Chb Quiz
24
2
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Name: _____________________________ Class (No.): __________(_____) Teacher: Mr Luca Lee YL
Motivation of learning logarithms:
In chapter 6, we learnt how to solve exponential equations like
2! = 8,
5"!#$ = 125,
10%!&" = 100000,
etc.
The RHS of the equation can always be written as a power of the base of the LHS, using the
examples above,
2! = 2' ,
5"!#$ = 5' ,
10%!&" = 10(
then we can compare the powers to find out the value of -, i.e.,
- = 3,
2- + 1 = 3,
4- − 2 = 5
This raises the following question:
Q: What if the RHS cannot be expressed as a power of the base of the LHS?
A: We will use logarithms
7.1 Common Logarithms (Base 10)
7.1.1 Definition of Common Logarithms
We will first learn about the case of base 10.
Definition:
Definition:
If - = 10) , then log - = 5
If log - = 5, then - = 10)
Examples:
Examples:
(a) 1000 = 10'
(a) log 1000 = 3
⟹ log 1000 = 3
(b) 10 = 10$
103 3
log
⟹ log 10 = 1
(c) 0.01 = 10&"
⟹ log 0.01 = −2
⟹ 1000 = 10'
(b) log 10 = 1
⟹ 10 = 10$
(c) log 0.01 = −2
⟹ 0.01 = 10&"
In other words, we have the following relation:
! = 10!
log ! = (
1
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Mgx
Motivation:
O
y
Consider - = 10) , since 10) is always positive for all values of 5, then - > 0, i.e., -,
(i)
again, is always positive.
X log
For example,
10% = 10000,
10&" = 0.01,
o
Xlog negative
etc.
Since we know that - is always positive, then log - is undefined for - ≤ 0.
(ii)
Since 10 = 10$ , then log 10 = 1.
(iii)
Since 1 = 10* , then log 1 = 0.
(iv)
From the examples, we see that log 10% = 4, log 10&" = −2 etc. In general, log 10, = @
for any real number @.
Properties of log derived from its definition:
(i)
log - is only defined for - > 0, i.e., log 0 and log(negative) are undefined.
-
log 0 is undefined, log(−2) is also undefined.
(ii)
log 10 = 1
(iii)
log 1 = 0
(iv)
log 10, = @, where @ is any real number.
Example 7.1 (Refer to Example 7.1 from 4B07)
Find the values of the following common logarithms without using a calculator.
(a)
(b)
log 10 000
Solution:
Togo4 4
(c)
log 0.000 01
log10 5
log 7
$
$***
8
5
(a) Because 10% = 10 000, therefore log 10 000 = 4.
(b) Because 0.000 01 = 10&( , therefore log 0.000 01 = −5.
$
$
$
(c) Because $*** = $*! = 10&' , therefore, log 7$***8 = −3.
Exercise 7.1a
Find the values of the following common logarithms without using a calculator.
(a)
log 100 000
5
(b)
log 0.0001
4
(c)
log 7
$
8
$* ***
4
2
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.1b
Find the values of the following common logarithms without using a calculator.
(a)
log 10 000 000
(b)
log 0.000 001
(c)
1
æ
ö
logç
÷
è 100 000 000 ø
big
1000 103
Exercise 7.1c [Important Example]
Given that B and C are real numbers, find the values of the following common logarithms without
using a calculator.
loga10 9
(a)
(b)
log 109 9
9
40
log 10 m
(c)
log 1.0
9
m
mtg
æ 1 ö
logç n ÷
è 10 ø
logion
logion
n
M
Example 7.2 (Refer to Example 7.2 from 4B07)
Find the values of the following unknowns correct to 3 significant figures.
(a)
10) = 3
(b)
log - = 2.3
Solution:
(a) Using the definition of log:
10) = 3
5 = log 3
Therefore, 5 = 0.477 (3 s.f.)
(b) Using the definition of log:
log - = 2.3
- = 10".'
Therefore, - = 200 (3 s.f.)
3
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.2a
Find the values of the following unknowns correct to 3 significant figures.
(a)
10) = 18
(b)
log - = 1.2
109 18
Y
log x
logis
X
1.2
01.2
Exercise 7.2b
Find the values of the following unknowns correct to 3 significant figures.
(a)
10 y =
1
8
Y
toga
I
10
log x = -1.8
(b)
X
logIg
1 g
10
l 8
7.1.2 Properties of Common Logarithms
Properties of log:
(i)
log(%&) = log % + log &
(ii)
log * " + = log % − log &
(iii)
c # ) = - log %, where . is a real number
log(%
!
