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5

Exponential and Logarithmic Functions

5

Exponential and Logarithmic Functions

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

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Activity

Activity 5.1 (p.216)

1. 1

2. No

3. y increases as x

increases; the graph
lies above the x-axis; it
cuts the y-axis at (0, 1).
4.

x –1 0 1

y

3

1

1 3

5. Yes

Activity 5.2 (p. 222)

M N logM logN log

10

1

100 –1 2
100 100 2 2
1000

10

1

3 –1
10

100

1

1 –2

100

1

1000 –2 3

2. log MN = log M + log

N

3. log













N

M

= log M –
log N

4. log M2 = 2 log M

Activity 5.3 (p.238)

1. 1

2. No

3. y increases as x

increases; the graph
lies on the right-hand
side of the y-axis; it
cuts the x-axis at (1,0).
4.

x 0.3 0.6 1

y –0.5 –0.2 0

5. They are both defined
for x > 0 and cut the x
-axis at (1,0); y

increases as x

increases in both
graphs; they both lie
on the right-hand side
of the y-axis.

Follow-up

Exercise

p. 205

1.

8

81

8

81

8

81 )

(

2 )

(

3 )

2 (

)

3 (

6
6
12
6
12

3
2
3

4
3
4
3
2

x





2.









4
4

1
4
4

1
1
3
)
2
(
2
1
2

3
2

a

b

a

b

a

b

a

b

a

b

a





























p. 207

1. ∵ 10  10  10 =
1000

103 = 
1000

∴ 3 1000 10

2. ∵ (–6)  (–6)  (–

6) = –216

(–6)3
= –216

∴ 3  216 6

3. ∵ 11  11 = 121

112 = 121

∴

11
121



4. ∵ 4  4  4  4 =

256

44 = 
256

∴

4
256
4





5. ∵

81

1

3

1

3

1

3

1

3

1 



81
1
3
1 4









∴

3
1
81

4 

6. ∵

32

1

2

1

2

1

2

1

2

1

2

1 



32
1
2
1 5










∴

2
1
32

5 

p. 210

1.

 

3
4

4
3
1
4

3 ( )

x
x

x




2.

3
5
3
5

3
1
5

3 5

1 )

(

1





x

3.

256

1 )

4 (

1 )

4 (

)

4 (

]

)

4 [(

)

64 (

4
4

3
4
3

3
4
3
3


































4.

216
125
6
5
6
5
36

3
2
3
2
2



















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5.

 

4
1
4
1
1

4
3
4
3

3
4
1
3
4

1 )

(

x






6.

20
11
20
11
5
4
4
1
5
4
4
1

5
1
4

4
1

5 4

1 )

(

1 a








p. 213

1. 3x + 2 – 3x = 216

(32)(3x) – 3x = 216

3x(32 – 1) = 216
8(3x) = 216

3x = 27

3x = 33

∴ x =3

2. 52x – 24(5x) = 25

(5x)2 – 24(5x) – 25 = 0

Let y = 5x, the equation

becomes

y2 – 24y – 25 = 0
(y – 25)(y + 1) = 0

y = 25
or y =
–1

∴ 5x = 25 or 5x =

–1 (rejected)
5x = 52

∴ x =2

3.














16

2

0

3

y
x

)

2
(

)
1
(




From (1),
3x + 1 = 3y

∴ x + 1 = y

From (2),
2x + y = 24

∴ x + y = 4
Consider the
simultaneous
equations:








4
1

y
x

)
4
(

)
3
(




By substituting (3) into

(4), we have

x + (x +1) = 4
2x = 3

x =

2

3

By substituting x =

2

3

into (3), we have
y =

2

3

+ 1 =

2

5

∴ The solution is x =

2

3

, y =

2

5

.

p. 220

1. (a) ∵ 0.01 =
10–2

∴ log 0.01 =



(b) ∵ 100 000 =
105

∴ log 100 000 =

2. (a) ∵ 10x = 20

∴ x = log 20
=

30
.

1 (cor. to 3 sig.

fig.)

(b) ∵ 10x = 0.8

∴ x = log
0.8

0969
.
0

 (cor. to

3 sig. fig.)
(c) ∵ 10x = 5

∴ x = log 5
=

699
.

0 (cor. to 3

sig. fig.)

p. 221

1. ∵ 92 = 81

∴ log9 81 =2
2. ∵ 43 = 64

∴ log4 64 =3
3. ∵ 112 = 121

∴ log11 121 =2
4. ∵ 25 = 32

∴ log2 32 =5

p. 225

1. 2 log 2 + log 5 = 2

log

2

21 + log 5
= log
(

2

21 )2 + log 5

= log
2 + log 5

= log
(2  5)

= log
10

2. log2 8 – log2 16 = log2
23 – log

2 24
= 3
log2 2 – 4 log2 2

= 3 – 4
=



3.

5

4

2 log

5

2 log

4

2 log

32 log

16 log

5
4



4.

4

1

5 log

2

5 log

2

1

5 log

25 log

5 log

2
2
1



5.

4

2

1

2 log

2

1 log

2 log

2

1 log

2 log

2

2
1





























x

6.

5

3

2

5

2

3 log

2

1

2 log

2

1

2 log

2

1 log

2 log

2

1 log

2 log

2
1
2

2
2

















































x

7. log 9 = log

6

54

= log 54 – log 6
=y x

p. 229

1. 3x – 1 = 7
log 3x – 1 = log 7
(x – 1)log 3 = log 7

x – 1 =

3
log

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x =

3
log

7
log

+ 1

=2.77

(cor. to 3 sig. fig.)
2. 52x = 8

log 52x = log 8

2x log 5 = log 8
2x =

log

8
log

x =

5
log

8
log



=0.646

(cor. to 3 sig. fig.)

3. 6x = 3x +
2

log 6x = log

3x + 2

x log 6 = (x

+ 2) log 3

x log 6 = x

log 3 + 2 log 3

x log 6 – x log 3 = 2
log 3

(log 6 – log 3) x = 2
log 3

x =

3
log
6
log

3
log
2



17
.

3 (cor. to 3 sig. fig.)

4. log (3x – 2) = 2
log (3x – 2) = log 100

3x – 2 = 100
3x = 102

x =34

5. log (3x + 1) – log (x –
2) = 1

















2

1

3 log

x

= log 10

2

1

3 



x

= 10

3x +
1 = 10(x – 2)

3x +
1 = 10x – 20

7x = 21

x =3

6. Let y = log (x – 1),
then the equation

[log (x – 1)]2 + 2 log (x

– 1) + 1 = 0 becomes

y2 + 2y + 1 = 0

(y + 1)2 = 0
y = –1

∴ log (x – 1) = –1

x – 1 =

10

1

x =
10

p. 233

1. Let I1 and I2 be the
sound intensities of the
concert and

that of the football
match respectively.
By the definition of
sound intensity level, we
have

95 = 10 log













0
1

I

and 80 = 10 log













I

∴ 95 – 80 = 10 log













0
1

I

– 10 log















0
2

I

15 = 10







































0
2
0

log

log

I

2

3

= log













2
1

I

2
1

I

= 2
3

10

= 31.6
(cor. to 3 sig. fig.)

