Mathematics for IB Diploma Applications and Interpretation HL 2 Worked Solutions [1 ed.] 9781925489873

Citation preview

MATHEMATICS

Applications and Interpretation HL

2

MichaellHaese

MarlqHumphries}

(ChrisiSangwin [NgodVo)

for use with

IB Diploma Programme Mark Humphries Ngoc Vo

Joseph Small Michael Mampusti

dINAOM

HAESE

SNOILNTOS

%

MATHEMATICS: APPLICATIONS AND INTERPRETATION Mark Humphries

B.Sc.(Hons.)

Ngoc Vo Michael Mampusti

B.Ma.Sc. B.Ma.Adv.(Hons.), Ph.D.

Joseph Small

HL WORKED

SOLUTIONS

B.Ma.Sc.

Published by:

Haese Mathematics

152 Richmond Road, Marleston, SA 5033, AUSTRALIA

Telephone:

Email: Web:

+61 8 8210 4666

[email protected] www.haesemathematics.com

National Library of Australia Card Number & ISBN

978-1-925489-87-3

© Haese & Harris Publications 2020

First Edition

2020

Artwork by Brian Houston and Bronson Mathews.

Typeset in Australia by Charlotte Frost and Deanne Gallasch. Typeset in Times Roman 10. This book is available on Snowflake only.

This book has been developed independently of the International Baccalaureate Organization (IBO). This book i in no way connected with, or endorsed by, the TBO. This book is copyright. Except as permitted by the Copyright Act (any fair dealing for the purposes of private study, rescarch, criticism or review), no part of this publication may be reproduced. stored in a retricval system,

or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise,

without the prior permission of the publisher. Enquiries to be made to Hacse Mathematics.

Copying for educational purposes: Where copies of part or the whole of the book are made under Part VB of the Copyright Act, the law requires that the educational institution or the body that administers it has given a

remuneration notice to Copyright Agency Limited (CAL). For information, contact the Copyright Agency Limited.

Acknowledgements: While cvery attempt has been made to trace and acknowledge copyright, the authors and

publishers apologise for any accidental infringement where copyright has proved uniraceable. They would be pleased to come to a suitable agreement with the rightful owner. Disclaimer: All the intemet addresses (URLS) given in this book were valid at the time of publication. While the authors and publisher regret any inconvenience that changes of address may cause readers, no responsibility for any such changes can be accepted by either the authors or the publisher.

FOREWORD This book gives you fully worked solutions for every question in Exercises, Review Sets, Activities, and Investigations (which do not involve student experimentation) in cach chapter of our textbook

Mathematics: Applications and Interpretation HL.

Correct answers can sometimes be obtained by different methods. In this book, where applicable, each

worked solution is modelled on the worked example in the textbook.

Be aware of the limitations of calculators and computer modelling packages. Understand that when your

calculator gives an answer that is different from the answer you find in the book. you have not necessarily

made a mistake, but the book may not be wrong either.

Please contact us if you notice any errors in this book.

MH

e-mail: web:

[email protected] www.haesemathematics.com

JS

NV

MM

TABLE OF CONTENTS

Chapter 1

EXPONENTIALS

Chapter 2

LOGARITHMS

Chapter 3

Chapter 4

68

APPROXIMATIONS

AND

ERROR

99

Chapter 5

LOANS AND ANNUITIES

114

MODELLING

141

Chapter 6

DIRECT AND INVERSE VARIATION

180

Chapter 7

BIVARIATE STATISTICS

207

NON-LINEAR MODELLING

260

VECTORS VECTOR APPLICATIONS

297

COMPLEX NUMBERS

434

Chapter 12

MATRICES

504

Chapter 13

EIGENVALUES AND EIGENVECTORS

555

Chapter 14

AFFINE TRANSFORMATIONS

616

Chapter 15

GRAPH THEORY

671

Chapter 16

VORONOI

735

Chapter 17

INTRODUCTION TO DIFFERENTIAL CALCULUS

777

Chapter 18

RULES OF DIFFERENTIATION

807

Chapter 19

PROPERTIES OF CURVES

Chapter 20

APPLICATIONS

875 964

Chapter 8

Chapter 9

Chapter 10 Chapter 11

Chapter 21

DIAGRAMS

INTRODUCTION

OF DIFFERENTIATION TO INTEGRATION

386

1015

Chapter 22

TECHNIQUES FOR INTEGRATION

1041

Chapter 23

DEFINITE INTEGRALS

1076

Chapter 24

KINEMATICS

1135

Chapter 25

DIFFERENTIAL EQUATIONS

1190

COUPLED

1232

Chapter 26 Chapter 27

DIFFERENTIAL EQUATIONS

DISCRETE RANDOM

VARIABLES

1282

Chapter 28

THE NORMAL DISTRIBUTION

1332

Chapter 29

ESTIMATION AND CONFIDENCE INTERVALS

1380

Chapter 30

HYPOTHESIS TESTING

1418

Chapter 31

x? HYPOTHESIS TESTS

1468

Chapter 1

EXPONENTIALS

2'x 2 —gtt3

X

—27%

=93

h (V2 3_= (27)(o518

B

72-5>

g

2

¢ 167 = (2)%

=2°

=2

=8

2 3 d 257 = (5%)7 =5

=

=k27

i

~1

= (273

=i 2

g 9 =372 =33

L4z

=271

=125 3

S

Chapter 1 (Exponentials)

Exercise IB_

7

EXERCISE 1

) 11

a

22xz? — ,T

2

a

3

b

1

zZxz2

=27 &

P +222+1) =2?x2®+a? x2?+a? x1

1o

2zt ?

b

or

VT

27(2"+1) =2"x2"4+2"x1

=27 42"

:

d

B B 2 ,z]xzu +z2xT =z

zxz ?

=l :z

0

=2°+22' +2* €

_3

¢

T(T7+2)

=T"xT+T" )

%2

+z

=z+1

e

=

3*(2-37)

—3Tx3

v:;:;z

33

*3

f

"

=z?

=2(3)~

£

27%(2°+5)

e B

h

3§

=145277) 1

i,

x’z

i

*XT

+T

=5

I

a

s 1 =z?+4+z7+1 1 1 oz %222 —z+527) : 1

=2

*x2?—z P

2XT

+5%)

x5% 4577 x 5%

3B +5+377)

=3 X3 +F X543 x3F :3h+5(3,)+30

=3 +5(3%) +1

'

*xzl4+z? x527

1

:

2227 3272

L e

— gt _g(g2e) 90 — 9% 397 1

=227 — 27 +52° 3

11

X222 +122 X3z

=5+1

1

+T

57%(5%

=27 427 +20

k

1

=5"+50

2 i@’ +a+a?)

_13

Xz? +x?

=2"+20+3

=20 45(277)

i

Loz

=72 +2z' +32°

=230=1 g

iz L + 32 _i7) 2% (22 +227

1

=222 —2? +5

3

a2

(2-DE'+3)

=2"x2"42"x3-1x2"-3

=22 4+ 2(2%) -3

oy

getl 3

b

(3 +2)(3"+5)

=3

x3"+3"x5+2x37+10

=3% +7(3%) + 10

_i

?

Chapter 1 (Exponentials)

Exercise 1B

(5" ~2)(5" ~ 1)

=5"%x5"—5"

d

(2°+3)?

=(2°)2 2"+2x x3+3%

x4-2x5"+8

=5% —6(5%) +8

=22 46(2°) +9 (1t +7)?

@ -1?