1
109M
alogM
Proof:
(i)
Let - = log E and 5 = log F. Then E = 10! and F = 10) .
EF = 10! ⋅ 10)
EF = 10!#)
log(EF) = - + 5
log(EF) = log E + log F
(ii)
Let - = log E and 5 = log F. Then E = 10! and F = 10) .
E 10!
=
F 10)
E
= 10!&)
F
E
log H I = - − 5
F
E
log H I = log E − log F
F
4
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
(iii)
Let - = log E. Then E = 10! .
E, = (10! ),
E, = 10,!
log(E, ) = @log(E, ) = @ log E
Special case when J = KL: log 10, = @ log 10 = @, because log 10 = 1.
DO NOT make the following mistakes:
(i)
log(- + 5) ≠ log - + log 5
(ii)
log(- − 5) ≠ log - − log 5
(iii)
log(-5) ≠ (log -)(log 5)
(iv)
log 7)8 ≠ ./0 )
(v)
(log E), ≠ @ log E
!
./0 !
4kgMa
4logM 4log N
Example 7.3 (Refer to Example 7.3 from 4B07)
Ei
Find the values of the following expressions without using a calculator.
(a)
log 125 + log 8
(b)
fr8
3 log 2 − log 80
Solution:
log
(a) Using the properties,
log
f
./0 √(
(c)
./0 $"(
logfo
log to
f
I
log 125 + log 8 = log(125 × 8)
I
log $ + log & = log($&)
= log 1000
= log 10'
=3
log(10" ) = ,
(b) Using the properties,
3 log 2 − log 80 = log 2' − log 80
, log $ = log $"
= log 8 − log 80
8
= log H I
80
1
= log H I
10
$
log $ − log & = log . /
&
= log 10&$
= −1
log(10" ) = ,
5
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
(c) Using the properties,
$
!
log √5
log 5"
=
log 125 log 5'
From Chapter 6: √, = ,"
1
log 5
=2
3 log 5
Remember:
log $
$
≠ log . /
log &
&
=
log $" = , log $
1
6
Exercise 7.3a
Find the values of the following expressions without using a calculator.
(a)
log 4 + log 25
log
(b)
log 90 − 2 log 3
./0 2
(c)
./0 "3
4 25
log100
a
log90 log 3
log
E
1953
FEI
L 1
31
Exercise 7.3b
Find the values of the following expressions without using a calculator.
(a)
a
æ3ö
log 6 - logç ÷
è5ø
log 6 bog 3
log
E
log 10
s
log 40 + 2 log 5
(b)
(c)
log 7
log 3 7
log 40 log 5
b
log to logs
log
40 5
log low
3
59,77
3
6
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
log E
loga logb loge
Exercise 7.3c
Find the values of the following expressions without using a calculator.
log 3 - log 12 - log 25
(a)
IgE log25
log Eg
on
2 log 2 + log 4 - log 8
(b)
log 32
log 3 16
log 2 logy dogs
log
2
4 8
On
Gta
log
(c)
21
Example 7.4 (Refer to Example 7.4 from 4B07)
Simplify
#
$
./0 ! ! #./04 5
./0 !
, where - > 0 and - ≠ 1.
Solution:
General step:
-
Always use log E, = @ log E to “lower” the power.
Alternatively,
1
log - ' + log 7-8
log -
=
log - '
log - &$
+
log -
logMa
alogM
1
1
log - ' + log 7 8 log 7- ' × 8
- =
log log -
=
3 log - − log log -
=
log - "
log -
=
2 log log -
=
2 log log -
=2
=2
Exercise 7.4a
7
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Simplify the following expressions, where - > 0 and - ≠ 1.
(a)
./0 ! % &./0 !
./0 ! !
./0 ! %
(b)
./0 √!#./0 !
Exercise 7.4b
Simplify
log x 2
3 log x - log 3 x
75
I
44É
, where x > 0 and x ¹ 1 .
LEIF
É
4g
I
1
8
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.4c
Simplify
XYZ
log x + log y
2
, where x > 0, y > 0 and xy ¹ 1 .
log( xy 2 )
2
4
log x't logy
log x t log y
4
s
i
qq.IE
x
Est
ygg
Example 7.5 (Refer to Example 7.5 from 4B07)
Given that log 3 = - and log 5 = 5, express the following in terms of - and/or 5.
(a)
log 15
Solution:
(b)
(c)
log 30
2
110910
General Steps:
-
I
log 18
KEY
Express the numbers, 15, 30 and 18, using 3, 5 and/or 10.