∴ The sound
intensity of the
concert is 31.6
times to that of the
football match.
2. Let I be the original

sound intensity
produced by the radio,
then the sound

intensity produced
after adjusting the
volume is 1.25I.

∵ The original sound

intensity level
produced by the
radio is 30 dB.

∴ 30 = 10 log













I

Let dB be the sound
intensity level
produced by the radio
after adjusting the

volume.

∴  = 10 log













25 .

1 I

– 30 = 10 log













25 .

1 I

– 10 log













I

– 30 = 10







































0
0

log

25 .

1 log

I

– 30 = 10 log 1.25

= 30 + 10 log
1.25

= 31.0 (cor. to 3
sig. fig.)

∴ The sound
intensity level
produced by the
radio after
adjusting the
volume is 31.0 dB.

p.235

1. Let E1 and E2 be the
relative energies
released by the
earthquakes measured
6 and 3 respectively.
By the definition of the
Richter scale, we have

6 = log E1 and
3 = log E2
i.e. E1 = 106 and

E2 = 103

∴ 3

2
1

10 

E

= 103
= 1000

∴ The strength of an
earthquake
measured 6 is
1000 times to that
measured 3.

2. Let E1 and E2 be the
relative energies
released by the
earthquakes in Nantou
and Turkey

respectively.

By the definition of the
Richter scale, we have

7.3 = log E1 and
6.3 = log E2
i.e. E1 = 107.3 and

E2 = 106.3

∴ 6.3

3
.
7

2
1

10 

E

= 10

∴ The strength of
the earthquake in
Nantou is 10 times

to that in Turkey.

Exercise

Exercise 5A (p.210)

Level 1

1.

7
1
3

7 3

(

)

x



2.

 

3
8

8
3
1
8

3 ( )

x
x
x




3.

 

2
5
2
5

5
2
1
5

1 )

(

1





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4.
5
2
5
2
5
1
2
5 2

1 )

(

1





x

5.
125
343
5
7
5
7
25

49
3
2
3
2
2
3






























6.
27
8
3
2
3
2
81
16
3
4
3
4
4
3






























7.
64
27
4
3
4
3
16
9
3

2
3
2
2
3






























8.
3
4
4
3
4
3
64
27
64
27
1
1
3
1
3
1
3
1
3
1

































































9.
1

2
3
1 














a

= (a–1+3
– 2)–1

= (a0)–1
= a0
=1
10.
1
4
3
3
4  














n

m

n

m

=

[(m–4 – 3)(n –3 – 4)]–1
= (m–7n –7)–
1

=m7n7
11.
3
1
2
1
2
2
1
4
3
1
2
1
2
4

)

(

  





























m

n

m

n

m

n

m

m2nm–1n–3

m2 – (–1)n1 – (–3)

=
4
3n
m
12.
































2
1
2
2
1
4
)
3
(
3
)
3
(
1
2
1
2
4
3
3

)

(

)

(

a

b

a

b

a

b

a

b

a3b–9a–2b–1

a3 – (–2)b–9 – (–1)

a5b–8

=
8
5

b

a

13.
6
13
6
11
3
2
2
3
2
1
3

4
3
2
2
1
2
3
3
4
3
2
2
1
2
4
3
2
3
2
3
2
2
1
2
4
3
3
2
b
a

b
a
b
a
b
a
b
a
b
a
b
a
b
a






















14.
3
2
4
1
4
1
3
2
)
1
(
4
3
1
3
1
1
4
3
3
1
)
2
(

2
1
)
2
(
2
1
4
3
3
1
2
2
1
2
1
4
3
3
1
a
b
b
a
b
a
ab
b
a
b

a
b
a
b
a
b
a

































15.
2
2
2
2
2
1
2
2

)

(

q
p
q
p
q
p
q
p

a

   



0

2

1

0
2
2
2



















q

p

a

q
p
q
p

∵

q = 2(2 + p)
when p = 2, q = 2(2 +
2) = 8.

when p = 3, q = 2(2 +
3) = 10.

∴ A possible
solution is p = 2, q = 8 or p

= 3, q = 10.

(or any other
reasonable answers)
16.
y

x
y
x
y
x
y
x

a























1
1

)

(

)

(

1

1
1


















y
x
a
a
a
a
a
y
x
y
x
∵

y = 1 – x

When x = 2, y = 1
– 2 = –1.

When x = 3, y = 1
– 3 = –2.

∴ A possible
solution is x = 2, y

= –1or x = 3, y = –
2. (or any other
reasonable
answers)

Level 2

17.

12
5
2
1
4
3
3
2
2
1
4
3
3
2
4
1
2
4
3
2
3
1
4 2
4
3
2
3

)

(

)

(

)

(

x















18.
60
61
60
61
5
3

4
1
3
2
5
3
4
1
3
2
5
1
3
4
1
3
2
5 3
4
1
3
2

1 )

(

x




















19.
60
19
60
19
3
2
5

2
4
3
3
2
5
2
4
3
3
1
2
2
5
1
4
3
3 2
2
5
4
3

1 )

(

)

(

)

(

p
















20.
5
13
5
3
2
1
2

5
5
3
2
1
2
5
5
3
1
2
1
2
1
5
5
3

1 (

)

(

)

</div>
(6)<div class='page_container' data-page=6>

12
23
12
23
)
3
2
(
4
3

2
1
3
2
4
3
2
1
3
2
4
1
3
2
1
3
2
4 3
2
1

1 )

(

t






























22.
12
11

3
1
2
1
4
3
3
1
2
1
4
3
9
1
3
2
1
4
3
9 3
4
1
2
4
3

)

(

)

(

q















23.

b

a

b

a

ab

b

a

b

a

ab

b

a

b

a

2
2
1
2
1
1
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1



























24.
6
1
6
1
6
1
6
1
1
3
1
2
1

1
2
1
3
1
3
1
2
1
2
1
3
1

1 n

m

n

m

n

m

mn

n

m

n

m



































25.
4
5
4
1
4
1
4
5

1
4
1
2
1
)
2
3
(
4
1
2
1
1
2
3
4
1
4
1
2
1
2
1
2
3
4
1
4
2

1
2
4
3
4

2 )

2 (

)

(

16 a

b

a

b

a

b

a

b

a

ab

a

ab










































26.
6
13
6
1
2
1
3
2
2
6
1
3
2
3
1
2
1
6
1
3
2
3
2
2
2
3
1
2

1
3
1
3
2
3
3
1
2
3
1
3
2
3
2

9

3 )

(

)

3 (

)

27 (

b

a

b

a

b

a

b

a

b

a

b

a

ab

a

b

a

ab

a

b


























27.
3
8
12
17
3
8
6
1
12
19
6
1
)
2
(
3
2
)
3
4
(
4
1
2
3
1
2

1
2
3
4
2
3
2
4
1
3
1
2
1
2
3
4
2
3
2
4
1
3
1
2
3
2
3
2
4
1

4

2 )

(

2

1

2

1

2 n

m

n

m

n

m

n

m

n

m

n

m

n

m

n

m

n

m











































































28.
60
29
3
2
60

29
3
2
5
2
3
1
4
1
)
3
1
(
3
2
1
1
5
2
3
1
3
1
3
2
4
1
5
2
3

1
3
1
2
3
4
1
5
2
3
1
3 2
4
1

4

1

4 )

