3 b (2a5)4 c

d e (a2b)3 f (3a3b2)2

g (2m3n5)4 h i

j k l

5a 2m10n5 is the simplified form of:

b The value of 4 − (5a)0 is:

6 Evaluate each of the following.

a 23 × 22 × 2 b 2 × 32 × 22 c (52)2

d e (23 × 5)2 f

g h (33 × 24)0 i 4(52 × 35)0


a (am + n)p b c

d (xy)3z e ab × (pq)0 f ma × nb × (mn)0

A m5n3 × 2m4n2 B C (2m5n2)2

D 2n(m5)2 × n4 E

A −1 B 9 C 1 D 3 E 5


1c4a7

3a3-------- 21b6

7b2----------- 48m8

12m3-------------

m7n3

m4n2------------ 2x4y3

4x4y--------------

20m5n3 p4

16m3n3 p2------------------------- 14x3y4z2

28x2y2z2----------------------


2

a4---

0


3 m2

3------

4

2n4

3--------

2

3m2n4

------------- 3 a2

b3-----

2

5m3

n2----------

4 7x2y5--------

3 3a5b3--------

4

mmultiple choiceultiple choice

6m10n4

3n------------------

2m5

n3----------

2

35 46×34 44×----------------- 3

5---

3

44 56×43 55×-----------------

a2

b3-----

x n3m2

npmq------------



Negative indicesSo far we have dealt only with indices that are positive whole numbers or zero. Toextend this we need to consider the meaning of an index that is a negative whole

number. Consider the expression . Using the Second Index Law, = a3 − 5

= a−2

Writing terms in the expanded notation we have: =

=

=

By equating the results of simplification, using the two methods, we have: a−2 = .

In general terms, = (1 = a0)

= a0 − n (using the Second Index Law)

= a−n

Seventh Index Law: a−n =

The convention is that an expression should be written using positive indices and weuse the Seventh Index Law to do this.

a3

a5----- a3

a5-----

a3

a5----- a a a××

a a a a a××××----------------------------------------

1a a×------------

1a2-----

1a2-----

1an----- a0

an-----

1an-----

Express each of the following with positive indices:

a x−3 b 2m−4n2 c

THINK WRITEa Write the expression. a x−3

Apply the Seventh Index Law. =

b Write the expression. b 2m−4n2

Apply the Seventh Index Law to write the expression with positive indices.

=

c Copy the expression and rewrite the fraction, using a division sign.

c = 4 ÷ a−3

Apply the Seventh Index Law to write the expression with positive indices.

= 4 ÷

To divide the fraction use ‘× and tip’. = 4 ×

= 4a3

4a 3–-------

1

21x3-----

1

22n2

m4--------

14

a 3–-------

21a3-----

3a3

1-----

4WORKEDExample


C h a p t e r 8 E x p o n e n t i a l f u n c t i o n s 263Worked example 4c shows the method used to demonstrate the converse of the Seventh

Index Law = an. Try this as an exercise for yourself!

All laws discussed in the previous section are applicable to the terms with negativeindices.

1a n–-------

Simplify each of the following, expressing the answers with positive indices.

a a2b−3 × a−5b b c

THINK WRITE

a Write the expression. a a2 b−3 × a−5b

Apply the First Index Law. Multiply numbers with the same base by adding the indices.

= a2 + (−5)b−3 + 1

= a−3b−2

Express the answer with positive indices.

=

b Write the expression. b

Apply the Second Index Law. Divide numbers with the same base by subtracting the indices.

=

=

Express the answer with positive indices.

=


Apply the Sixth Index Law. Multiply the indices of both numerator and denominator by the index outside the brackets. Remember that 2 = 21.

=

Express all numbers and pronumerals with positive indices.

=

Simplify. =

2x4 y2

3x y5--------------- 2m3

n 2–----------

2–

1

2

31

a3b2-----------

12x4y2

3xy5--------------

22x4 1– y2 5–

3---------------------------

2x3y 3–

3----------------

32x3

3y3--------

12m3

n 2–----------

2–

22 2– m 6–

n4-----------------

31

22m6n4------------------

41

4m6n4---------------

5WORKEDExample



Numbers in index form can be easily evaluated, if they are expressed with positiveindices first. Consider the following example:

A calculator can be used to evaluate numbers with a negative index. The base is alwaysentered before the index.

Evaluate 6 × 3−3 without using a calculator.

THINK WRITE

Write the multiplication. 6 × 3−3

Express 3−3 with a positive index. = 6 ×

Multiply the numerator of the fraction by the whole number.

=

Evaluate the denominator. =

Cancel by dividing both numerator and denominator by the same number.

=

1

2133-----

3633-----

46

27------

529---

6WORKEDExample

Evaluate 2−5 using a calculator.

THINK WRITE/DISPLAYUsing a scientific calculator:

Press .

(Note that the button could be or on your calculator.)

2−5 = 0.031 25

Using a graphics calculator:

Press .

1

2 yx +/− 5 =yx xy

ax

2

2 ^ (−) 5 ENTER

7WORKEDExample

remember1. A number with a negative index can be expressed with a positive index using

the following identities.

(a) a−m = (b) = am

2. All index laws apply to terms with negative indices.3. Always express answers with positive indices unless required otherwise.4. Numbers and pronumerals without an index are understood to have an index of 1.

1am------ 1

a m–--------

remember


C h a p t e r 8 E x p o n e n t i a l f u n c t i o n s 265

Negative indices

1 Express each of the following with positive indices.

a x−5 b y−4 c 2a−9

d a−3 e 3x2y−3 f 2−2m−3n−4

g 6a3b−1c−5 h i

j k l

2 Simplify each of the following, expressing the answers with positive indices.

a a3b−2 × a−5b−1 b 2x−2y × 3x−4y−2 c 3m2n−5 × m−2n−3

d 4a3b2 ÷ a5b7 e 2xy6 ÷ 3x2y5 f 5x−2y3 ÷ 6xy2

g h i

j (2a3m4)−5 k 4(p7q−4)−2 l 3(a−2b−3)4

m n o

3 Evaluate each of the following without using a calculator.

a 2−3 b 6−2 c 3−4

d 3−2 × 23 e 4−3 × 22 f 5 × 6−2

g h i

j k l

4 Evaluate each of the following, using a calculator.

a 3−6 b 12−4 c 7−5

d e f (0.045)−5

5

a Which of the following, when simplified, gives ?

b When simplified, is equal to:

A B C

D E

A B C D E

8BWWORKEDORKEDEExample

4

Mathcad

Negativeindices

45---

1a 6–------- 2

3a 4–-----------

EXCEL Spreadsheet

Index laws

6a3b 2–----------- 7a 4–

2b 3–----------- 2m3n 5–

3a 2– b4------------------


5

6m4n2n3m6--------------- 4x2y9

x7y 3–-------------- 2m2n 4–

6m5n 1–------------------

2 p2

3q3---------

3– a 4–

2b 3–-----------

2 6a2

3b 2–-----------

3–


6

62 3–------- 4 3 3–×

2 3–----------------- 1

3--- 5 2– 34××

160 24×82 2 4–×-------------------- 53 250×

252 5 4–×---------------------- 34 42×

123 150×-----------------------


712---

8– 34---

7–

mmultiple choiceultiple choice3m4

4n2----------

3m 4– n 2–

4-------------------- 3 2 2– m4 n 2–××× 3n 2–

2 2– m 4–-----------------

22n 2–

3 1– m 4–----------------- 3m4 22n 2–×

3a 2– b 7– 34---a 4– b6÷

4

a6b

13------------- 9b

4a6

-------- 9a2

4b-------- 4a

2

b13

-------- 4a2

b--------


MA

TH

SQUEST

C H A L L

EN

GE

MA

TH

SQUEST

C H A L L

EN

GE


c When (2x6y−4)−3 is simplified, it is equal to:

d If is equal to , then x and y (in that order) are:

A B C D E

A −3 and −6 B −6 and −3 C −3 and 2 D −3 and −2 E −2 and −3

2x18

y12

---------- x18

8y12

---------- y12

8x18

---------- 8y12

x18

---------- x18

6y12

----------

2ax

by--------

3 8b9

a6--------

GAME

time

Exponential functions— 001

11 Simplify .

2 Simplify .

3 Simplify .

4 Simplify .

5 Evaluate .

6 Evaluate .

7 Express 3n−4 with a positive index.

Simplify questions 8–10 expressing youranswer with positive indices.

8

9

10

8a2bc3 5b2c5×

24d3e4 f 5

56e2 f-----------------------

9g7h( )3

5i j4

ijk3----------

2

9l0 4m6( )0×

43 162×45 82×

--------------------

6 p 3– q 2–

36 pq 3–-------------------

r 3– s4 r 5– s 3–×

3t 3– u4

18t 1– u 3–---------------------

3–

1 One million can be written as 1000 × 1000. a List another two ways in which one million can be written as the

product of two numbers containing the digits 1 and 0.b Find a way of writing one million as the product of two numbers,

neither of which contains the digit 0.

2 Imagine a piece of paper that you tear in half, put the two piecestogether, tear those in half, put the 4 pieces of paper together, tear thosein half, and so on. After how many tears would you have about amillion pieces of paper?

3 Without using a calculator, which of these numbers is the larger: 230 or320?



Fractional indicesWe are now ready to look at indices that are fractional.

Numbers (or terms) with fractional indices can be written as surds, using thefollowing laws:

• =

• =

= To understand how these laws are formed, consider the following numerical examples.

Using the First Index Law = 41

(2 × 2 = 4) so = 4

If these two identities are true then = .Similarly:Using the First Index Law = 81

(2 × 2 × 2 = 8) so = 8

If these two identities are true then = .

This can be generalised to .

Now consider: = or =

= =

= =

The Eighth Law of Indices is .As can be seen from the above identities, the denominator of a fraction (n) indicates

the power or type of root. That is, n = 3 implies cube root, n = 4 implies fourth root,etc. Note that when n = 2 (square root), it is the convention not to write 2 at the square

root sign. That is, for , we write rather than .

a1n---

an

amn----

amn

an( )m

412---

412---

×4 4×

412---

4

813---

813---

813---

××83 83 83××

813---

83

a1n---

an=

amn----

am

1n---×

amn----

a1n--- m×

am( )1n---

a1n---

m

amn an( )m

amn----

amn an( )m= =

a12---

a a2

Evaluate each of the following without using a calculator.

a b

THINK WRITE

a Rewrite the number using the Eighth Law of Indices. a = Evaluate. = 3

b Rewrite the number using . b = Evaluate the square root. = 43

Evaluate the result. = 64

912---

1632---

1 912---

9

2

1 amn----

an( )m= 1632---

16( )3

2

3

8WORKEDExample



A number with a fractional index can be evaluated on a scientific calculator by directly

entering the fractional index using the function or by using the function.On a graphics calculator, the ^ button is still used to enter the fractional index or theMATH menu to calculate a surd.

abc--- x

Using a calculator, evaluate each of the following. Give the answer correct to 2 decimal places.

a b

THINK WRITE/DISPLAYa Method 1

Enter the base then the fractional index on a calculator and evaluate. On a graphics calculator, enter brackets around the fractional index.

Round to 2 decimal places.

a

Method 2Express as a surd. = Evaluate using your calculator.On a graphics calculator, enter the powerof the root, , then press , select5: , press and .

= 1.319 507 911≈ 1.32

Round to 2 decimal places.Note: Both methods are shown on the screen above right.

b Method 1Enter the base then the fractional index on a calculator and evaluate. On a graphics calculator, press

.

b

Round to 2 decimal places.

Method 2Express as a surd. = Evaluate using your calculator. On a graphics calculator, press , enter the power of the root, , press , select 5: and press . Close the brackets by pressing

and square the result by pressing and .

= 1.368 738 107≈ 1.37

Round to 2 decimal places.Note: Both methods are shown on the screen above right.

415---

327---

1

2

1 415---

415---

45

2

5 MATHx 4 ENTER

3

1

3 ^ ( 2 ÷ 7 )ENTER

2

1 327---

327---

37( )2

2(

7 MATH x

3)^ 2 ENTER

3

9WORKEDExample


C h a p t e r 8 E x p o n e n t i a l f u n c t i o n s 269All index laws discussed so far are valid for fractional indices. The following exampleillustrates this.

Simplify each of the following:

a b c

THINK WRITE

a Write the expression. a

Apply the First Index Law to multiply numbers with the same base by adding the indices.

=


Use the Fourth Index Law to multiply each index inside the brackets by the index outside the brackets.

=

Simplify the fractions. =


Use the Sixth Index Law multiplying the index in both the numerator and denominator by the index outside the brackets.

=

m15---

m25---

× a2b3( )16--- x

23---

y34---

----- 1

2---

1 m15---

m25---

×

2 m35---

1 a2b3( )16---

2 a26---b

36---

3 a13---b

12---

1x

23---

y34---

----- 1

2---

2x

13---

y38---

-----

10WORKEDExample

remember1. Fractional indices are those which are expressed as fractions.2. Numbers with fractional indices can be written as surds, using the following

identities:

3. All index laws are applicable to fractional indices.

a1n---

an=

amn----

amn an( )m= =

remember



Fractional indices


a b c

d e f

g h i

j k l

2 Using a calculator, evaluate each of the following. Give the answer correct to 2decimal places.

a b c

d e f

g h i


a b c

d e f

g h i


a b c

d e f


a b c

d e f

g h i

8C

Mathca

d

Fractional indices


816

12---

2512---

8112---

EXCEL

Spreadsheet

Index laws

813---

6413---

8114---

1614---

2532---

3632---

10072---

1634---

2723---


9

313---

512---

715---

819---

1238---

0.6( )45---

23---

32--- 3

4---

34--- 4

5---

23---


10aSkillSH

EET 8.14

35---

415---

× 218---

238---

× a12---

a13---

×

x34---

x25---

× 5m13---

2m15---

× 12---b

37---

4b27---

×

4y2– y29---

× 25---a

38---

0.05a34---

× 5x3 x12---

×

a23---b

34---

a13---b

34---

× x35---y

29---

x15---y

13---

× 2ab13---

3a35---b

45---

×

6m37--- 1

3---m

14---n

25---

× x3y12---z

13---

x16---y

13---z

12---

× 2a25---b

38---c

14---

4b34---c

34---

×

SkillSH

EET 8.23

12---

313---

÷ 523---

514---

÷ 122 1232---

÷

a67---

a37---

÷ x32---

x14---

÷ m45---

m59---

------

2x34---

4x35---

-------- 7n2

21n43---

----------- 25b35---

20b14---

-----------


C h a p t e r 8 E x p o n e n t i a l f u n c t i o n s 2716 Simplify each of the following.

a b c

d e f


a b c

d e f

g h i


a b c

d e f

g h i

9

a If is equal to , then m and n could not be:

b When simplified, is equal to:


a b c

d e f

g h i

A 1 and 3 B 2 and 6 C 3 and 8 D 4 and 9 E both C and D

A B C D E

x3y2 x43---y

35---

÷ a59---b

23---

a25---b

25---

÷ m38---n

47---

3n38---

÷

10x45---y 5x

23---y

14---

÷ 5a34---b

35---

20a15---b

14---

----------------- p78---q

14---

7 p23---q

16---

---------------

SkillSH

EET 8.3

234---

35---

523---

14---

715---

6

a3( )110------

m49---

38---

2b12---

13---

4 p37---

1415------

xmn----

np---

3mab---

bc---


10b, ca

12---b

13---

12---

a4b( )34---

x35---y

78---

2

3a13---b

35---c

34---

13---

5 x12---y

23---z

25---

12--- a

34---

b-----

23---

m45---

n78---

------ 2

b35---

c49---

----- 2

3---

4x7

2y34---

-------- 1

2---


a34---

mn----

a14---

amn----

bnp---

------ p

m----

SkillSH

EET 8.4

amp----

bnm----

------ apn---

bnm----

------ ampn

-------

bnm----

--------- ap

bm------ a

m2

np-------

bnm

p2-------

---------

GAMEtime

Exponentialfuntions— 002a8 b93 m164

16x4 8y93 16x8y124

WorkS

HEET 8.1

27m9n153 32 p5q105 216a6b183



Career profileJ I M E L L I OT — F l o o d E n g i n e e r

Qualifications:Diploma of Civil EngineeringBachelor of Engineering (Hons)Master of Engineering ScienceMaster of Business AdministrationMember Institution of Engineers (Australia)

Employer:Bureau of Meteorology

Company Web site:www.bom.gov.au

Civil Engineering was recommended as a natural progression from my interest/aptitude in mathematics and physical sciences. Hydrology and water resources were part of the latter section of my engineering studies and I was interested enough in the idea of mathematical modelling in that area to pursue a career in it.

As an engineer working in the area of flooding, it is important to be able to estimate how much water flows down a river during a flood event, particularly at the peak. Flood flows are normally presented as the volume of water per unit of time (for example, cubic metres per second). These flows can be difficult to measure directly, so it is common to have to make an estimate after the event. Debris and other markings are used to identify the maximum level reached by the flood. From these markings the slope of the

river during flood can be measured. The formula for relating the ‘slope’ of the flood to the velocity of flow is called Manning’s

formula. This formula is: v =

where C1 = 1.0 for SI unitsv = velocity of flow (m/s)S = slopeR = hydraulic radius (m)

= (= area/wetted perimeter)n = roughness coefficient

A difficult part of this process is the estimate of n, the roughness coefficient. There are various empirical relationships for doing this. For example, in gravel-bed rivers whereS > 0.002, an equation that can be used is:n = 0.32S0.38R–0.16 where R is the hydraulic radius (m), and S is slope.

Manning’s formula is used to estimate the velocity of flow of flood water at the peak of the flood. Multiplying this velocity by the average cross-sectional area provides an estimate of the peak flow rate.

A typical day’s work for me could involve using these estimates of peak flow to get some idea of the severity of the flooding and the damage caused. This would include understanding the relationships between heavy rainfall and flooding. Flood volumes are an essential input to this.

I was always fascinated by the neat, concise way mathematics could be used to describe things. I also enjoy its abstract nature.

Questions1. List a unit for flood flow.2. Describe how Jim can identify the

maximum level reached by a flood.3. What is the name of the formula used to

calculate the velocity of flow?4. Investigate what engineering courses are

available. How many different ‘types’ of engineering are there?

C1R23---

S12---

n------------------



Further use of index lawsSo far we have considered situations where one particular index law was used for sim-plifying expressions with indices. However, in most practical situations more than onelaw is needed to simplify the expression.

The following examples show simplification of expressions with indices, using sev-eral index laws.


a (2a3b)4 × 4a2b3 b c

THINK WRITE

a Write the expression. a (2a3b)4 × 4a2b3

Apply the Fourth Index Law. Multiply each index inside the brackets by the index outside the brackets.

= 24a12b4 × 4a2b3

Evaluate the number. = 16a12b4 × 4a2b3

Multiply coefficients and multiply pronumerals. Apply the First Index Law to multiply pronumerals with the same base by adding the indices.

= 16 × 4 × a14b7

= 64a14b7


Apply the Fourth Index Law in the denominator. Multiply each index inside the brackets by the index outside the brackets.

=


=

Use to express the answer with positive

indices.

=


Simplify each numerator and denominator by multiplying coefficients and then pronumerals with the same base.

=


=

Simplify the numerator using a0 = 1. =

=

7x y3

3x3 y2( )2---------------------- 2m5n 3m7n4×

7m3n3 mn2×-------------------------------------

1

2

3

4

17xy3

3x3y2( )2---------------------

27xy3

9x6y4--------------

37x 5– y 1–

9------------------

4 a m– 1am------= 7

9x5y-----------

12m5n 3m7n4×7m3n3 mn2×

-----------------------------------

26m12n5

7m4n5------------------

36m8n0

7---------------

46m8 1×

7-------------------

6m8

7----------

11WORKEDExample



When more than one index law is used to simplify an expression the following stepscan be taken.1. If an expression contains brackets, expand them first.2. If an expression is a fraction, simplify each numerator and denominator then divide

(simplify across then down).3. Express the final answer with positive indices.The following example illustrates the use of index laws for multiplication and divisionof fractions.

Note that the whole numbers could be cancelled in step 3.


a b

THINK WRITE

a Write the expression. a

Remove the brackets in the numerator of the first fraction and in the denominator of the second fraction.

=

Multiply numerators and multiply denominators of the fractions. (Simplify across.)

=

Divide the numerator by the denominator. (Simplify down.)