(a) Since 15 = 3 × 5, then
log 15 = log(3 × 5)
= log 3 + log 5
=-+5
(b) Since 30 = 3 × 10, then
log 30 = log(3 × 10)
= log 3 + log 10
=-+1
(c) Note that 18 = 2 × 3 × 3, but 2 isn’t one of 3, 5, and 10. So, we have to express 2 in terms of 5
and 10 as well. Then 18 =
$*
(
×3×3
10
log 18 = log H × 3 × 3I
5
2
= log 10 − log 5 + log 3 + log 3
= 1 − 5 + 2-
9
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
7
110910
Exercise 7.5a
Given that log 2 = @ and log 3 = R, express the following in terms of @ and/or R.
(a)
(b)
log 18
T92
log
log 3
10910
3
5
12
log E
1093 log 10
log
1
29
log 15
log 3
10910
21092
log
log 2
2b
t
log 15
to
log 22 10
log 9 2
log9 1092
Fogg
2
21093
log 40
b
log18
a
(c)
log 40
b
al
l
t
2
a
logoff llog.IM
Exercise 7.5b
Given that log 2 = a and log 7 = b , express the following in terms of a and/or b.
(a)
10914
a
log
æ4ö
logç ÷
è7ø
1
44
b
F
21092 log
La
bi
log
t
log
E
(c)
o
1097
1097
1
5
1095
t
log
logtt log 10 logz
log I
o
log 35
log 7
a
b
logs
ogl
log 35
o
logs log
log 4 lost
log 2 log
æ1ö
logç ÷
è5ø
(b)
t
l
a
log
o
log 10
log
I
lot
t
1092
log 2
a
10
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Further Practice P.7.10 Q2
Simplify the following expressions, where - > 0, 5 > 0 and -5 " ≠ 1.
(a)
% ./06√!)7
./0(!) % )
(b)
rxyF xy
')
3 log - − log(35) + log 7 ! ! 8
d
44ft
Iliff
logy
log x
4 Ilogxtlogy
log x t 2 logy
zlogx
logy
logxtalogy
a
9
3logx
310gX
7011
Togi
log3g
logx logby
log
log
log I
Exercise 7A Q45
s
f
log 33
log
E
log
IT
01
Given that 2 = 10: and 12 = 10; , express log 3 in terms of S and T.
log
2
2
1093
10
105
12
r
log12
log
5
E
log 12 log 22
S
Zr
11
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Selected Exercises:
Further Practice P.7.10
Exercise 7A
Level 1: Q8-13, Q24-25, Q27-28
Level 2: Q36-44.
End of Section Exercises:
Q1.
Q2.
Find the values of the following common logarithms without using a calculator.
(a)
log10 7
(b)
log10 -9
(c)
log10 3.4
(d)
log 10 000 000
(e)
log 0.000 001
(f)
æ1ö
logç ÷
è 10 ø
(g)
æ 1 ö
÷÷
logçç
è 10 000 ø
(h)
log 100 000
(i)
log 0.001
(j)
log 4 100
(k)
(l)
log 7 '
10 y = 6
8
(b)
10 y =
2
3
log x = 0.1
(c)
Find the values of the following expressions without using a calculator.
æ1ö
3 log 2 + log ç ÷
è8ø
(a)
log 2 + log 50
(b)
log 60 - log 6
(d)
æ1ö
log ç ÷ - 2 log 5
è4ø
(e)
log 27
log 3
(g)
æ 1 ö
log 8 - log ç
÷
è 125 ø
(h)
1
log 64 + 2 log 5
3
(i)
log 9 - log 6 - log15
æ2ö
è3ø
(k)
log 4 - log 8
log 8
(l)
log 3 + log 4
log 9 + 4 log 2
(j) 3 log ç ÷ - log 8 + log 27
Q4.
$
√$*****
Find the value of the following unknows correct to 3 significant figures.
(a)
Q3.
&
log √1000
(c)
(f)
log 8
log 16
Simplify the following expressions, where - > 0 and - ≠ 1.
(a)
log x 4
log x 6
(b)
log x 3 - log x 2
2 log x
(e)
log x + log x 3
log x - log x
(c)
æ 1 ö
log x 2 + log ç 3 ÷
èx ø
log x
(d)
æ 4ö
2 log(5 x) + logç 3 ÷ + log x
èx ø
12
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
(f)
Q5.
æxö
log( xy ) + logçç ÷÷ - 2 log x
è yø
log( x 2 y 3 ) - 2 log( xy )
log 3 y
(g)
Given that log 2 = @, express the following in terms of @.