4 (

64 y

x

y

x

y

x

y

x

y

x

xy

y

x

y

x

xy

y

x

y

x

xy



















Exercise 5B (p.214)

Level 1

1.
2
2
3
2
3
2
3
2
2

5 )

5 (

5

125

5 

x

x
x
∴

2 x

= 2

x =4

2.
4
4
3
3
3

2 )

2 (

16

8 

x
x
x

∴ x =4

3.
2
1
2
1
2

6

36

6 



x
x
∴

2 x

+1= 2

x =2

4.
2
3
2
5
2

5
3
2
5
2
3

2 )

2 (

)

2 (

32

4 

x
x
x
∴

3 2x

= 2

x =3

5. 3x + 2 – 3x = 8

(32)(3x) – 3x = 8

3x(32 – 1) = 8
8(3x) = 8

3x = 1

3x = 30

∴ x =0

6. 3(3x) – 3x – 1 = 216
3(3x) – 3x(3–1) = 216
3x(3 –3–1) = 216
3x(3 –

3

1

) = 216

3x













3

8

= 216
3x = 81

3x = 34

∴ x =4

7. 2x + 2 + 2x = 10

(22)(2x) + 2x = 10

2x(22 + 1) = 10
5(2x) = 10

2x = 2

2x = 21

∴ x =1

8. 42x = 16(2x)

(22)2x = 24(2x)

24x = 24 + x

∴ 4x = 4 + x

x = 34

9. 23x = 8(4x)

23x = 23(22)x

23x = 23(22x)

23x = 23 + 2x

∴ 3x = 3 + 2x

x =3

10. 9(3x + 1) = 272x

32(3x + 1) = (33)2x

3x + 3 = 36x

∴ x + 3 = 6x

x = 53

Level 2

11. 4x + 1 + 4x – 4x – 1 = 19
(4)4x + 4x – (4– 1)4x =

4x(4 + 1 – 4– 1) = 
19

4x(5 –

4

1

) =

4x(

4

19

</div>
(7)<div class='page_container' data-page=7>

4x = 4

∴ x =1

12. 6x + 2 – 2(6x + 1) – 12(6x)

= 2

(62)6x – 2(6)6x – 12(6x)

= 2

6x(36 – 12 – 12)

= 2

6x(12)

= 2

6x

6

1

6x

= 6–1

∴ x

=1

13. 22x – 3(2x) + 2 = 0

(2x)2 – 3(2x) + 2 = 0

Let y = 2x, the equation

becomes

y2 – 3y + 2 = 0
(y – 1)(y – 2) = 0

y = 1 or

y = 2

∴ 2x = 1 or 2x =

2x = 20 or 2x =

∴ x =0 or x =

14. 32x – 12(3x) + 27 = 0

(3x)2 – 12(3x) + 27 = 0

Let y = 3x, the equation

becomes

y2 – 12y + 27 = 0
(y – 3)(y – 9) = 0

y = 3 or

y = 9

∴ 3x = 3 or 3x =

3x = 31 or 3x =

∴ x =1 or x =

15. 5(52x) – 26(5x) + 5 = 0

5(5x)2 – 26(5x) + 5 = 0

Let y = 5x, the equation

becomes
5y2 – 26y + 5 = 0
(5y – 1)(y – 5) = 0

y =

5

1

ory = 5

∴ 5x =

5

1

or 5x =

5x = 5–1 or 5x =

∴ x =1 or

x =1

16. 24x – 20(22x) + 64 = 0

(22x)2 – 20(22x) + 64 = 0

Let y = 22x, the

equation becomes

y2 – 20y + 64 = 0
(y – 4)(y – 16) = 0

y = 4
or y =
16

∴ 22x = 4 or 22x =

22x = 22 or 22x =

∴ 2x = 2 or 2x

= 4

x =1 or x

17.












1

25

125

5

2
2

y
x

)
2
(

)
1
(




From (1),
5x + 2y = 53

∴ x + 2y = 3
From (2),

252x + y = 250

∴ 2x + y = 0

y = –2x

Consider the
simultaneous
equations:











x
y

y
x

2
3
2

)
4
(

)
3
(




By substituting (4) into
(3), we have

x + 2(–2x) = 3
–3x = 3

x = –1
By substituting x = –1
into (4), we have

y = –2(–1)
= 2

∴ The solution is x =
–1, y = 2.

18.












4

27

3

3x y
y
x

)
2
(

)
1
(






From (1),
3x + y = 33

∴ x + y = 3
From (2),

43x – y = 41

∴ 3x – y = 1
Consider the
simultaneous
equations:











1
3

3
y
x

y
x

)
4
(

)
3
(




(3) + (4),

(x + y) + (3x – y) = 3 +
1

4x = 4
x = 1
By substituting x = 1

into (3), we have

1 + y = 3

y = 2

</div>
(8)<div class='page_container' data-page=8>

1, y = 2.
19.













3

9

729

3

2
2

y
x

)
2
(

)
1
(




From (1),
32x + y = 36

∴ 2x + y = 6
From (2),

(32)x – 2y = 3

32x – 4y = 31

∴ 2x – 4y = 1
Consider the

simultaneous
equations:











1
4
2

6
2

y
x

)
4
(

)
3
(




(3) – (4),

(2x + y) – (2x – 4y) = 6
– 1

5y = 5

y = 1
By substituting y = 1
into (3), we have

2x + 1 = 6

x =

2

5

∴ The solution is x =

2

5

, y = 1.

20.













625

5

4

2

3 2

y
x

)
2
(

)
1
(




From (1),
23x – 2y = 22

∴ 3x – 2y = 2
From (2),

5x + y = 54

∴ x + y = 4

y = 4 – x

Consider the
simultaneous
equations:











x
y

y
x

4
2
2
3

)
4
(

)
3
(




By substituting (4) into
(3), we have

3x – 2(4 – x) = 2
5x = 10

x = 2
By substituting x = 2
into (4), we have

y = 4 – 2 = 2

∴ The solution is x =
2, y = 2.

Exercise 5C (p.226)

Level 1

1. ∵ 1 000 000 =
106

∴ log 1 000 000 =

2. ∵ 625 = 54

∴ log5 625 =4

3. ∵

16

1

= 4–2

∴ log4













16

1



4. ∵

100

1

= 10–2

∴ log













100

1



5. ∵ 10x = 120

∴ x = log 120
=2.08

(cor. to 3 sig. fig.)
6. ∵ 10x = 88

∴ x = log 88
=1.94

(cor. to 3 sig. fig.)

7. log 200 – log 2 = log

2

200

= log
100

= log
102

= 2 log
10

8. log 4 + log 25 = log(4

 25)

= log
100

= log
102

= 2 log
10

9. log3













12

1

+ log3 108

= log3 (

12

1

 108)
=
log3 9

=
log3 32

=
2 log3 3

10.

3

2

3 log

2

3 log

27 log

9 log

3
2



11.

2

3

4 log

2

4 log

3

4 log

16 log

64 log

2
3



12.

4

5 log

2

1

5 log

2

5 log

5

1 log

5 log

25

1 log

2
2
2
1
2

2
2

2
1
2

2
2





































13. log 6 = log (2  3)

= log 2 + log 3
=xy

14. log 18 = log

2
1
2

)

3

2 (



= log (

2

21 

= log 21

2

log 3

2

1

log 2+

log 3

= xy

2
1

15. Let loga 8 = x, logb 8 =
y.

Then ax = 8 and by = 8.