=)

=(3°)?-2x3"x1+12

=3 -2(3") +1

+2x 4" x T+17

=477 1 14(4%) + 49

(F +2)(T —2)

h

(2 +3)2-3)

=277 -3 =229

=4-9

i

(e =

k

=

77%)2

=P

=2 RTR T+

=7 _2x 7P +77% =7 _2477%

5% 45

=57 X 57+ 5 =5°(5"+1)

1

(777)?

b

s g —5x5"—5 5(5" — 1) x;m -

x

y

2 +4+I—24

-2 =52-2x5x27"+(277)? =25-10(27") +27%

=3"(3%+1)

™ =TT =T +7*

6" 6

4 16

3n+2 4 gn

“©

=343

=6x6"" -6

=66

=42x4"-16

=16x4"-16 =16(4" - 1)

—1)

gntl | gn—t

92 _ on+3

=

2

=10(3")

I

8

x 2P

=2"x2"-2"x8

=27(2" -8)

="t

—on 1y

4fl+1

— (@21 gmnmt

x 22 4ot gqon-l

=5(2"1)

9°—4

4725

—

= (3 +2)(3° - 2)

=(2°+5)(2° - 5)

(21)2

— o2

g1

—92n—1

3 93 4 g2n—1

=271 g4 221 =9(2*"1) 16— 9°


+6(2) +9

= (3 +27)(3" - 27)

= (5+2)(5-2) 97 +10(3%) +25

h

=(3%)2+10(3%) +25

4462749 =(2*+3)?

{a® +6a+9=(a+3)*}

47— 14(27) +49 =(2%) —14(2%) +49

=@-7? {a® - 14a+49 = (a—T7)%}

= (3 +5)* {a® +10a + 25 = (a+5)%}

257 —4(5%) +4

=(5%)2 - 4(5%) +4

=(5"-2)*

{a® —da+4=(a—2)%}

(252 —22 -2 =(2°+1)(2°-2)

b

£ -7(2%) +12

d

{a?—a-2=(a+1)(a—2)} = (22 -7(2%) +12 {a® - Ta+12=(a—3)(a—14)}

47 4+9(2%) +18 = (292 +9(2°) +18 =(2°+3)(2° +6)

{a®*+9a+18 = (a+3)(a+6)}

f

=(2°)2 -2 -20

=(3+3)(3*-2)

{a*+a—6=(a+3)(a—2)}

=(2"-3)(2"-4) 47 —2720

(3°)2+3°-6

97 +9(3%) +14

=(3°)2+9(3°) + 14

=(+2)(3+7)

=(2°+4)(2" - 5) {a® —a—20=(a+4)(a—5)}

{a®>+9a+14=(a+2)(a+7)}

h

=(3"+5)(3"-1)

25" 45 -2 =) +5" -2 =(5*+2)(5° - 1)

{a® +4a—5=(a+5)(a—1)}

{®+a-2=(a+2)(a-1)}

97 +4(3%) -5

=(3)+43) -5

497 — 7" 112 =(7)?-17(7") +12

= (7" —4)(7" —3)

{a® —Ta+12=(a—4)(a—3)} 12" 6"

e12"6"

_2"6"

_(12\"

—on

—on

=

=3

b

X

-

-

2910%

= 10"

4

2

2a

L@

=107

a

9

10

Chapter | (Exponentials) Exercise 1B 60

cC

3

6"

"

2bgh

3

= T

=

=3

=3

—3b

6\b

77

or

5°7%

g

7z

am\®

24k

T

(4"

('zn)

-0

_(1\"

F

6

or

8_“

_

=)

30

=3

(gye 3

sntl

h

_

=

=(%)(2\k

_ 5!

=)

=9

—

_

e s

W

=

_ 293¢

=5* or

"

=!