(a)
log 16
(b)
æ 1 ö
logç ÷
è 32 ø
(e)
log 5
(f)
log 0.4
(c)
(d)
log 8
log 200
Q6. Given that log 3 = x and log 5 = y, express the following in terms of x and y.
(a)
log 75
(b)
æ 3 ö
logç ÷
è 25 ø
(c)
log 45
Q7. Given that log 2 = x and log 3 = y, express the following in terms of x and y.
(a)
log 24
(b) log 45
(c)
æ 27 ö
logç ÷
è 5 ø
Q8. Given that log 3 = x and log 4 = y, express the following in terms of x and y.
(a)
log 36
(b)
log 1.2
(c)
log18
7.2 Logarithm Equation
General step to solving logarithm equations:
Case 1: Use the properties of log, e.g.
log / = / = 10#
and other properties as well.
Case 2: Compare the log, e.g.
log / = log 2
/=2
Example 7.6 (Refer to Example 7.6 from 4B07)
Solving the following logarithm equations.
(a)
log(3- + 1) = 2
(b)
log - − log(- − 3) = −1
Solution:
(a)
This belongs to case 1
log(3- + 1) = 2
3- + 1 = 10"
3- = 99
Remember:
log(2 + 3) ≠ log 2 + log 3
- = 33
13
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
(b)
This belongs to case 1
log - − log(- − 3) = −1
log 7
8 = −1
-−3
= 10&$
-−3
1
=
- − 3 10
10- = - − 3
9- = −3
-=−
1
(rej. )
3
$
However, log - is undefined for - ≤ 0, we have to check whether the solution - = − ' will
$
$
lead to log(negative). For example, log 7− '8 and log 7− ' − 38 = log 7−
$*
'
8 are both the
$
case of log(negative). Therefore, we need to reject - = − .
'
Hence, log - − log(- − 3) = −1 has no solution.
Exercise 7.6a
Solve the following logarithm equation.
(a)
(b)
log(3- − 8) = 1
log 3 87 1
3
8
3
x
10
18
6
b
log(- − 5) − log - = 1
log
x
log
5
log
L
x
XE
I
XII
10
x 5
10X
5
9x
5g
ve
No solution
log
g
b
undefined
14
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.6b
Solve the following logarithmic equations.
(a)
log(1 - 3 x) = -1
log11 3 1
1 311
x
1
10
31
log(2 x - 1) - log( x + 4) = 1
(b)
II
log
1
I
IF
0.9
0.311
2
lol
1
10
Sx
41
4
x
4
log
4
co
4
4
40
cry
b undefended
No solution
Example 7.7 (Refer to Example 7.7 from 4B07)
Solve the logarithm equation log(- − 2) + log 3 = log(2- + 5).
Solution:
This belongs to case 2:
log(- − 2) + log 3 = log(2- + 5)
log[3(- − 2)] = log(2- + 5)
3(- − 2) = 2- + 5
Remember:
log 2 + log 3 ≠ log(2 + 3)
3- − 6 = 2- + 5
- = 11
We need to check whether - = 11 satisfies the original equation, i.e., we check if log(negative) will
happen. In this case, log(11 − 2) = log 9 and log(2(11) + 5) = log 27 are both well-defined.
15
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.7a
Solve the logarithm equation log(2- − 3) = log(- + 3) − log 2.
Exercise 7.7b
Solve the logarithmic equation log(3 x - 2) - log 5 = log(6 - x) .
3
log
3g
3
x
log
2
6
x
30 5x
Sx
32
x
44
Exercise 7.7c
sx.
Solve the logarithmic equation log( x - 2) + log( x + 1) = 2 log
log
x 2
x 2
x
Log x2
xx
XT
x
X
x2
2
X 2
0
x
No solution
2
veg
16
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Motivation:
The most important application of log is to solve exponential equations. We can do so by taking
log on both sides.
General step to solving exponential equations with log:
To solve @ ! = R, we take log on both sides:
24 5
2 4
2 2
X
use
log
@! = R
log(@ ! ) = log R
- log @ = log R
2
-=
4 log $ = log $(
log R
log @
Example 7.8 (Refer to Example 7.8 from 4B07)
Solve the following exponential equations and give your answers correct to 3 significant figures.
(a)
2!&$ = 5
3! = 4!#$
(b)
Solution:
3! = 4!#$
2!&$ = 5
log 3! = log 4!#$
log(2!&$ ) = log 5
- log 3 = (- + 1) log 4
(- − 1) log 2 = log 5
- log 3 − - log 4 = log 4
log 5
-−1=
log 2
-=
I
-(log 3 − log 4) = log 4
log 5
+1
log 2
-=
log 4
log 3 − log 4
- = −4.82 (3 s. f. )
- = 3.32 (3 s. f. )
Exercise 7.8a
Solve the following exponential equations and give your answers correct to 3 significant figures.