∵ loga 8 logb 8 = 1

∴ xy = 1

x =

y
1

Let a = 2, then
2x = 8

2x = 23
x = 3

∴ y =

3

1

∴ 3

8



b

b = 83
= 512

Let a =

2

1

, then

3
2
1
2

1
8
2
1





























x

x
x

∴

3

1 



y

∴

512
1
8
8

3
3
1








b
b

∴ A possible
solution is a = 2, b

= 512
or a =

2

1

, b =

512

1

. (or any
other reasonable
answers)
16.

b
a
log
log

= 3
log a = 3 log b

</div>
(9)<div class='page_container' data-page=9>

∴ a = b3

∴ A possible
solution is a = 8, b

= 2 or a = 27, b =
3. (or any other
reasonable
answers)

Level 2

17. 3 log 5 + log













25

2

log 53 + log













25

2

log















25

2

5

=
log 10

18. log(100 10)– log

)
10
10

( = log












10
10

10
100

= log 10
=1

19. log4 2 – log4 18 + 2 log4













2

3

= log4













18

2

+ log4
2

2
3








= log4




















2
3
9
1

= log4















4

9

1

= log4













4

1

= log4 4–1
= –log4 4
=1

20.

4

11

2 log

2

11

2 log

2

5

2 log

3

2 log

3

2 log

)

2 log(

2 log

4 log

32 log

8 log

2
5
2

2
1
5
3

















21.

5

1

2 log

5

2 log

32 log

2 log

5 .

0

4 log

10

2

5 log

5 .

0 log

4 log

10 log

2 log

5 log

5 .

0 log

4 log

2

1

2 log

2

5 log

5
2

2
2





























22.

2

1

10 log

2

1

10 log

1

100 log

10 log

25 .

0

25 log

100

125

8 log

25 .

0 log

25 log

100 log

125 log

8 log

25 .

0 log

5 log

10 log

2

5 log

8 log

25 .

0 log

5 log

2

5 log

3

8 log

2
2



































23.

5

2

10

4 log

)

4

6 (

log

4 log

6 log

4 log

4 )

log

2 (

3 log

4 log

3 log

2
4

















x

24.

1

5 log

)

3

8 (

log

5 log

3 log

8 log

)

3

2 (

log

3 )

log

2 (

4 log

3 log

2 log

3 log

4 log

2

3
3

3
3
3

3
3

3
2

3
3
3























x

25.

6 log

3

1 log

2 log

3

1 log

2
3

1
4

4
3

4
4



















x

xy

y

x

xy

y

x

26.

6 log

)

12

6 (

log

12 log

6 log

)

log

4 (

3 )

log

3 (

2 log

3 log

2

2
2

4
3































x

xy

y

x

xy

y

x

27. log 200 = log (4  5 

10)

= log 4 + log 5
+ log 10

=xy1

28. log 320= log

2
1

320

2

1

log 320
=

2

1

log (43

 5)

2

1

(log 43
+ log 5)

2

1

(3 log

4 + log 5)
=
)
3
(
2

1 x y



Exercise 5D (p.229)

Level 1

</div>
(10)<div class='page_container' data-page=10>

log 2x = log 10
x log 2 = 1

x =

2
log

=3.32 (cor.
to 3 sig. fig.)

2. 32x – 1 = 6
log 32x – 1 = log 6
(2x – 1) log 3 = log 6

2x – 1 =

3
log

6
log

2x =

3
log

6
log

+ 1

x =

2
1
3
log

6
log












=1.32

(cor. to 3 sig. fig.)
3. 41 – 3x = 5

log 41 – 3x = log 5

(1 – 3x) log 4 = log 5
1 – 3x =

4
log

5
log

3x = 1 –

4
log

5
log

x =

3
4

log

5
log

1 










0537
.
0

 (cor. to 3

sig. fig.)

4. 2(3x + 2) = 5
log [2(3x + 2)] = log
5

log 2 + log 3x + 2 = log
5

(x + 2) log 3 = log
5 – log 2

x + 2 =

3
log

2
log
5
log 

x =

3
log

2
log
5
log 

– 2
=

.
1

 (cor. to 3 sig.

fig.)

5. 9x + 1 = 8x

log 9x + 1 = log 
8x

(x + 1) log 9 = x log
8

x log 9 + log 9 = x

log 8

(log 8 – log 9) x = log
9

x =

9
log
8
log

log



7
.
18

 (cor. to 3 sig.

fig.)

6. 42x =

63x – 1

log 42x =

log 63x – 1

2x log 4 =
(3x – 1) log 6

2x log 4 =
3x log 6 – log 6

(3 log 6 – 2 log 4) x =
log 6

x =

4
log
2
6
log
3

6
log



688
.

0 (cor. to 3 sig.

fig.)

7. log (6x – 4) = 2
log (6x – 4) = log 100

6x – 4 = 100
6x = 104

x = 3
1
17

8. log3 (4x – 1) = 3
log3 (4x – 1) = 3 log3 3
log3 (4x – 1) = log3 33

4x – 1 = 27
4x = 28

x =7

9. log (3x – 2) = 0
log (3x – 2) = log 1

3x – 2 = 1
3x = 3

x =1

10. log2 (3x + 1) = 4
log2 (3x + 1) = 4 log2 2
log2 (3x + 1) = log2 24

3x + 1 = 16
3x = 15

x =5

Level 2

11. log (8x – 2) – log 3 = 1
log (8x – 2) =
log 10 + log 3

log (8x – 2) =
log (10  3)

8x – 2 =
30

8x =
32

x =

12. log (2x – 3) + log 2 = –
1

log [(2x – 3)  2] =
–log 10

log (4x – 6) =
log 10–1

4x – 6 =
10–1

4x =

10

1

+ 6

4x =

10

61

x =
40

13. log4 (x + 3) + log4 (x –
3) = 2

log4 [(x + 3)(x –
3)] = 2log4 4

log4 (x2 –
9) = log4 42

x2 –
9 = 16

x2 = 25
x = 5
or

x = –5 (rejected)

∴ x =5

14. log3 (4x + 9) – log3 (3x
– 2) = 1

log3

















2

3

9

4 x

= log3 3

2

3

9

4 



x

= 3
4x + 9 = 9x – 6
5x = 15

x =3

15. log (x – 1) + log (2x –
1) = 1

log [(x – 1)(2x –
1)] = log 10

log (2x2 – 3x + 
1) = log 10

2x2 – 3x +
1 = 10

2x2 – 3x –
9 = 0

(x – 3)(2x +
3) = 0

x = 3
or

x =

2

3

</div>
(11)<div class='page_container' data-page=11>

∴ x =3

16. Let y = log2 (2x – 1),
then the equation

[log2 (2x – 1)]2 – 3 log2
(2x – 1) + 2 = 0 becomes

y2 – 3y + 2 = 0
(y – 1)(y – 2) = 0

y = 1 or

y = 2
For y = 1, log2 (2x – 1)
= 1

2x – 1
= 2

x

2

3

For y = 2, log2 (2x – 1)
= 2

2x – 1
= 4

x

2

5

∴ x = 23 or 25

17. Let y = log (x + 1),

then the equation

[log (x + 1)]2 – log (x +
1) – 12 = 0 becomes

y2 – y – 12 = 0
(y + 3)(y – 4) = 0

y = –3 or

y = 4
For y = –3, log (x + 1)
= –3

x + 1
=

1000

1

x

1000

999 

For y = 4, log (x + 1) =
4

x + 1 =
10 000

x =
9999

∴ x = 1000999 or

9999

18. Let y = log (x – 1),
then the equation

[log (x – 1)]2 – 3 log (x
– 1) + 2 = 0 becomes

y2 – 3y + 2 = 0
(y – 1)(y – 2) = 0

y = 1 or

y = 2
For y = 1, log (x – 1) =
1

x – 1 =

x =
11

For y = 2, log (x – 1) =
2

x – 1 =
100

x =
101

∴ x =11 or 101

19.