=)

9

w

4n

T anpn

35°

_

an

=

(7)

_gb

357

e

d

= =g

(8

i

8

6 4 am

a

—

om

2n 4 19n

—

b

on

o1

an

_ gmgm 4 om

_2n 4 onen

_amam

_2m@Em 1) T

_2M(146m) T

_2n(an 42 T

-

om

-

3" 1

4

19t

3¢

3¢

on

-

=1+6"

.

e

142n _6npenan

g

T T 1+om

—5 _ six5_5" =

B

h

q4n —

g

=

=2

on

g

n(m —1) o

=4

=271

omen

o 1 5% x5 — 5" 4

_en142m)

sn+l _gn

on

=442

T+

S

8


0, the graph lies above the asymptote. k0.

16

Chapter 1 (Exponentials)

b

i

Exercise 1E

Each point will become 1 times its previous distance from the y-axis.

q

If 0 < g 1, g =1, the graph remains unchanged. il

the graph becomes steeper, and if

The graph is stretched horizontally with invariant y-axis and scale factor l.

q

iii The horizontal asymptote y = 0 will remain the same.
2.

y =27 y = 10"

a#1,

have

above

horizontal asymptote and are increasing.

as

a>0,

y =

the

0

y = 2%

corresponds to €, and corresponds to B.

the graphs of y =a "

and

y = (—1~)

a

are

Chapter 1 (Exponentials) y=-5%

has

k

1.

the graph lies below the horizontal asymptote y =0 and is decreasing. y = —5"

d y=(4)"

corresponds to E.

has k>0

and 0—3}.

Chapter 1 (Exponentials) e

Exercise 1E

y=2-2°

i When z=0,

y=2-2°

il The horizontal asymptote is y = 2.

—2-1

=1

the y-intercept is 1.

il

When

When

z=2,

y=2-2% =2-4 ==2 z=-2, y=2-2"2 =2-= 1

v

=8_1

i~3

=1 1 v f

The domainis

{z |z €R}.

Therangeis

{y|y3}.

21

22

Chapter 1 (Exponentials)

g

Exercise 1E

y=3-277 i

When

=0,

y=3-2°

il

=3-1

The horizontal asymptote is y = 3.

=3

the y-intercept is 2.

iiil When

z=2,

y=3-272 =35

iv

1

=12 _1

i1 _u 1

When z=-2,

y=3-2"02 =322

=3-4 =-1 v

The domainis

{x |z € R}.

Therangeis

{y|y < 3}.

ix3T41

hy

i When

o=0,

y=—-§x3"+1 -3 x1+1

i The horizontal asymptote is y = 1.

2

the y-intercept is . ili

v

When

z =2,

When

z=-2,

y=-1x3"?+1

The domain is {x | x € R}.

The rangeis

{y|y < 1}.

Chapter 1 (Exponentials) Exercise IE a

y=kx2+c Substituting

(0, —5)

into the equation gives

—5=kx2"+c o k+e=—5

Substituting

.. (1)

(2, 10)

into the equation gives

10=kx2+c ©.

dk+c=10

Now

.. (2

(2)— (1) gives

3k=15 . k=5 andso c=-10.

b y=5x27-10 When

10

a

z=6,

y=5x2°—10

=5x64—10 =310

f(0)=35-ad" 35-1 =25 the y-intercept is 2.5.

Pis (0,25). b

The point

(—1,2)

lies on the graph.

o f(-1)=2

© 35-a'=2

- a=15

¢

1

f(z)=3.5—1.5"%

has horizontal asymptote

a y=kx2+c Substituting

(0, 4)

into the equation gives

4=kx2+c

kt+e=4 Substituting

s Now

.. (1)

(2, 7)

into the equation gives

T=kx2+c

dk+e=T7

(2)—(1)

.2

gives

3k=3

. k=1

andso

the exponential model is y = 2% + 3.

c=3.

y =3.5.

23

24

Chapter| (Exponentials) b

Exercise 1E

y=kx3+c Substituting

(—1, 3)

into the equation gives

3=kx3'+¢

tkte=3 Substituting

.. ()

(1, —13)

into the equation gives

-13=kx3"+c © 3k4+e=-13 ... (2) Now

(2)—(1)

gives

$k=-16 . k=-6

andso

c=5.

the exponential model is y = —6 x 37 + 5.