(a)
3!#$ = 2
(b)
4! = 5!&$
4
1094
x
logy
log 4
log
1095
x
1
5
1095
x 1
logs
21095
1095
gl
1
1095
17
g
g
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.8b
Solve the following exponential equations and give your answers correct to 3 significant figures.
(a)
x 3
6 x -3 = 12
log6
X
3
7 x +1 = 3 x -1
(b)
10912
1091
log 6
x
19
x
4.39
3
3s
f
LITEX
Exercise 7.8c
Solve the exponential equation 2 5 x • 3 2 x = 60 and give your answer correct to 3 significant figures.
log 60
191251.34
log 25
5 1092
x
51092
log 3
log 60
t
2x
logs
log 60
t
21093
log 60
t
x
5
x
3
0.723
093
13 s f
18
3153
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
55
1
Further Practice P.7.15 Q2
' !
(a)
7(8 = 4
log
3
log
1094
E
Log 4
151
1094
log
log4
1095
3
its
45g
Solve the exponential equation 3
3
213
3 3
2.3
3
8
3
3
3
x 1093
= 8.
While having
8
x
Sub
547
54
710
0
I
20
35
or
log
102
54
y
x
f
log x
2
into
toy
t
3
log
2
logy
x
4
3
24
0
x
log 1 591
log
Ex
can't take log
log(- − 55) = 2
Solve the simultaneous equation [
.
log - − log 5 = 1
log x
1093
8
311
Exercise 7B Q36
+ 2(3
!&$ )
1095
44 3159
Exercise 7B Q35
!#$
152 225
3157
4! = 3(5! )
(b)
75
102
2011
2004
1
logy
g
L
Ey
10
x
Loy
19
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7B Q37
log 5 = log(2- ) − 1
Solve the simultaneous equation \
.
log - = 2 log ]5 + 1
logy
log
logy log
x
log
2x
I
log x
1
E
1
10
toy
2X
x
5g
Compare
Lug log
sub
zlogFM
109
log
log x
log yet
log Y
into
II
0
0
1
x
by
Ytl
4
44
Selected Exercises:
Further Practice P.7.15
7
Exercise 7B
yet
Level 1: Q7-13, Q17-19.
541
Level 2: Q22-23, Q26-27, Q28-29, Q32-35.
End of Section Exercises:
Q1.
Solve the following logarithm equations.
(a)
log( x - 6) = 0
(b)
log(9 x - 8) = 1
(c)
log 5 + log( x + 2) = 2
(d)
log (4 x - 3) - log 5 = -1
(e)
log(4 x - 3) = log 9
(f)
log 8 + log 2 = log( x + 2)
(g)
log(7 x + 11) = 2 log 5
(h)
log (5 x - 1) - 3 log 4 = 0
(i)
log(3x - 1) - log( x - 1) = log 5 (j)
log (2 x - 7) - log 3 = log( x - 3)
(k)
1
log(8 x - 4) = 1
2
(l)
log( x - 7) + log 2 = 1
(m)
log(3x - 4) - log 2 = 3
(n)
log(2 x - 1) - log 7 = log 9
(o)
log( x - 2) = log( x + 1) - 1
(p)
log( x + 8) + log 3 = log(5 x - 2)
(r)
log( x + 7) - log(4 x + 3) = log 4 - log 6
(q)
log(3x) + 4 log 2 = log(8 x + 5)
20
x
5g
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Q2.
Solve the following exponential equations.
x
Q3.
(a)
4 =7
(b)
æ2ö
ç ÷ =6
è3ø
(e)
4 x -1 = 5
(f)
3 x • 8 x = 12
(i)
7 x +1 + 7 x = 15 (j)
x
(c)
5 2 x -3 = 9
(d)
5 x = 6 x -1
(g)
6 x = 3 x +1
(h)
7 x = 10(2 x )
4 x + 2 - 2(4 x +1 ) = 18
In each of the following, express 5 in terms of -.
(a)
7.3
log(3x + 2 y ) = 1
(b)
2 log x - 5 log y = 0
Applications of Common Logarithm
7.3.1 Sound Intensity Problem [mainly for physics students]
Concept:
The sound intensity level (f) is measured in decibels (dB). It is defined as
a
f = 10 log H I
a*
where
-
a is the sound intensity (in W/m" ) measured, and
-
a* is the lowest sound intensity that can be heard by humans, which equals to
a* = 10&$" W/m" .