=

Express the answer with positive indices. =


Remove the brackets. =

Change the division sign to multiplication and tip the second fraction (× and tip).

=

Multiply numerators and multiply denominators. (Simplify across.)

=

Cancel common factors and divide pronumerals with the same base. (Simplify down.)

=

Simplify and express the answer with positive indices.

=

5a2b3( )2

a10---------------------- a2b5

a3b( )7----------------× 8m3n 4–

6mn2( )3--------------------- 4m 2– n 4–

6m 5– n---------------------÷

15a2b3( )2

a10--------------------- a2b5

a3b( )7----------------×

225a4b6

a10----------------- a2b5

a21b7-------------×

325a6b11

a31b7-------------------

4 25a 25– b4

525b4

a25-----------

18m3n 4–

6mn2( )3-------------------- 4m 2– n 4–

6m 5– n--------------------÷

28m3n 4–

216m3n6--------------------- 4m 2– n 4–

6m 5– n--------------------÷

38m3n 4–

216m3n6--------------------- 6m 5– n

4m 2– n 4–--------------------×

448m 2– n 3–

864mn2-----------------------

5m 3– n 5–

18-----------------

61

18m3n5------------------

12WORKEDExample



Further use of index laws


a b

c d

e f

g h

i


a b c

d e f

g h i


a b c

d e f

g h i

remember1. Simplification of expressions with indices often involves application of more

than one index law.2. If an expression contains brackets, they should be removed first.3. If the expression contains fractions, simplify across then down.4. When dividing fractions, change ÷ to × and tip the second fraction (× and tip).5. Express the final answer with positive indices.

remember

8DWWORKEDORKEDEExample

11a

Mathcad

Further useof index

laws3a2b2( )3 2a4b3× 4ab5( )2 3a3b6×

2m3n 5– m2n 3–( ) 6–× 2 pq3( )2 5 p2q4( )3×

2a7b2( )2 3a3b3( )2× 5 b3c 2–( )3 3 bc5( ) 4–×

6x12---y

13---

4x34---y

45---

12---

× 16m3n4( )34---

m12---n

14---

3

×

2 p23---q

13---

34---–

3 p14---q

34---–

13---–

×


11b 5a2b3

2a3b( )3------------------- 4x5y6

2xy3( )4------------------- 3m2n3( )3

2m5n5( )7-----------------------

4x3y10

2x7y4----------------

6 3a3b 5–

2a7b4( ) 3–------------------------ 3g2h5

2g4h--------------

3

5 p6q13---

2

25 p12---q

14---

23---

-------------------------- 3b2c3

5b 3– c 4–------------------

4– x13---y

14---z

12---

2

x23---y

14---–z

13---

32---–

-----------------------------


11c 2a2b 3a3b4×4a3b5

--------------------------------- 4m6n3 12mn5×6m7n6

-------------------------------------- 10m6n5 2m2n3×12m4n 5m2n3×-----------------------------------------

6x3y2 4x6y×9xy5 2x3y6×-------------------------------- 6x3y2( )4

9x5y2 4xy7×-------------------------------- 5x2y3 2xy5×

10x3y4 x4y2×----------------------------------

a3b2 2 ab5( )3×6 a2b3( )3 a4b×------------------------------------- p6q2( ) 3– 3 pq×

2 p 4– q 2– 5 pq4( ) 2–×------------------------------------------------ 6x

32---y

12---

x45---y

35---

×

2 x12---y

15---

3x12---y

15---

×

----------------------------------------




a b c

d e f

g h i


a b

c d

e f

g h


a

b

7 Evaluate the following for x = 8. (Hint: Simplify first.)

8 a Simplify the following fraction.

b Find the value of y if the fraction is equal to 125.

9

Which of the following is not the same as ?

A B C

D E


12a a3b2

5a4b7-------------- 2a6b

a9b3------------× 2a6( )2

10a7b3----------------- 4ab6

6a3------------× m4n3( )2

m6n( )4------------------- m3n3( )3

2mn( )2-------------------×

2m3n2

3mn5---------------

3 6m2n4

4m3n10------------------× 2xy2

3x3y5--------------

4 x3y9

2y10----------

2× 4x 5– y 3–

x2y2( ) 2–-------------------- 3x5y6

2 2– x 7– y------------------×

5 p6q 5–

3q 4–----------------- 5 p6q4

3 p5---------------

2–× 2a

12---b

13---

6a13---b

12---

--------------4a

14---b

12---

b14---a

--------------------× 3x23---y

15---

9x13---y

14---

-------------- 4x12---

x34---y

--------×


12b 5a2b3

6a7b5-------------- a9b4

3ab6------------÷ 7a2b4

3a6b7-------------- 3ab

2a6b4--------------

3÷

4a9

b6--------

3 3a7

2b5--------

4÷ 5x2y6

2x4y5( )2--------------------- 4x6y( )3

10xy3-------------------÷

x5y 3–

2xy5-------------

4– 4x6y 10–

3x 2– y2( ) 3–--------------------------÷ 3m3n4

2m 6– n 5–-------------------- 2m4n6

m 1– n---------------

2–÷

4m12---n

34--- 6m

13---n

14---

8m34---n

12---

----------------÷ 4b3c13---

6c15---b

-------------- 1

2---

2b3c15---–

32---–

÷

52 2×( )0 5 3– 20×( )5× 56 2 1–×( ) 3–÷

23 33×( ) 2– 26 39×( )0

26 3 2–( ) 3–×-----------------------------÷

2x( ) 3– x2---

2× 2x

23( )4------------÷

a2y 9by× 5ab( )y×ay( )3 5 3by( )2×

---------------------------------------------


4xy( )32---

WorkS

HEET 8.28x

32---y

32--- 4xy( )3 64x3y3

2x3y3( )12---

32( ) 1–--------------------- 4xy

12---

2xy2( )12---

×



Exponential functions and their graphs

Relationships of the form y = ax are called exponential functions with base a, where ais a real number not equal to 1, and x is the index or exponent. Exponent is anotherword for index or power.

(Note that in this chapter we will take only positive values of a.)The name ‘exponential’ is used, as the independent variable x is the exponent.

The graph in worked example 13 has several important features.

• The graph passes through (0, 1). That is, the y-intercept is 1. The graph of any equa-tion in the form y = ax will pass through this point.

• The graph passes through the point (1, 2). All graphs of the form y = ax will passthrough the point (1, a).

• y > 0 for all values of x. You will notice that for negative values of x, the graph getsvery close to but will never touch the x-axis. When this occurs, the line which thegraph approaches is called an asymptote. The equation of the asymptote for y = ax isy = 0.

Complete the table of values below and use it to plot the graph of y = 2x.

THINK WRITE

Substitute each value of x into the function y = 2x to obtain the corresponding y-value.

Plot each point generated on a set of axes.