¢

y=kx2"%+c Substituting

(-2, 9)

into the equation gives

9=kx2"4¢ =kx2¥+ec L dkte=9 .. (1) Substituting

(1, —5)

into the equation gives

—5=kx2"'+c

lkte=-5 Now

(1)—(2)

..

gives

k=14 .

k=4

the exponential model is

@

|

The

graph

lies

c=-T.

y =4 x 277 — 7.

above

the

horizontal

asymptote.

k must be positive.

il

The horizontal asymptote lies below the T-axis.

¢ must be negative.

y=kx4"

+c

Substituting (1, —4)

into the equation gives

—4

=k x4' 4+¢

s dkte=—4 .. (D) Substituting (2, 2) into the equation gives 2=k x 4%+ ¢ 16k+c=2 ... (2) gives

12k=6

- k=

i &

(2)— (1)

&

Now

£

b

m_

12

andso

the exponential function is y = 4 x 47 — 6.

Chapter 1 (Exponentials)

=0,

y=3x4"—6

1-6

=

x

ol

¢ When

=

5 T the y-intercept is —4%.

d The horizontal asymptote of y =4 x 4% —6 is y=—6. 13

f(r)=kxa"+c The graph has horizontal asymptote The y-intercept is 10, so

.

y =2,

so

c¢=2.

f(0) = 10

kxa®+2=10 . k=8

The graph passes through

(5, 258),

so

f(5) = 258

. 8xa°+2=258

. 8a° =256 s G082

;. a=2

So, the exponential function is f(z) =8 x 27 +2.

14 fz)=kxa*+c The graph has horizontal asymptote The y-interceptis 1, so

y =4,

so

c=4.

f(0) =1

L kxa®+4=1 . k=-3

The graph passes through (L. 2), so f(1) =2 . —3xa'+4=2

a . a=}

)

So, the exponential function is f(z) = —3 x ()" +4.

15

f(z) =ka"+c f)=11 - kate=11 f2)=17 f3)=29 Now

(2)—(1)

.

ka’+c=17 ka®+c=29

gives

.. (1) .. (2

3)

ka®>—ka=6 o kala—1)=6

and (3)—(2) gives ka® —ka® =12

.. (4)

. ka*(a—1)=12

. alka(a—1)] =12 6a=12 s a=2

{using (4)}

Exercise IE

25

26

Chapter 1 (Exponentials)

Exercise

Substituting a = 2 into (4) gives Substituting @ =2 and k=3

1E

k(2)(1

k=3

into (1) gives

3(2)+c= 11 .

..

c=5

the exponential function is f(x) = 3 x 2% + 5.

f0)=3x2045=8 the y-intercept is 8.

16

f(z)=ka " +c f-2)=21 ka®+e=21 f(Hy=0 ©okal4e=0 f@)=-3 L ka?+e=-% Now

(1)—(2)

gives

.. (1) .. (2) ..(Q3)

ka® —ka ' =21

ka® — k= 21a . k(a®—1)=21a 2la 2 ln—'L:,_] and

(2)—(3)

gives

ka™'

- (4)

—

.. (5)

.

+

2la

Equating (4) and (5) gives

3a?

Fo1-

Using technology,

26D

a =2

{a>0}

Substituting a =2 into (4) gives k= 2‘3152)1 S k=6 Substituting

a =2

and

k =6

2

into (1) gives

6(2)? +c¢ =21 . 24+c¢=21 Loe=-3

So, the exponential function is f(z) =6 x 27% — 3.

a

f(0)=6x20-3

=3

the y-intercept is 3. b

The horizontal asymptote is y = —3.

1

a

Exercise 1F

il

and

ii

|

when

2" =06

when

Wegraph

Y,

z~16. z~—0.7.

=2%

and

Yy

=3

on the same set of axes, and find their

point of intersection.