Example 7.9 (Refer to Example 7.9 from 4B07)
Given that the sound intensity produced by a jackhammer is 10&' W/m" , find its sound intensity
level.
Solution:
Recall that a* = 10&$" W/m" .
The corresponding sound intensity level
= 10 log b
10&'
c dB
10&$"
= 10 log 102 dB
ß log 102 = 9 log 10 = 9
= 10 × 9 dB
= 90 dB
21
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Sound
Exercise 7.9a
level P dB
intensity
&<
"
Given that the sound intensity of a light rainfall is 10
p
p
lo log
to
log
40 dB
E
if
20
SoundIntensity
W/m , find its sound intensity level.
10
W m
w m
dB
Exercise 7.9b
Given that the sound intensity produced by a new model vacuum cleaner is 10&( W/m" , find its
sound intensity level.
B
to
log
p to
dB
I
1Er dB
Example 7.10
Given that the sound intensity level of a diesel engine is 75 dB, find its sound intensity. (Given your
answer in scientific notation and correct to 3 significant figures.)
Solution:
This is a problem of solving logarithm equation. We need to use the formula for sound intensity level
to obtain a logarithm equation.
Let a W/m" be the required sound intensity.
a
10 log H I = f
a*
a
I = 75
10&$"
a
log H &$" I = 7.5
10
a
= 103.(
10&$"
10 log H
a = 103.( × 10&$"
a = 10&%.(
a = 3.16 × 10&(
∴ The sound intensity of the diesel engine is 3.16 × 10&( W/m" .
22
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.10a
If the sound intensity level of a conversation is 54 dB, what is its sound intensity? (Give your answer
in scientific notation and correct to 3 significant figures.)
54
10
54
log
105
105 4
log
Fiz
It
10
I
12
1
2.51
10
7
3
s
f
Exercise 7.10b
It is given that the sound intensity level produced by a crying baby is 46 dB. Find its sound intensity.
(Give your answer in scientific notation and correct to 3 significant figures.)
23
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.10c
It is given that the sound intensity level of traffic in rush hour on Nathan Road and Jordan Road are
88 dB and 80 dB respectively. How many times of the sound intensity of traffic on Jordan Road is the
sound intensity on Nathan Road? (Give your answer correct to 3 significant figures.)
Lee
In
W Im
Is
I
p cong
situ
be the sound
intensity
Ly
J
88
fo
N
on
10
log
1010g
tf
Fr
Q
8
to
0.8
log
0.8
of
1008
31
log Ttv
In
log In
log 10
log
I
to
log
log Ig
Ev
log 10
log Ey
31
6 31
3s
f
24
12
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
7.3.2 Richter Scale
Concept:
The Richter Scale measures the magnitude of an earthquake. The magnitude (E) of an
earthquake measured on the Richter scale is defined as
2
E = log h + j
3
where
-
h is the energy in joules (J) released in an earthquake, and
-
j is a constant.
Example 7.11 (Refer to Example 7.11 from 4B06)
The magnitudes of the Eastern Sichuan Earthquake (2008) and the Pacific coast of Tohoku
Earthquake (2011) were measured 7.9 and 9.0 respectively on the Richter scale. How many times of
the energy released in the Eastern Sichuan Earthquake was the energy released in the Pacific coast
of Tohoku Earthquake? (Give your answer in 3 significant figures.)
Solution:
Let h$ and h" be the energy released (in J) in the Eastern Sichuan Earthquake and the Pacific coast
of Tohoku Earthquake respectively.
-
We need to use the Richter scale formula to obtain a pair of logarithm equations.
2
7.9 = log h$ + j … … (1)
3
i
2
9.0 = log h" + j … … (2)
3
Since we would like to know the times h" is larger than that of h$ , we would like to compute the
value of
=%
=#
. From the equations,
2
2
9.0 − 7.9 = log h" + j − H log h$ + jI
3
3
2
2
1.1 = log h" − log h$
, log $ − , log & = ,(log $ − log &)
3
3
$
= , log . /
2
h"
&
1.1 = log H I
where , is a constant.
3
h$
log H
h"
I = 1.65
h$
h"
= 10$.>(
h$
= 44.7 (3 s. f. )
∴ The energy released in the Pacific coast of Tohoku Earthquake was 44.7 times that in the Eastern
Sichaun Earthquake.
25
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.11a
The magnitudes of the Great Chilean Earthquake (1960) and the San Francisco Earthquake (1906)
were measured 9.5 and 7.8 respectively on the Richter scale. How many times of the energy
released in the San Francisco Earthquake was the energy released in the Great Chilean Earthquake?