Join with a smooth curve.

x −4 −3 −2 −1 0 1 2 3 4

y

1 x −4 −3 −2 −1 0 1 2 3 4

y 1 2 4 8 16116------ 1

8--- 1

4--- 1

2---

2

3

x

y

0

y = 2x

3 421–1–2–3–4

2468

1012141618

13WORKEDExample



Use your graphics calculator to plot the graph of y = 2x.

1. Press .

2. For Y1 enter 2^X.

3. Press and adjust the settings.

4. Press to see the graph of y = 2x.

5. Now on the same axes graph y = 3x. Press andfor Y2 enter 3^X.

6. Press to see both y = 2x and y = 3x on thesame set of axes.

When the exponential function is of the form y = kax (ory = k × ax), the coefficient k has the effect of dilating thegraph, without changing the basic shape or the locationof the asymptote. This can be seen in the followingworked example.

Graphics CalculatorGraphics Calculator tip!tip! Drawing exponential functions

Y=

WINDOW

GRAPH

Y=

GRAPH

a Plot the graph of y = 3 × 2x for −3 ≤ x ≤ 3.b State the y-intercept.c Write the equation of the horizontal asymptote.

THINK WRITE

a Prepare a table of values taking x-values from −3 to 3. Fill in the table by substituting each value of x into the given equation.

a

Draw a set of axes on graph paper to plot the points from the table and join them with a smooth curve.

1x −3 −2 −1 0 1 2 3

y 1 3 6 12 2438--- 3

4--- 1

2---

2

x

y

0

y = 3 × 2x

y = 0

321

2468

1012141618202224

–1–2–3

14WORKEDExample



The worked example below considers the effect of a negative in the exponent.

THINK WRITE

b Locate where the curve cuts the y-axis. Alternatively, find the y-value for x = 0 in the table.

b The y-intercept is 3.

c Find an imaginary line to which the curve gets closer and closer but does not cross. As it is a horizontal asymptote, the equation will be of the form y = constant.

c The equation of the asymptote is y = 0.

Plot the graph of y = 3−x for −3 ≤ x ≤ 3, clearly showing the y-intercept and the horizontal asymptote.

THINK WRITE

Draw up a table of values.

Substitute the values of x into the equation to find the corresponding y-values.

Draw a set of axes, plot the points generated from the table and join with a smooth curve.

1x −3 −2 −1 0 1 2 3

y 27 9 3 1 13--- 1

9--- 1

27------

2

3

x

y

0

y = 3–x

y = 0

321

468

10121416182022242628

–1–2–3

2

15WORKEDExample

remember1. Relationships of the form y = ax, where a ≠ 1 are called exponential functions

with base a.2. To obtain the graph of an exponential function, construct a table of values first

and then plot the points from the table and join them with a smooth curve. Alternatively use a graphics calculator or a computer graphing package.

3. An asymptote is a line which the graph approaches but never cuts.

remember



Exponential functions and their graphs

You may use a graphics calculator or computer graphing package to assist you in this exercise.

1 Complete the table below and use the table to plot the graph of y = 10x.

2 Plot the graph of each of the following exponential functions.a y = 4x b y = 5x c y = 6x

3 On the one set of axes draw the graphs of y = 2x, y = 3x and y = 4x.

4 Use your answer to question 3 to describe the effect of increasing the value of a onthe graph of y = ax.

5 a Plot the graph of y = 2 × 3x for −3 ≤ x ≤ 3.b State the y-intercept.c Write the equation of the horizontal asymptote.

6 Complete the following table of values and then plot the graph of y = 2x, y = 3 × 2x,and y = × 2x on the same set of axes.

7 Study the graphs in question 6 and state the effect that the value of k has on graphswith equation y = k × ax.

8 Plot the graph of y = 2−x for −3 ≤ x ≤ 3, clearly showing the y-intercept and thehorizontal asymptote.

9 On the one set of axes sketch the graphs of y = 3x and y = 3−x.

10 Use your answer to question 9 to describe the effect of a negative index on the graphof y = ax.

11 a Complete the table of values below and use the points generated to sketch thegraph of y = x.

b By writing with a negative index, show algebraically that the functions y = x

and y = 2−x are identical.

12 Draw the graphs of y = (1.2)x, y = (1.5)x and y = (1.8)x.

13 a Draw the graph of y = 10 × (1.3)x.b State the y-intercept.c Write the equation of the horizontal asymptote.

x −4 −3 −2 −1 0 1 2 3 4

y

x −3 −2 −1 0 1 2 3

2x

3 × 2x

× 2x

x −3 −2 −1 0 1 2 3

y

8EMathca

d

Exponential graphs


13SkillSH

EET 8.5

EXCEL

Spreadsheet

Graphs of the formy = a ¥ b2 + c


14

GCpro

gram

Exponential graphs 1

5---

15---


15

12---

12--- 1

2---


C h a p t e r 8 E x p o n e n t i a l f u n c t i o n s 28114 Myung-Hye invests $1000 at 10% p.a. interest compounded annually. This investment

can be represented by the function A = 1000 × (1.1)n, where A is the amount to whichthe investment grows and n is the number of years of the investment. a Prepare a table of values for 0 ≤ n ≤ 6. Substitute integer values of n into the equa-

tion and use a calculator to find corresponding values of A.b Plot the points generated by the table, clearly labelling the axes. Join the points

with a smooth curve.c Use the table of values or the graph to find the value of the investment after 3 years.

15 Kevin buys a car for $40 000. The car depreciates at therate of 15% p.a. The value, $V, of the car after n yearscan be given by the equation V = 40 000 × (0.85)n.a Prepare a table of values for 0 ≤ n ≤ 5. Substitute

integer values of n into the equation and use a cal-culator to find corresponding values of A. Roundanswers to the nearest whole number as required.

b Plot the points generated by the table, clearlylabelling the axes. Join the points with a smoothcurve.

c Describe what is happening to the value of the caras n increases.

d Find the value of the car after 5 years. Give theanswer to the nearest dollar.

Further exponential graphsIn the previous exercise we saw the effect of changing the value of a in y = ax and changing the value of k in y = kax. Now we will consider the effect of changing some of the other factors in an exponential equation.1 Use your graphics calculator to draw the graph of y = 3x.2 Now use the calculator to draw the graphs of each of the following on the one

set of axes.a y = 3x b y = 3x + 2 c y = 3x − 3

3 For the graphs drawn in part 2, state the equation of the horizontal asymptote.4 Use your answers to parts 2 and 3 to state the effect that changing the value of

c has on the graph of y = 3x + c.5 On the same set of axes draw the graphs of:

a y = 4x b y = 4x + 1 c y = 4x − 3.6 Use your answer to part 5 to state the effect that changing the value of b has on

the graph of y = 4x − b.Now let’s look at sketching some exponential graphs. Remember that a sketch graph shows the basic shape of the relationship and its key features. A sketch graph is not drawn by plotting points from a table of values.7 Sketch graphs of each of the following on the one set of axes, showing the

y-intercept and the horizontal asymptote.a y = 2x and y = 2x + 4 b y = 2x and y = 2x − 1c y = 2x and y = 2x + 4 d y = 2x and y = 2x − 1

Check your answers with a graphics calculator or computer graphing package.