1
log
log

32
2

y
x

)
2
(

)
1
(




From (2),

log x = log y + log 10
log x = log 10y

x = 10y

……(3)

By substituting (3) into
(1), we have

10y – 2y = 32
8y = 32

y = 4
By substituting y = 4
into (3), we have

x = 10(4) = 40

∴ The solution is x =
40, y = 4.

20.










1
log

log

1
x
y

y
x

)
2
(

)
1
(




From (2),

log y = log x – log 10
log x = log y + log 10
log x = log 10y

x = 10y

……(3)

By substituting (3) into
(1), we have

10y – y = 1
9y = 1

y =

9

1

By substituting y =

9

1

into (3), we have

x = 10(

9

1

) =

9

10

∴ The solution is x =

9

10

, y =

9

1 Exercise 5E (p.236)

Level 1

1. Let I1 and I2 be the
sound intensities in
restaurants A and B

respectively.
By the definition of
sound intensity level, we
have

60 = 10 log













0
1

I

and 75 = 10 log













0
2

I

∴ 75 – 60 = 10 log













0
2

I

– 10 log















0
1

I

15 = 10







































1
0

log

log

I

2

3

= log













1
2

I

1
2

I

= 2
3

10

= 31.6
(cor. to 3 sig. fig.)

∴ The sound
intensity in
restaurant B is
31.6 times to that
in restaurant A.
2. Let I1 and I2 be the

sound intensities after
the party and during
the party respectively.
By the definition of
sound intensity level, we
have

35 = 10 log













0
1

I

and 67 = 10 log













0
2

I

∴ 35 – 67 = 10 log













0
1

I

– 10 log















I

–32 = 10







































0
2
0

log

log

I

5

16 

log















2
1

I

2
1

I

5
16

10



= 0.000
631 (cor. to 3 sig. fig.)

∴ The sound
intensity after the
party is 0.000 631
times to that
during the party.
3. Let I1 and I2 be the

sound intensities
before adjustment and
after adjustment
respectively.
Then the sound
intensity levels before
and after adjustment
are 10 log















0
1

I

and 10 log















0
2

I

dB
respectively.

∵ The sound
intensity level
increases by 5 dB
after the

adjustment.

∴ 5 = 10 log













I

– 10 log













0
1

I

5 = 10







































0
1
0

log

log

I

2

1

= log













1
2

</div>
(12)<div class='page_container' data-page=12>

1
2

I

= 2
1

10

= 3.16 (cor. to
3 sig. fig.)

∴ The sound
intensity after
adjustment is 3.16
times to that
before adjustment.
4. Let E1 and E2 be the

relative energies
released by the
earthquakes measured
8 and 2 respectively.
By the definition of the
Richter scale, we have

8 = log E1 and
2 = log E2
i.e. E1 = 108 and

E2 = 102

∴ 2

2
1

10 

E

= 106

∴ The strength of an
earthquake
measured 8 is 106
times to that
measured 2.
5. Let E1 and E2 be the

relative energies
released by the
earthquakes in the
Philippines and
Indonesia
respectively.
By the definition of the
Richter scale, we have:

For Philippines,
For
Indonesia,

7.3 = log E1
6.0 = log E2

E1 = 107.3
E2 = 106.0

∴ 6.0

3
.
7

2
1

10 

E

= 101.3
= 20.0 (cor. to
3 sig. fig.)

∴ The strength of the
earthquake in the

Philippines is 20.0
times to that in
Indonesia.

Level 2

6. Let I be the sound
intensity of the

gathering when the
master of ceremony
does not speak, then
the sound intensity is
2I when the master of
ceremony speaks.

∵ The sound
intensity level in
the gathering is 30
dB.

∴ 30 = 10 log













I

Let  dB be the sound
intensity level when
the master of ceremony
speaks.

∴  = 10 log













2 I

– 30 = 10 log













2 I

– 10 log













I

– 30 = 10







































0
0

log

2 log

I

– 30 = 10 log 2
= 30 + 10 log 2

= 33.0 (cor. to 3
sig. fig.)

∴ The sound
intensity level is
33.0 dB when the
master of
ceremony speaks.
7. Let I be the original

sound intensity of the

Hi-Fi, then the sound
intensity is 0.6I after
adjustment.

∵ The original sound
intensity level is
90 dB.

∴ 90 = 10 log













I

Let  dB be the sound
intensity level of the
Hi-Fi after adjustment.

∴  = 10 log













6 .

0 I

– 90 = 10 log













6 .

0 I

– 10 log













I

– 90 = 10







































0
0

log

6 .

0 log

I

– 90 = 10 log 0.6

= 90 + 10 log
0.6

= 87.8 (cor. to 3
sig. fig.)

∴ The sound
intensity level of
the Hi-Fi is 87.8
dB after
adjustment.
8. Let I be the original

sound intensity of the
noise, then the sound
intensity is 1.1I after
adjustment.

The sound intensity

levels before and after
adjustment are 10 log













I

dB and 10 log













1 .

1 I

dB
respectively.

∴ 10 log















1 .

1 I

– 10 log















I

= 10







































0
0

log

1 .

1 log

I

= 10 log 1.1
= 0.414 (cor. to 3
sig. fig.)

∴ The increase in the
sound intensity
level is 0.414 dB.
9. Let E be the relative

energy released by an
earthquake measured
5.8.

By the definition of the
Richter scale, we have

5.8 = log E
E = 105.8

∴ The relative
energy released by

an earthquake that
is 20 times to that
measured 5.8
= 20  105.8

∴ The magnitude of
the earthquake on
the Richter scale
= log (20  105.8)

=7.10 (cor. to
3 sig. fig.)

10. Let E be the relative

energy released by an
earthquake measured
7.2.

By the definition of the
Richter scale, we have

7.2 = log E
E = 107.2

∴ The relative
energy released by
an earthquake that
is

8

1

times to
that measured 7.2
=

8

1

 107.2

∴ The magnitude of

the earthquake on
the Richter scale
= log (

8

1



107.2)

=6.30 (cor. to
3 sig. fig.)