(Give your answer correct to 3 significant figures.)
95
78
1.7
3 log E
Goal
tk
33 log E t K
log
É
O
log Er
E
E
255
log
E
355
13s
f
Exercise 7.11b
In an underground atomic bomb test, the energy released by an atomic bomb will cause an
earthquake measured 6.0 on the Richter scale. How many atomic bombs of this kind are involved if a
bomb test causes an earthquake measured 7.5 on the Richter scale? (Give your answer correct to
the nearest integer.)
26
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
7.3.3 Logarithmic Transformation
Motivation:
In application, we would like to convert nonlinear relation into linear relation.
Q: Why?
A: Because linear graphs are easy to interpret and to look at.
When we take log on both sides, we will transform the curved (nonlinear relation) into a straight
line (linear relation) with a “different” x- and y-axes.
Case 1: Converting l = mn? .
Steps:
(1) Take log on both sides and simplify the expression.
log 5 = log o@ !
log 5 = log o + log @ !
log 5 = - log @ + log o
y Kat
logy log ka
logy logk t logat
logy x loga t logk
(2) The vertical axis is no longer the 5-axis, it becomes the log 5-axis.
(3) Interpretation:
On the right-hand side, think about log 5 = B- + p.
-
B is the slope. In this case, slope = log @.
-
p is the intercept on the vertical axis, in this case, vertical intercept = log o.
o
-
Remark: you may call it the log 5-axis, then p is the “log 5-intercept”.
Each point on the coordinate plane on the RHS has coordinates (-, log 5).
o
Example: If (1,2) is a point on the plane on the RHS, then
(-, log 5) = (1,2) ⟹ - = 1 and log 5 = 2
27
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Example
Let 5 = 5(3! ). Rewrite the expression in the form log 5 = - log @ + log o. Write down the slope and
the intercept on the vertical axis.
Solution:
For this kind of question, we take log on both sides:
log 5 = log 5(3! )
Then we use the properties of log to simplify the expression:
log 5 = log 5 + log 3!
= log 5 + - log 3
= - log 3 + log 5
By considering log 5 = - log 3 + log 5 as a straight line,
(i)
the slope is log 3
(ii)
and the intercept on the vertical axis is log 5.
Exercise
$
Let 5 = " (5! ). Rewrite the expression in the form log 5 = - log @ + log o. Write down the slope
and the intercept on the vertical axis.
5
y
log y
log It log
logy
x
y
t
5
Gatta
intercept
log I
log
t
log
2
I
2
77
28
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise
Let 5 = 9(7&"! ). Rewrite the expression in the form log 5 = - log @ + log o. Write down the slope
and the intercept on the vertical axis.
Exercise
Let 5 = 2(5'!#$ ). Express log 5 in terms of -. Write down the slope and the intercept on the
vertical axis.
29
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Example 7.12a [Converting l = mn? ] (Refer to Example 7.12 from 4B07)
Let 5 = @R ! , where @ and R are positive constants.
The figure shows the graph of log 5 against -. The graph
passes through (0,2) and (5,1.5).
(a) Express log 5 in terms of @, R and -.
(b) Find the values of @ and R, correct to 3 significant figures
if necessary.
Solution:
(a) It is given that 5 = @R ! , we take log on both sides and get
log 5 = log @R !
= log @ + - log R
= - log R + log @
(b) By considering log 5 = B- + p, we have
-
slope = log R, i.e.,
It is given that the straight line passes through (0,2) and (5,1.5), therefore, the slope is
2 − 1.5
0−5
log R =
log R = −0.1
R = 10&*.$
R = 0.794 (3 s. f. )
-
and the intercept on the vertical axis is log @
The intercept is (0,2), therefore,
log @ = 2
@ = 10"
= 100
30
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.12a
The figure shows the graph of log 5 against -. The graph
passes through (0,1) and (2,2). It is given that 5 = @R ! ,
where @ and R are positive constants.
(a) Express log 5 in terms of @, R and -.
(b) Find the values of @ and R, correct to 3 significant figures
if necessary.
a
b
logy
log
logb
x
log Eggs
a
a
If
I
Intercept
co
b
lot
3.16
3s
f
31
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Case 2: Converting l = mq@ .
Steps:
vs
case
ye ab
logy xlogbtloga
(4) Take log on both sides and simplify the expression.
logy log x
log 5 = log o + log - A
logy logkenlogx
log 5 = C log - + log o
logy nlogxt
logk
The vertical axis is no longer the 5-axis, it becomes the log 5-axis.
log 5 = log o- A
(5)
Fox
The horizontal axis is no longer the --axis, it becomes the log --axis.
y
mx to
stope
IEingen
(6) Interpretation:
On the right-hand side, think about log 5 = B log - + p.