2 important events ofComplete the table of values for the

two exponential functions and plot their graphs.The values of the functions and the letters

beside the x-values give thepuzzle code.

–3y = 2x:

y

x

y = ( )x + 5:1–2 y

x

A –2 D –1 E 0 G 1 H 2 I 3 Y 4 L

–3

y

x

N –2 O –1 P 0 R 1 S 2 T 3 U 4 V

1–1–2

2

4

6

8

10

12

14

16

18

20

–4

–2–3 0 2 3 4

1–2

5 8 13 8 9 7 6 2 9 5 5 9 7 13

1 4 5 13 5 2 9 5

4 1 2 5 13 8 6 9 16 51–4

1—16 5

1—16

1–4

1–2

1–2

1–2

1–2

1–2

1–4

1–4

1–2

1–2

1–2

1–2

1–2

1–2

1–2

1–4

1–2

1–8

1–8

1–4

1–2

1–8


Ch 08-p282 Page 282 Wednesday, November 21, 2001 4:19 PM

MA

TH

SQUEST

C H A L L

EN

GE

MA

TH

SQUEST

C H A L L

EN

GE


1 Simplify .

2 Simplify .

3 Simplify .

4 Simplify .

5 Evaluate when x = 3.

6 Draw the graph of y = 5x.

7 Draw the graph of y = 10x.

8 Draw the graph of y = 5 × 2x.

9 Draw the graph of y = × 3x.

10 Draw the graph of y = 2−x.

2

5ab 2–( ) 3– 25 a4b2( )14---

×

36e4d14---–

81e8d 4–( )14---

---------------------------

3 j2k3( ) 2–

6 j3k 2–( )2----------------------- 3 j

6k13---

-------- 1–

÷

18hi12---–

27h2i---------------- 54h

52---i32---

72h12---–i12---

------------------×

3x 2– 27x3------

2× 9x3

3 3x( )2----------------÷

14---

1 A touring bus averages 80 km/h, but in foggy conditions, the driverslows down to an average of 20 fewer km/h. How many minutes laterthan usual will the bus arrive at its next stop 60 km away if there is fog?

2 A square is cut into two sections by a straight line passing through thecentre of the square.

Show that no matter which line is drawn, the areas of the two sectionsare always equal.

3 What are the possible integer values of a and b such that ab = ba?

4 What are the possible integer values of a, b and c such that (ab)c = ? abc



Modelling exponential growth and decay

Exponential functions can be used to model many real situations involving naturalgrowth and decay. Exponential growth is when a quantity grows by a constant per-centage in each fixed period of time. Examples of exponential growth include growthof investment at a certain rate of compound interest and growth in the number of cellsin a bacterial colony.

Exponential decay is when a quantity decreases by a constant percentage in eachfixed period of time. Examples of exponential decay include yearly loss of value of anitem (called depreciation) and radioactive decay.

Both exponential growth and decay can be modelled by exponential functions of thetype y = kax (y = k × ax). The difference is in the value of the base a. When a > 1, thereis exponential growth and when 0 < a < 1 there is exponential decay.

The value of k corresponds to the initial quantity that is growing or decaying. Forexample, an initial number of bacteria in a culture dish, the population of a city at thebeginning of a certain time interval, or the original value of an investment.

The number of bacteria, N, in a Petri dish after x hours is given by the equation N = 50 × 2x.

a Find the initial number of bacteria in the Petri dish.

b Find the number of bacteria in the Petri dish after three hours.

c Draw the graph of the function of N against x.

d Use the graph to estimate the length of time that it will take for the number of bacteria to triple.

THINK WRITE

a Write the formula. a N = 50 × 2x

The initial number of bacteria in the Petri dish is N = 50 × 2x when x = 0. So substitute x = 0 into the given formula and evaluate. (Notice that this is the value of k for equations of the form y = k × ax.)

When x = 0, N = 50 × 20

= 50 × 1= 50

Write the answer in a sentence. The initial number of bacteria in the Petri dish is 50.

1

2

3

16WORKEDExample

MQ10 VIC ch 08 4 Tuesday, November 20, 2001 11:23 AM


THINK WRITE

b To find the number of bacteria in the Petri dish after 3 hours substitute x = 3 into the formula and evaluate.

b When x = 3, N = 50 × 23

= 50 × 8

= 400

Write the answer in a sentence. After 3 hours there are 400 bacteria in the Petri dish.

c Draw a set of axes, labelling the horizontal axis x and the vertical axis N.

c

Plot the points generated by the answers to parts a and b.

Calculate the value of N whenx = 1 and x = 2 and plot the points generated.

At x = 1, N = 50 × 21 At x = 2, N = 50 × 22

= 50 × 2 = 50 × 4

= 100 = 200

Join the points plotted with a smooth curve.

d Triple 50 is 150, so draw a horizontal line fromN = 150 to the curve and from this point draw a vertical line to the x-axis.

d

The point on the x-axis will be the estimate of the time taken for the number of bacteria to triple.

The time taken will be approximately1.6 hours.

1

2

1

2

3

4

x

N

0

N = 50 × 2x

321

200

300

400

500

100

1

x

N

0

N = 50 × 2x

321

200

300

400

500

100

2



A new computer costs $3000. It is estimated that each year it will be losing 12% of the previous year’s value.a Find the value, $V, of the computer after the first year.b Find the value of the computer after the second year.c Find the equation which relates the value of the computer to the number of years, n, it

has been used.d Use your equation to find the value of the computer in 10 years time.

THINK WRITEa State the original value of the computer. a V0 = 3000

Since 12% of the value is being lost each year, the value of the computer will be 88% or (100 − 12)% of the previous year’s value. Therefore, the value after the first year (V1) is 88% of the original cost.

V1 = 88% of 3000= 0.88 × 3000= 2640

Write the answer in a sentence. The value of the computer after 1 year is $2640.

b The value of the computer after the second year, V2, is 88% of the value after the first year.

b V2 = 88% of 2640 = 0.88 × 2640 = 2323.2

Write the answer in a sentence. The value of the computer after the second year is $2323.20.

c The original value is V0. c V0 = 3000The value after the first year, V1, is obtained by multiplying the original value by 0.88.

V1 = 3000 × 0.88

The value after the second year, V2, is obtained by multiplying V1 by 0.88, or by multiplying the original value, V0, by (0.88)2.

V2 = (3000 × 0.88) × 0.88= 3000 × (0.88)2

The value after the third year, V3, is obtained by multiplying V2 by 0.88, or V0 by (0.88)3.

V3 = 3000 × (0.88)2 × 0.88 = 3000 × (0.88)3

By observing the pattern we can generalise as follows: the value after the nth year, Vn, can be obtained by multiplying the original value, V0, by 0.88 n times; that is, by (0.88)n.