Revision Exercise 5

(p.244)

Level 1

1. (a)

5
7

7
5
1
7

)

(

)

(

x

</div>
(13)<div class='page_container' data-page=13>

(b)

3
4
3
4

3
1
4

3 4

1 )

(

1





x

2. (a)

36
25
25
36
5
6
5
6
125
216
125

216

1
1
2

1
3
2

1
3
2
3

























































































(b)

3
7
7
3

7
3
343

27
343

1
1
3
1
3

1
3
1













































































3. (a)

b

c

a

c

b

a

c

b

a

c

b

a

4
3
2

4
1
3
2

3
1
12
3
2

3 2 3 12

(

)







(b)

6
7
6
7
2
3
3
1
2
3
3
1

2
1
3

6
1
2
3

6 2

1 )

(

)

(

x






(c)

3
50
3
50
3

4
18

3
4
18

3
1
4
3
15

3 4

3
3
5

1 )

(

)

(

x




















(d)

a
a

b
a

b
a
b
a

ab
b

a

ab
b

a
b
a
ab
ab

b
a

)
(

]
)
[(
)

(
)
(

2
1
2
1
2
1
2
1

2
1
2
1
2

1
2
1

2
1
2

1
1
1
2
1

2
4
1
1

2
1
1
2
1
2
4
1
2
1
2































4. (a)

3
4

3
1
4
3

2 )

2 (

16

2 

x

∴ x = 34

(b)

6
6
2
1
2

6
2
1
2

2 )

2 (

2 ]

)

2 [(

64

4 

x
x
x

x

∴ x =6

(c) 5x + 1 – 2(5x) =

5

3

5(5x) – 2(5x) =

5

3

3(5x) =

5

3

5x =

5

1

5x = 5–1
x =



(d) 2x + 2 – 2x = 24

22(2x) – 2x = 24

(22 – 1)2x = 24

3  2x = 24

2x = 8

2x = 23
x =3

5. (a) ∵

000
10

= 10–4

∴ log











000
10

=4

(b) ∵ 0.000 000
1 = 10–7

∴ log 0.000 000
1 =7

(c) ∵ 64 = 82

∴ log8 64 =2
(d) ∵

2
1

5

1





∴

2
1
5

1
log5 








6. (a) ∵ 10x = 45

∴ x = log 45
=

65
.

1 (cor. to 3 sig.
fig.)

(b) ∵ 102x = 20

∴ 2x = log
20

x =

651
.

0 (cor. to 3 sig.
fig.)

7. (a) log 400 – log 8 +
log 2 = log

8

400

+
log 2

= log 50 + log 2
= log (50  2)

= log 100

(b) log 125 + log 80 =
log (125  80)

=
log 10 000

=
log 104

(c) log6 108 + log6 2 =
log6 (108  2)

=
log6 216

=
log6 63

(d) log9 9 + log9













81

1

= log9















81

1

9

= log9













9

1

= log9 9–1
=1

(e)

2
4
log
2

4
log
4

4
log

4
log
16

log
256
log

2
4

</div>
(14)<div class='page_container' data-page=14>

(f)

8

2

1

4

5 log

2

1

5 log

4

5 log

625 log

2
1
4



8. (a) 5x = 10

log 5x = log 10
x log 5 = 1

x =

5
log

=1.43

(cor. to 3 sig. fig.)
(b) 2x – 1 = 6

log 2x – 1 = log 
6

(x – 1) log 2 = log
6

x – 1 =

2
log

6
log

x =

2
log

6
log

+ 1

58
.

3 (cor. to 3 sig. fig.)

(c) 32x + 3 = 33
log 32x + 3 = 
log 33

(2x + 3) log 3 =
log 33

2x + 3 =

3
log

33
log

2x =

3
log

33
log

– 3

x =

0913
.

0 (cor. to 3 sig.

fig.)

(d) log (3x + 1) = 2
log (3x + 1) = log
100

3x + 1 = 100
3x = 99

x =

(e) log















1

2 x

= –
1

log















1

2 x

log













10

1

2 x

– 1 =

10

1

2 x

=

1.1

x =

2
.
2

(f) log2 (6x – 4) = 5
log2 (6x – 4) = log2
25

6x – 4 = 32
6x = 36

x =6

9. (a) log 144 = log (24

 32)

= log 24 + 
log 32

= 4 log 2 +
2 log 3

y

x 2

4 

(b) log 288 = log

2
1

288

2

1

log (25 32)
=

2

1

(log 25 + log 32)
=

2

1

[5

log 2 + 2 log 3]
=
)
2
5
(
2

1 x y



10. Let I1 and I2 be the
sound intensities of
groups A and B

respectively.
By the definition of
sound intensity level, we
have

45 = 10 log













0
1

I

and 53 = 10 log













0
2

I

∴ 53 – 45 = 10 log













0
2

I

– 10 log















0
1

I

8 = 10







































0
1
0

log

log

I

5

4

= log













1
2

I

1
2

I

= 5
4

10

= 6.31
(cor. to 3 sig. fig.)

∴ The sound
intensity of group

B is 6.31 times to
that of group A.
11. Let I1 and I2 be the

sound intensities of the
cassette player before
and after adjustment.

∵ The sound
intensity level
increases by 8 dB.

∴ 10 log















0
2

I

–

10 log















0
1

I

= 8







































0
1
0

log

log

I

= 8

log















1
2

I

5

4

1
2

I

= 5
4

10

I2 = 5
4

10

I1

∴ The percentage
increase of the
sound intensity:
=

%

100 



I

%

100

10

1
1
1
5
4





I

= 531% (cor.
to 3 sig. fig.)
12. Let E1 and E2 be the

relative energies
released by the
earthquakes measured
7.5 and 6.5

respectively.

By the definition of the
Richter scale, we have

7.5 = log E1 and
6.5 = log E2
i.e. E1 = 107.5 and

E2 = 106.5

∴ 6.5

5
.
7

2
1

10 

E

= 10

∴ The strength of an
earthquake
measured 7.5 is 10
times to that
measured 6.5.
13. Let E1 and E2 be the

relative energies
released by the
earthquakes in
Indonesia and Pakistan
respectively.

By the definition of the
Richter scale, we have:

For Indonesia,
For Pakistan,

7.5 = log E1
4.6 = log E2

E1 = 107.5
E2 = 104.6

∴ 4.6

5
.
7

2
1

10 

E

= 102.9
= 794 (cor. to
3 sig. fig.)