[EXE]:Show coordinates

¥1=2°(x)

v2-3

"

We graph

%=1.584902501

Wegraph

Y; =2%

B [EXETshon coordinates ¥1=2%(x)

[T 7309055042 ¥=0.0. v

Yo =20

B Texcishon coordimates y2=20" = T

e it amiopaoms_vzo

on

The solution is z ~ 4.32.

We graph

Y; = 4%

and

v

Yz =100

on

the same set of axes, and find their point of intersection. @

wisecr|

msect

The solution is 2 ~ —0.737.

the same set of axes, and find their point

of intersection.

m

[

_1¥=3

and

Yy = 0.6

point of intersection.

/

The solution is z ~ 1.58.

Y, = 2%

on the same set of axes, and find their

44‘_”““

a

27

From the graph:

i 2" =3

b

Chapter 1 (Exponentials)

[EXE]:Show coords

=3.321926005_

¥=100

The solution is 2 ~ 3.32.

Chapter | (Exponentials)

¢

Wegraph

Y; =3%

and

Y, =30

We graph Y; = (1.2)% and Y2 =3 on

on

the same set of axes, and find their point of intersection.

= o [x=3. 085003274

e

Exercise 1F

the same set of axes, and find their point

of intersection.

——wTSECT 5

o0

The solution is z ~ 3.10.

The solution is z ~ 6.03.

Wegraph Y; = (1.04)% and Y» =4.238

We graph Y; = (0.9)% and Y =0.5 on

on the same set of axes, and find their

the same set of axes, and find their point

The solution is z ~ 36.8.

The solution is z =~ 6.58.

point of intersection.

a Wegraph Y; =3x2X

and Y, =93 on

the same set of axes, and find their point

of intersection.

The solution is = ~ 4.95.

¢ We graph

Y; =8x 3%

and

Yo =120

point of intersection.

v-120

The solution is z ~ 2.46.

b

We graph Y = 40x(0.8)X and Y5 = 10

on the same set of axes, and find their point of intersection.

The solution is = ~ 6.21.

on the same set of axes, and find their

— 5 [1=2.400073821

of intersection.

We graph Y2

=

34

Y; = 21 x (1.05)%

and

on the same set of axes, and

find their point of intersection.

et The solution is z ~ 9.88.

Chapter1 (Exponentials) Exercise IF e

We graph Y; =500 x (0.95)% and

We graph

Y2 = 350

Y> = 470 on the same set of axes, and find their point of intersection.

on the same set of axes, and

find their point of intersection.

Y; = 250 x (1.125)

[=5.350815043 The solution is z ~ 6.95.

g

Wegraph

Y; =3%+5

and

Y =50

point of intersection.

[B_[ExEl:Show coordinates

and

Y2 = 30 on the same set of axes, and find their point of intersection.

[x=7. 228289519 o, Oy-s0 The solution is x ~ 7.23.

When z=-2,

y=5 5=4xa2-7 12=4a"2 a® = La=m

{a >0}

and

Y> = 80 on the same set of axes, and find their point of intersection.

The solution is =~ 1.71.

Y; =20+ 80 x (0.75)%

a>0

Y, = 60+ 10 x (1.5)X

P

v=50

y=4xa"-7,

We graph

— 5Tt x=1.700511201 "¥=80

The solution is z ~ 3.46.

i We graph

y=470

[8_[EXE]:Showa coordinates B < B oo vo=so

——= o —+—+—tmeect|

[x=3.464973821

and

The solution is z =~ 5.36.

on the same set of axes, and find their

a

29

1 :

Chapter 1 (Exponentials) Exercise 1F b

,1/:4x(fi)’—7 The z-intercept occurs where

y =0

8 /l>0,

d>0

As the intensity of light diminishes as the depth increases, we would expect that

0 < a < 1.