Vn = 3000 × (0.88)n

d Substitute n = 10 into the equation obtained in part b to find the value of the computer after 10 years.

d When n = 10, V10 = 3000 × (0.88)10

= 835.50

Write the answer in a sentence. The value of the computer after 10 years is $835.50.

1

2

3

1

2

1

2

3

4

5

1

2

17WORKEDExample


C h a p t e r 8 E x p o n e n t i a l f u n c t i o n s 287Sometimes the relationship between the two variables closely resembles an exponentialpattern, but cannot be described exactly by an exponential function. In such cases, partof the data are used to model the relationship with exponential growth or the decayfunction.

The population of a certain city is shown in the table below.

Assume that the relationship between the population, P, and the year, x, can be modelled by the function P = kax, where x is the number of years after 1975. The value of P must be multiplied by 1000 in order to find the actual population.a State the value of k, which is the population,

in thousands, at the start of the period.b Use a middle point in the data set to find the

value of a, correct to 2 decimal places. Hence, write the formula, connecting the population, P, with the number of years, x, since 1975.

c For the years given, find the size of the population using the formula obtained in part b. Compare it with the actual size of the population in those years.

d Predict the population of the city in the years 2005 and 2010.

Continued over page

Year 1975 1980 1985 1990 1995 2000

Population (× 1000)

128 170 232 316 412 549

THINK WRITEa From the given table, state the value of k

that corresponds to the population of the city in the year 1975.

a k = 128

b Write the given formula for the population of the city.

b P = kax

Replace the value of k with the value found in a.

P = 128 × ax

Using a middle point of the data, replace x with the number of years since 1975 and P with the corresponding value.

Middle point is (1985, 232).When x = 10, P = 232, so

232 = 128 × a10.

1

2

3

18WORKEDExample



THINK WRITE

Solve the equation for a. a10 =

a10 = 1.8125a = a = 1.0613...

Round the answer to two decimal places.

a ≈ 1.06

Rewrite the formula with this value of a.

So P = 128 × (1.06)x

c Draw a table of values and enter the given years, the number of years since 1975, x, and the population for each year, P. Round values of P to the nearest whole number.

c

Comment on the closeness of the fit.

The numbers for the population obtained using the formula closely resemble the actual data.

d Find the value of x, the number of years after 1975.

d For the year 2005, x = 30.

Substitute this value of x into the formula and evaluate.

P = 128 × (1.06)30

P = 735.166 87. . .

Round to the nearest whole number.

P ≈ 735

Answer the question in a sentence. The predicted population for 2005 is 735 000.

Repeat for the year 2010. For the year 2010, x = 35. P = 128 × (1.06)35

P = 983.819 . . . P ≈ 984 The predicted population for 2010 is 984 000.

4232128---------

1.812510

5

6

1Year 1975 1980 1985 1990 1995 2000

x 0 5 10 15 20 25

P 128 171 229 307 411 549

2

1

2

3

4

5

rememberIn the function y = kax:1. k represents the initial amount or quantity.2. a is the base.

If a > 1, the function represents exponential growth.If 0 < a < 1, it represents exponential decay.

3. To find the value of a:(i) in the case of exponential growth add the % increase to 100% and change

the resulting percentage into a decimal(ii) in the case of exponential decay subtract the % decrease from 100% and

change the resulting percentage into a decimal.

remember



Modelling exponential growth and decay

1 The number of micro-organisms, N, in a culture dish after x hours is given by the equa-tion N = 2000 × 3x.a Find the initial number of micro-organisms in the dish.b Find the number of micro-organisms in a dish after 5 hours.c Draw the graph of N against x.d Use the graph to estimate the number of hours needed for the initial number of

micro-organisms to quadruple.

2 The value of an investment (in dollars) after n years is given by A = 5000 × (1.075)n.a Find the size of the initial investment.b Find the value of the investment (to the nearest dollar) after 6 years.c Draw the graph of A against n.d Use the graph to estimate the number of years needed for the initial investment to double.

3a The function P = 300 × (0.89)n represents an:

A exponential growth with the initial amount of 300B exponential growth with the initial amount of 0.89C exponential decay with the initial amount of 300D exponential decay with the initial amount of 0.89E exponential decay with the initial amount of 300 × 0.89

b The relationship between two variables, A and t, is described by the functionA = 45 × (1.095)t, where t is the time, in months, and A is the amount, in dollars.This function indicates:A a monthly growth of $45 B a monthly growth of 9.5 centsC a monthly growth of 1.095% D a monthly growth of 9.5%E a yearly growth of 9.5%

4 A new washing machine cost $950. It is estimated that each year it will be losing 7 percent of the previous year’s value.a Find the value of the machine after the first year. b Find the value of the machine after the second year.c Find the equation that relates the value of the machine, $V, to the number of years,

n, that it has been used.d Use your equation to find the value of the machine in 12 years time.

5 A certain radioactive element decays in such a way that every 50 years the amountpresent decreases by 15%. In 1900, 120 mg of the element was present. a Find the amount present in 1950.b Find the amount present in the year 2000.c Find the rule that connects the amount of the element present, A, with the number of

50-year intervals, t, since 1900.d Find the amount present in the year 2010. Round your answer to 3 decimal places.e Graph the function of A against t.f Use the graph to estimate the half-life of this element (that is, the number of years

needed for half the initial amount to decay).

8F

EXCEL Spreadsheet

Graphs ofthe form

y = a ¥ bx + c


16



17 SkillSH

EET 8.6Ski

llSHEET 8.7



6 When a T-shirt made of a certain fabric iswashed, it loses 2% of its colour.a Find the percentage of colour that

remains after: i two washingsii five washings.

b Write a function for the percentage ofcolour, C, remaining after w washings.

c Draw the graph of C against w.d Use the graph to estimate the number

of washings after which there is only85% of the original colour left.

7 The population of a certain country isshown in the table below.

Assume that the relationship between thepopulation, P, and the year, n, can bemodelled by the formula P = kan, where nis the number of years since 1960.a State the value of k.b Use the middle point of the data set to

find the value of a rounded to twodecimal places. Hence, write the for-mula that connects the two variables,P and n.

c For the years given in the table, find the size of the population, using your formula.Compare the numbers obtained with the actual size of the population.

d Predict the population of the country in the year 2005.

8 The temperature in the room, recorded at 10-minute intervals after the airconditionerwas turned on, is shown in the table below.

Assume that the relationship between the temperature, T, and the time, t, can be mod-elled by the formula T = cat, where t is the time, in minutes, since the airconditionerwas turned on.a State the value of c.b Use the middle point in the data set to find the value of a to 2 decimal places.c Write the rule connecting T and t.d Using your rule, find the temperature in the room 10, 20, 30 and 40 minutes after

the airconditioner was turned on and compare your numbers with the recorded tem-perature. Comment on your findings. (Give answers correct to 1 decimal place.)