∴ The strength of the
earthquake in
Indonesia is 794
times to that in
Pakistan.
14. (a) log (4x – 2) – log

2x = y

</div>
(15)<div class='page_container' data-page=15>















x

2

4

= y

x

1

2 

= 10y

2x –
1 = x10y

x (2 –

10y) =1

x =2 10y
1



(b) When y = 0, x =

10

2

1 

2

1 

= 1
When y = 1, x =

10

2

1 

8

1 

∴ A possible
solution is x =
1, y = 0 or x =

8

1 

y = 1. (or any
other
reasonable
answers)

Level 2

15. (a)

6
1
6
1

3
1
6
1

3
1

3
1
2
1

3
1

3
1
2
4
1
3
1
3

2
4

1 )

(

)

(

a























(b)

2
1
2
1

10
1
5
3

5
2
2
1
5
3

5
1
2

1
3
5
3

5 2

6 3

5
3

1 )

(

)

(

p























16. (a)

4
7

3
1

)
2
1
(
4
5
3
2
1

2
1
3
2
4
5

2
1
3
2
2
1
4
3
4
3

4
1

2
1
3
2
4
3
2
1
4
3
4
1

2
1
3
2
4
3
2
1
4
1
3

2
1

3
2

4
3

4 3

)

(

y

x

y

x

y

x

xy

y

x

y

x

y

x

y

x

y

x

y

xy

y

x

y

xy










































(b)

12
19
6
7

2
1
12
13
2
1
3
5

2
1
2
1
12
13
3
5

2
1
2
1
)
3
1
(
4
3
)
3

2
(
1

2
1
2
1
3
1
3
2
4
3

2
1
2
1
1
3
1
3
2
4
3

2
1
2

1
1
3
1
2
4
3

2
1
1

3 2

4
3

)

(

]

)

[(

)

(

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

ab

b

a

b

a

ab

b

a

b

a

ab

b

a

b

a

ab





































































17. (a) 5x + 1 + 5x – 5x – 1
= 29

(5)5x + 5x – (5– 1)5x

= 29

5x(5 + 1 – 5– 1)
= 29

5x(6 –

5

1

) = 29

5x(

5

29

) = 29

5x = 5

∴

x =1

(b) 3x + 3 – 2(3x + 2) + 
3x = 30

(33)3x – 2(32)3x +

3x = 30

3x(27 – 18 +

1) = 30
3x(10) = 30

3x = 3

∴

x =1

(c) 22x – 5(2x) + 4 =

(2x)2 – 5(2x) + 4 =

Let y = 2x, the

equation becomes

y2 – 5y + 4 = 0
(y – 1)(y – 4) = 0

y = 1

ory =
4

∴ 2x = 1 or

2x = 4

2x = 20 or
2x = 22

∴ x =0 or

x =2

(d) 42x – 10(4x) + 16

= 0

(4x)2 – 10(4x) + 16

= 0

Let y = 4x, the

equation becomes

y2 – 10y + 16 = 0
(y – 2)(y – 8) = 0

y = 2
ory =
8

∴ 4x = 2 or

4x = 8

22x = 21 or
22x = 23

2x = 1 or
2x = 3

∴ x = 21 or

x = 23

18. (a)












1

16

64

4

2
2

y
x

)
2
(

)

1
(




From (1),
4x + 2y = 43

∴ x + 2y = 3
From (2),

162x – y = 160

∴ 2x – y = 0

y = 2x

Consider the
simultaneous
equations:










x
y

y
x

2
3
2

)
4
(

)
3
(




By substituting (4)
into (3), we have

x + 2(2x) = 3
5x = 3

x =

5

3

</div>
(16)<div class='page_container' data-page=16>

5

3

into (4), we have

y = 2(

5

3

)
=

5

6

∴ The solution
is x =

5

3

, y =

5

6

(b)














9

3

y
x

y
x
y
x

)
2
(

)
1
(




From (1),
3x + y = (32)2x – y

3x + y = 34x – 2y

∴ x + y = 4x –

2y

y = x

From (2),
93x + y = 91

∴ 3x + y = 1
Consider the
simultaneous
equations:









1
3x y

x
y

)
4
(

)
3
(




By substituting (3)
into (4), we have

3x + x = 1
4x = 1

x =

4

1

By substituting x

4

1

into (3), we
have y =

4

1

∴ The solution
is x =

4

1

, y =

4

1

.

19. (a)

2

1 

log 625 –
log 4 = log 2

625

 –
log 4

= log

625
1

– log 4

= log

4

25

1

= log

100

1

= log 10–2
=2

(b)

7

1

2 log

7

2 log

128 log

2

1 log

128 log

64

32 log

128 log

64 log

32 log

7
1











(c)

1

9 log

9

1 log

9 log

243

27 log

9 log

243 log

3 log

9 log

243 log

3 log

3 













(d)

2 log

4

2 log

4 log

2 log

4

9

36 log

2 log

)

2

6 (

9 log

36 log

2 log

2 )

2 log

4 (

5 .

1

3 log

36 log

2 log

5 .

1

3 log

2

36 log

2 log

2

16 log

5 .

1

2
4

















20. (a)

4 log

)

log(

log

2
2
2
4
2
2
8
4

8
4





ab

b

a

ab

b

a

(b)

3

16 log

3 log

16 log

3 log

)

8 (

log

3 log

8 log

2 log

4

3
4
2













x

(c)

4

3 log

3

2 log

2

1 log

)

log(

log

)

log(

)

log(

log

3
2
2
1

3
1
3
1
3
1

2
1
2
1
2
1

3
1
3

1
3
1

2
1
2

1
2
1

3
1
3

2
1
2

3
3
















































x

xy

y

x

xy

y

x

xy

y

x

xy

y

x

xy

y

x

xy

y

x

xy

(d)

1 log

1
2
2

2
2
2

2
2





































a

b

a

ab

b

a

b

a

ab

b

a

21. (a) log (3x – 2) + log
2 = 1

log [(3x – 2) 

2] = log 10

(3x – 2) 

2 = 10

3x –
2 = 5

</div>
(17)<div class='page_container' data-page=17>

x = 37

(b) log 2x + 6 log 2 =
log 96

log 2x + log 26 = 
log 96

log (2x 26) =

log 96

2x  26 = 
96

128x =
96

x =
4

(c) log3 (4x – 2) – log3
(x + 1) = 1

log3

















1

2

4 x

= log3 3

1

2

4 



x

= 3
4x – 2 = 3(x + 1)
4x – 2 = 3x + 3

x =5

(d) log5 (3x + 3) –
log5 (x – 1) = 2 log5

log5

















1

3 x

= log5 2 2
1

)

5 (

1

3 



x

= 5

3x + 3 = 5(x – 1)
3x + 3 = 5x – 5
2x = 8

x =4

(e) Let y = log3 (2x +

3), then the equation

[log3 (2x + 3)]2 – 4
log3 (2x + 3) + 3 = 0
becomes

y2 – 4y + 3 = 0
(y – 1)(y – 3) = 0

y = 1
ory =
3

For y = 1, log3 (2x
+ 3) = 1

2x + 3 = 3
2x = 0

x = 0

For y = 3, log3 (2x
+ 3) = 3

2x + 3 = 27
2x = 24

x = 12

∴ x =0 or

22. (a)











2
2
3

2
log
2
log

y
x

)
2
(

)
1
(




From (1),
log x + log y2
= log 100

log xy2
= log 100

xy2
= 100

……(3)

From (2),
3x = 2 + 2y

x =

3

2 

y

……(4)
By substituting (4)
into (3), we have















3

2 y

y2 = 100
2y2 + 
2y3 = 300

y3 + y2 – 
150 = 0

(y – 5)(y2 + 6y +

30) = 0

y = 5 or

y =

)

1 (

2 )

30 )(

1 (

4

6 





2

84

6 



(rejected
)

By substituting y

= 5 into (4), we have

x =

3 )

5 (

2 

= 4

∴ The solution
is x = 4, y = 5.
(b)








1
2
log
log

1
)
5
6
log(

xy
y
x

)
2
(

)
1
(






From (1),
6x – 5y = 10

6x = 10
+ 5y

x =

6

5

10 

y

……(3)
From (2),

log

2 xy

=

log 10

2 xy

= 10

xy = 20

……(4)
By substituting (3)
into (4), we have















6

5

10 y

y = 20
10y + 5y2
= 120

y2 + 2y – 24 
= 0

(y – 4)(y + 6)
= 0

y

= 4 or y

= –6

By substituting y

= 4 into (3), we have

x =

6 )

4 (

5

10 

= 5
By substituting y

= –6 into (3), we have

x =

6 )

6 (

5

10 



3

10 

∴ The solution
is x = 5, y = 4 or x =

3

10 

, y = –6.
23. Let I be the original

sound intensity, then

the increased sound
intensity is 1.8I.
The change of the

corresponding
sound intensity
level







































0
0

log

10

8 .