40

Chapter | (Exponentials)

b

Exercise 1G.2

When d=1, L=95 © 10xa' =95 .


0 So, the domainis {z | > 0} and the rangeis il

As

z— 0%,

f(z) — —oc,

As

z—

f(x)— oc,

f(0) When

o0,

so the vertical asymptote is x = 0. so there is no horizontal asymptote.

is undefined, so there is no y-intercept. f(z) =0,

logz=2

v @=10

oz =100 So, the z-intercept is 100.

{y |y € R}.

Iny=3-=z

- oy=etT @) =T

79

80

Chapter 2 (Logarithms)

b

Exercise 2D

f(z) =log(z+1) i

log(x+1)

is defined when

z+1>

So, the domain is {x |z > —1} il

As z — —1%, As

x — o0,

f(z) — —oc, f(x)— oo,

f(0) =log1 =0, When

f(z)=0,

0, thatis, when

= > —1

and the range is {y |y € R}. so the vertical asymptote is = = —1.

so there is no horizontal asymptote.

so the y-intercept is 0. log(z+1)=0

L z+1=10° . z=0

So, the z-intercept is 0.

¢

f(z)=2logz+1 i il

logz is defined when

= >0

So, the domain is {x | z > 0} and the range is {y | y € R}. As

= — 0%,

f(z)— —oc,

so the vertical asymptote is = = 0.

As & — o0, f(z) — oo, so there is no horizontal asymptote. f(0) is undefined, so there is no y-intercept.

When f(z) =0,

2logz+1=0

- 2logr=—1 - logz=-%

s

LT=10T7 = s So, the z-intercept is V‘E'

Chapter 2 (Logarithms)

d

Exercise 2D

f(z)=1-logz i

logx is defined when

z > 0

So, the domain is {x | 2 >0} ii

As

2 — 0%,

f(xr) — oc,

As

x—

f(x)— —o00,

00,

and the range is {y |y € R}.

so the vertical asymptote is = = 0. so there is no horizontal asymptote.

£(0) is undefined, so there is no y-intercept.

When

f(z)=0,

1—logz=0

logz =1

So, the a-intercept is 10. i

x =10

y

y=1-loga

3

a

For

y=Inz,

when

y =0,

Inx

z=1 Ais

y=Inz

asits z-intercept is 1,

so B must be y = In(z — 2).

b

y =In(z —2)

is a horizontal translation of

y=Inz, 2 units to the right. y = In(z +2) is a horizontal translation of y=Inz,

2 units to the left.

y=In(z+2)

81

82

Chapter 2 (Logarithms) Exercise 2D ¢

y=Inz As

has domain

x — 0%,

y—

y=In(z —2)

{x |z >0}

—o0,

so

has domain

As

x— 2T,

y — —o0,

As

¢ — 2%,

y=Inz

has vertical asymptote

{z |z >2}. so

y=In(z—2)

has vertical asymptote

y=1In(z +2) has domain {z|a > —2}. 4

y=In(z?),

=2z

y—

z = 0.

—o0,

so

y=In(z+2)

z = 2.

has vertical asymptote

z =

>0

{mlnb=1In(®™)}

So, the y-values are twice as large for y = In(z?) as they are for y = Inx. Therefore, yes, Kelly is correct.

5

a y=In@®) = 3lna isa . vertical stretch ; of y=Inz ¥

b y — 1..(1)z — ()

with scale factor 3.

reflection of y =Inz ¥

y=In(z%

= ~Ing isa in the z-axis.

y=Inz

¢

y=In(z+e) of y=Inz,

y=Inz

is a horizontal translation

d

e units to the left. y=In(z +e)

l-ex~-172

e

For £>0,

y=In@a®) +2=2z+2

upwards. For z 0,

in

y=In(z—2)—3

is a

y=Inxz

2 3 )

through

translation of

Chapter 2 (Logarithms)

6

a ln(es?) =lne+In(z?) —1+42lne {z>0}

7

a When

=2,

y=b

Y

Exercise 2D

y=a+2logz

b=>5—1log2

=log(10°) —log2

= log(1232) = log 50 000

b When =2, y=b=1og50000 {froma} *. a+2log2 = log50000 . a=1log50000 — 2log2 = log 50 000 — log(2?)