Year 1960 1965 1970 1975 1980

Population (in millions)

118 130 144 160 178

Time (min) 0 10 20 30 40

Temperature 32 26 21 18 17


18

WorkS

HEET 8.3



A growing investmentAt the start of this chapter we met Bindi who was investing $20 000 in a fixed term deposit earning 6% p.a. interest. When Bindi has $30 000 she intends to put a deposit on a house.

1 Write an exponential function that will model the growth of Bindi’s investment.

2 Use your graphics calculator to graph this function.3 Find the length of time (correct to the nearest year) that it will

take for Bindi’s investment to grow to $30 000.4 Suppose that Bindi had been able to invest at 8% p.a. How

much quicker would Bindi’s investment have grown to the $30 000 she needs?

5 Calvin has $15 000 to invest. Find the interest rate at which Calvin must invest his money, if his investment is to grow to $30 000 in less than 8 years.

Qualifications:Bachelor of Commerce (Accounting/ Commercial Law)

I studied accounting in Year 11 and really enjoyed it. I was always much better with ‘numbers’ than ‘words’. Also, I find thata lot of accounting work involves investigating and solving problems whichI enjoy doing.

Some of the ways I use mathematics in my work are:• Working out loan repayments based on various

variables; that is, the amount of the loan, the interest rate, the repayment period and so on.

• Taxation calculations, including income tax payable and fringe benefit tax. These calculations involve applying different tax rates.

• Calculating shiftworkers’ penalty rates, overtime, holiday and sickness rates.During a typical day I may be involved in

salary calculations, adjustments, queries, account reconciliation, ‘testing’ of the accounting system, calculating cash flows and preparing management reports. I find mathematics challenging and practical and I use it every day.

Questions1. List three tasks Donna may need to work on

in her job.2. What tertiary courses are available to

become an accountant?3. Find out what subjects are prerequisites for

studying accounting at a tertiary institution.

Career profileD O N N A P U L L E N — A c c o u n t i n g M a n a g e r

MQ10 VIC ch 08 Page 291 Friday, November 30, 2001 10:40 AM


Copy the sentences below. Fill in the gaps by choosing the correct word or expression from the word list that follows.

1 To multiply numbers with the same base, the indices are : am × an = am + n.

2 To divide numbers with the same base, the indices are : am ÷ an = am − n.

3 To remove brackets, the indices inside the bracket by theindex outside the bracket: (am)n = am × n.

4 Any number that has an index of is equal to 1: a0 = 1.5 To remove a bracket containing a product or a fraction, multiply the index

of every inside the bracket by the index outside the bracket:

(ab)m = ambm or .

6 A number with a negative index is equal to the of the same

number with a positive index: .

7 Numbers with fractional indices can be expressed as .

8 The denominator of the fractional index indicates the or

type of root: .

9 Functions of the form y = ax are called functions with base a,where a is a real number, not equal to 1.

10 For the graph of y = kax, the larger the value of k, the the graph.11 If k < 0, the graph is in the x-axis.12 When a > 1, the function as x increases. The the

value of a, the steeper the graph.13 When 0 < a < 1, the function as x increases. The

the value of a, the less steep the graph.14 The graph of y = ax has a horizontal with equation y = 0.15 Exponential growth or decay can be modelled with the function y = k × ax.

i The value k represents the or quantity.ii If the base a > 1, the function represents exponential .iii If 0 < a < 1, the function represents exponential .

summary

ab---

m am

bm-------=

a n– 1an-----=

amn----

an( )mamn= =

W O R D L I S Taddedsmallerdecayzeroinitial amount

multiplyreciprocalincreaseslargerexponential

powersurdssubtractedreflectedasymptote

steeperfactorgrowthdecreases




a b

c d


a b

3 Simplify each of the following and express your answer with positive indices.

a bc


a b

c


a b

c

6 Evaluate each of the following, without using a calculator. Show all working.

a b

7 Simplify.

a b


a b

c

8A

CHAPTERreview

5x3 3x5y4× 35---× x2y6 26a4b6c5

12a3b3c3-----------------------

20m5n2

6------------------

3 14 p7

21q3------------

4

8A5a0 2a

3------

0– 12+ 3b( )0– 4b( )0

2-------------–

8B2a 5– b2 4a 6– b 4–× 4x 5– y 3– 20x12y 5–÷2m 3– n2( ) 4–

8B12---

3–2 3( ) 3–× 9

2---

2×

4 3– 58 2–-------× 5–

8C2a

45---b

12---

3a12---b

34---

× 5a34---b

25---

× 43x34---y

19---

16x45---y

13---

-----------------

4a13---

b3--------

12---

8C16

34---

8114---

×

6 1612---

×----------------------- 125

23---

2723---

–

12---

8Ca93 16a8b24 3 a5( )15

–+ 32x5y105 64x3y63+

8D5a 2– b( ) 3– 4a6b 2–×2a2b3 5 2– a 3– b 6–×----------------------------------------------- 2x4y 5–

3y6x 2–---------------- 4xy 2–

3x 6– y3----------------

3–×

2m3n4

5m12---n

--------------- 1

3--- 4m

13---n 2–

523---–

------------------ 1

2---–

÷



9 Simplify each of the following and then evaluate.

a b

10 If m = 2, find the value of:

11 For the exponential function y = 5x:a complete the table of values below

b plot the graph.

12 Draw the graph of y = 10 × 3x for −4 ≤ x ≤ 4.

13 Draw the graph of y = 10−x for −4 ≤ x ≤ 4.

14 a On the same axes draw the graphs of y = (1.2)x and y = (1.5)x.b Use your answer to part a to explain the effect of changing the value of a in the equation

of y = ax.

15 a On the one set of axes draw the graphs of y = 2 × 3x, y = 5 × 3x and y = × 3x.b Use your answer to part a to explain the effect of changing the value of k in the equation

y = kax.

16 a On the same set of axes sketch the graphs of y = (2.5)x and y = (2.5)−x.b Use your answer to part a to explain the effect of a negative index on the equation y = ax.

17 A radioactive substance decays so that its mass, m, in grams after t days is given bym = 3.5 × 2−0.2t. a Find the initial mass of the substance.b Find the mass of the substance after 4 days.c Graph the function.d Use the graph to estimate the number of days needed for the mass of the substance to

reduce to of the original mass.

18 A certain investment pays 6.5% p.a.a Construct the formula which gives the total value, A, of the investment after n years, if

the original amount invested was A0.b If $5000 was invested, find the total value after 7 years.c Graph the function.d Use your graph to find the number of years needed for the initial investment of $5000 to

double in size.

x −3 −2 −1 0 1 2 3

y

8D

3 56×( )12---

332---

× 5 2– 36 512---–

×

0

+× 6 3 2–×( ) 1–3

12---

613---

×

6

62– 3 3–( )0×------------------------------÷

8D6a3m 2b2m× 3ab( ) m–×

4b( )m 9a4m( )12---

×

---------------------------------------------------------

8E

8E8E8E

8E12---

8E

8F

110------

8F

testtest

CHAPTERyyourselfourself

testyyourselfourself

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