1 log

10 I

dB
= 10







































0
0

log

8 .

1 log

I

= (10 log 1.8) dB
= 2.55 dB (cor. to 3
sig. fig.)

∴ The corresponding
sound intensity
level is increased
by 2.55 dB.
24. (a) Let E be the

relative energy
released by an
earthquake
measured 2.
By the definition
of the Richter scale, we
have

2 = log E
E = 102

∴ The relative
energy
released by
the

earthquake in

City A is 1.5
times to that
measured 2
= 1.5  102

∴ The

magnitude of
the

earthquake in
City A on the
Richter scale
= log (1.5 

102)
=2.18

(cor. to 3 sig. fig.)
(b) Let E be the

relative energy
released by an
earthquake
measured 8.
By the definition
of the Richter scale, we
have

8 = log E

E = 108

</div>
(18)<div class='page_container' data-page=18>

the

earthquake in
City B is half
of that
measured 8
= 0.5  108

∴ The

magnitude of
the

earthquake in
City B on the
Richter scale
= log (0.5 

108)
=7.70

(cor. to 3 sig. fig.)

Multiple Choice

Questions (p.246)

1. Answer: B

4
3
2
3
4
3
2
3

4
1
3

3
2
1

4 3

)

(

)

(

)

(

x





2. Answer: C
log 0.18 = log













100

18

= log 18 – log
100

= log (2  32) 
– log 100

= log 2 + log
32 – log 100

= log 2 + 2 log
3 – log 100

2 

 b
a

3. Answer: C

9(32x) – 10(3x) + 1 = 0

9(3x)2 – 10(3x) + 1 = 0

Let y = 3x, the equation

becomes

9y2 – 10y + 1 = 0
(9y – 1)(y – 1) = 0

y =

9

1

y = 1

∴ 3x =

9

1

or 3x =

3x = 3–2 or 3x =

∴ x =2 or

x =0

4. Answer: A












1

2

27

3

2x y
y
x

)
2
(

)
1
(






From (1),
3x + y = 33

∴ x + y = 3
From (2),

22x – y = 20

∴ 2x – y = 0
Consider the
simultaneous
equations:











0
2

3
y
x

y
x

)
4
(

)
3
(




(3) + (4),

(x + y) + (2x – y) =
3 + 0

3x

= 3

x =
1

By substituting x =1
into (3), we have

1 + y = 3

y = 2

∴ The solution is x =
1, y = 2.

5. Answer: C
32x = 23y

log 32x = log 23y

log (32)x = log (23)y
x log 32 = y log 23
x log 9 = y log 8

9
log

8
log



y
x

x : y = loglog98

6. Answer: C
log3 (3x + 12) – log3
(2x + 1) = 1

log3















1

2

12

3 x

= log3 3

1

2

12

3 

x

= 3
3x + 12 = 3(2x + 1)
3x + 12 = 6x + 3
3x = 9

x =3

7. Answer: D

(log x)2 – 2 log x2 + 3 =
0

(log x)2 – 4 log x + 3 = 
0

Let y = log x, then the
equation (log x)2 – 4 
log x + 3 = 0 becomes

y2 – 4y + 3 = 0
(y – 1)(y – 3) = 0

y = 1 or

y = 3
For y = 1, log x = 1

x = 10
For y = 3, log x = 3

x = 1000

∴ x =10 or

1000

8. Answer: A
Since

x

2
1

log

is

undefined for x 0,

the graph of

x

y

2
1

log



does

not cut the y-axis.

∴ C and D cannot be
the answer.

When x =

2

1

,

2

1 log

2
1



y

= 1.

∴ The graph passes
through the point (

2

1

,

1).

∴ The answer is A.
9. Answer: B

When x = 0, y = a0 = 1.

∴ The graph of y =

ax passes through

the point (0, 1).

∴ III cannot be the
answer.

10. Answer: D
Since the graph cuts
the y-axis, it cannot be
a logarithmic function.

∴ A and C cannot be

the answer.

Read from the graph,
the function passes
through the point (1,
3).

The point (1, 3)
satisfies the function y

= 3x but does not

satisfy the function y =
10x.

∴ The answer is D.
11. Answer: B

f(5 +2x) f(5 – 2x) =
52(5 +2x) + 152(5 – 2x) + 1

=
510 +4x + 1+ 10 – 4x + 1

=
22

12. Answer: B

x4 log x = 10

log x4 log x = 1

4 log x log x = 1
(log x)2 =

4

1

log x =

2

1

or
log x =

2

1 

(rejected)

x = 2
1

10

= 10

HKMO (p. 248)

1. 4a = 10 and

25b = 10

log 4a = 1 and

log 25b = 1

a log 4 = 1 and

b

log 25 = 1

a

1

= log 4 and

b

1

= log 25

∴

a

1 b

1

= log 4
+ log 25

= log (4 

25)

= log 100
=2

2. ∵ logxt = 6

∴ x6 = t

∵ logyt = 10

∴ y10 = t

</div>
(19)<div class='page_container' data-page=19>

∴ z15 = t
(xyz)60 = x60y60z60

= (x6)10(y10)
6(z15)4

= t10t 6t4
= t20

∴ logxyzt20 = 60

20 logxyzt = 60

logxyzt = 3

∴ d =3

3. S = log1443 2 + log144
6 3

= log144 3

2

+ log144

6
1

3

= log144 6

1
2

2

 +

log144 6
1

3

6

1

log144 4 +

6

1

log144 3
=

6

1

(log144 4 +
log144 3)

6

1

log144 12

6

1

log144 2

144

6

1



2

1

</div>

ON TAP HKI PHAN MU LOGA

<i>5</i>

<i>Exponential and Logarithmic Functions</i>

<i><b>5</b></i>

<b> Exponential and Logarithmic Functions</b>

<b>Activity </b>

<b>Activity 5.1 (p.216)</b>

3

1

<b>Activity 5.2 (p. 222)</b>

10

1

10

1

100

1

100

1













<i>N</i>

<i>M</i>

<b>Activity 5.3 (p.238)</b>

<b> Follow-up </b>

<b>Exercise </b>

<b>p. 205</b>

8

81

8

81

8

81

)

(

2

)

(

3

)

2

(

)

3

(

<i>x</i>

<i>x</i>

<i>x</i>

<i>x</i>

<i>x</i>

<i>x</i>

<i>x</i>

<i>x</i>

















<i>a</i>

<i>b</i>

<i>b</i>

<i>a</i>

<i>b</i>

<i>a</i>

<i>b</i>

<i>a</i>

<i>b</i>

<i>a</i>

<i>b</i>

<i>a</i>