= log((20) =log 12500

8 fizx—e¥ a

gizm2-1

f is defined by - flisdefinedby

y = e** z=e%

© 2y=Inz

b

(9o f)(x) =g(f(z)) =g(e*)

=2e* —1 (go f)(x) is definedby

= (gof)"\(z)

isdefined by

y=2e*"—1

z=2¢* —1

oz 4+1=2e%

y_T+1 2v_zt

=1m@2z-1)

83

84

Chapter 2 (Logarithms)

Investigation 2

Logarithmic scales

|

INVESTIGATION 2 1

Ay 100

70

/L]

S

oo}—— e

’/

i

:

(]

=Y

20

2

As the y-values approach a power of 10, the minor tick marks become closer together. The spacing of the y-values 1, 2, ..... 9, and the y-values 10, 20, ...., 90, matches the spacing of the numbers on the slide rules in the Opening Problem.

3

From the graph, each distance measures ~ 4.7 cm. log 10 — log 1

log 20 — log2

log 50 — log 5

—10

= log(%)

=1

=log 10

=log10

=

=1

= log(%)

Chapter 2 (Logarithms)

&

log(5x 10%) = log5+log(10°)

and

Exercise 2E

log(10%%) = 2.5

=log5+2 =270

Since log(5 x 10%) > log(10*®),

1

a

5 x 10

is further to the right on the logarithmic scale.

P corresponds to the value

107" i = 5ok

Q corresponds to the value

10~! = 5

b

i log(k=)

=1

=3

*(W):f("’ 2

R

il

log(fi)

=log R

P

)

3

R 2

Qi -1

i 0

R 1

2

—log10

=logR—1
Cy,

vinegar is more acidic.

s Cv=10724

~0.00398

88

Chapter 2 (Logarithms)

Exercise 2E

il If C=200Cy,

iii

pH = —log(200Cy) = —1log 200 — log Cy = —1log 200 + 6.6 ~4.30

The H30™ concentration in vinegar ga is about

10724~ (=430) ~ 10190 x~ 79.2

times

greater than in the substance in ii.

d

If C>1,

then

logC >0

. pH=—logC

=h Fy

.

FA . log (F_o> =g =249

Ep _ 24

y

. FA=10*F, Now,

the brightness of an object varies inversely with the square of its distance from

the observer, so if star B is 2 times further away than star A but has the same apparent

magnitude, then star B must be 22 =4 times brighter than star A. Fy = 4F), 2.4, . the absolute magnitude of star Bis M = —2.5log (%’”’) 0

= —2.5log(4 x 10*4) = —2.5(log 4 + log(10*4))

= —2.5(log 4+ 2.4) ~ -7.51

b

From 2 d i, the observed flux density of the Sun is 10696 F;, If the Sun was 10 parsecs away from the Earth, then the observed flux density would be

T

—6

.

P

. the Sun has absolute magnitude ~

2

—2.5log (M)

14

0

~ —2.5log(2.35 x 1072:304)

~ —2.5(log 2.35 — 2.304)

~4.83

1

a

logvi0

b

=— log(10? 3

=loxtllF) a

log 27

3

a 32=10%*

~1.431

~ 1015051

¢

~log(107%)

—1 -3 b

log(0.58)




b

= i“(fs) B 91“(3 ) :1n(144x )

iln9-In2 1

S -2 —In3-In2

== 5

a

=log

a

log16 +2log3

b

16 +log(3?)

log36

b

= log(4 . x 9) = log(2?

= 1“(525) -u(Z)

log16 —2log 3

= log 16 — log(3?)

loghd