-
B is the slope. In this case, slope = C.
-
p is the intercept on the vertical axis, in this case, vertical intercept = log o.
o
-
Remark: you may call it the log 5-axis, then p is the “log 5-intercept”.
Each point on the coordinate plane on the RHS has coordinates (log - , log 5).
o
Example: If (1,2) is a point on the plane on the RHS, then
(log - , log 5) = (1,2) ⟹ log - = 1 and log 5 = 2
y kx
Example
#
!
Rewrite 5 = 2- into log 5 = C log - + log o. Find the values of C and o.
logy
Exercise
$
log 2
1092 t
5
flog
Ja
log k logs
K
2
'
Rewrite 5 = ' - &) into log 5 = C log - + log o. Find the values of C and o.
logy
log
XE
log's t log
log's
54 log
k
x
x
n
541
32
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Example 7.12b [Converting l = mq@ ]
The figure shows the graph of log 5 against log -.
The graph passes through (0, 4) and (6, 0.7). It is
n
given that y = kx , where o and C are constants,
and o > 0.
(a)
Express log 5 in terms of o, C and log -.
(b)
Find the values of o and C.
(c)
Find the value of - when 5 = 8500.
Solution:
(a) We take log on both sides.
5 = o- A
log 5 = log o- A
= log o + log - A
= C log - + log o
(b) By considering log 5 = B log - + p, we have
-
slope = C, i.e.,
4.7 − 4
6−0
7
=
60
C=
-
intercept on the vertical axis = log o, i.e.,
log o = 4
o = 10%
o = 1000
(c) First of all, we substitute the values of o and C back into log 5 = C log - + log o and get
log 5 =
7
log - + 4
60
When 5 = 8500,
7
log - + 4
60
7
log 8500 − 4 =
log 60
60
(log 8500 − 4)
log - =
7
- = 0.248 (3 s. f. )
log 8500 =
33
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.12b [Converting l = mq@ ]
The figure shows the graph of log 5 against log -. The graph
passes through (0, 1) and (3, 0.5). It is given that y = kx n ,
where o and C are constants, and o > 0.
(a)
Express log 5 in terms of o, C and log -.
(b)
Find the values of o and C.
(c)
Find the value of - when 5 = 100.
log y
a
b
nlogxt log k
h
C
If
log k
f
k
I
104
Lox
y
Sub y
100
100
LOX
lo
X
X
o
X
O
F
b
1
or
00000
P 7 6403 Me
D
8
10
6
1
a
logy
ax
x
blog
109924ft
3
log x
intercept
log
2
a
a
w
got
34
O
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Example 7.13 (Refer to Example 7.13 from 4B07)
The number of cells (F) on the -th day of an experiment is given by F = 1500 log(- + 82).
(a) Find the number of cells on the 6th day of the experiment, correct to the nearest integer.
(b) Given that the number of cells on the oth day of the experiment is 3000, find the value of o.
Solution:
(a) When - = 6,
F = 1500 log(6 + 82)
= 2917 (cor. to. the nearest integer)
∴ The number of cells on the 6th day is 2917.
(b) When - = o, we have F = 3000. Hence,
3000 = 1500 log(o + 82)
log(o + 82) = 2
o + 82 = 10"
o = 18
Exercise 7.13a
The number of visitors (F) to a tourist attraction on the sth day since opening is estimated by
F = 250 + 100 log(2s).
(a) Find the number of visitors on the 8th day.
(b) If the number of visitors on the oth day is 400, find the value of o.
(Given your answer correct to the nearest integer.)
35
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.13b
In a company, the estimated sales t (in thousand) of a new tablet on the -th day since it was
launched is given by S = 5 log(3 x - 2) + 10 .
(a)
Find the sales of the tablet on the 10th day after the launch, correct to the nearest integer.
(b) If the daily sales of the tablet reach 22 000, a bonus will be given to all employees. When will
this happen?
a
5
X
5
3
log
10
5
s
17
log
222
15
22
2.4
to
10
the
nearest
17000
integer
3
log
2
10
2
2
84.3
X
on
5
3
3
2
Sfm thousands
log 13
102 4
30
Cor
The sales
b
10
2
the
85th
day
36
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Example 7.14 (Refer to Example 7.14 from 4B07)
A sum of money ($u) is deposited in a bank at an interest rate of 4% p.a. compounded yearly. Find
the minimum number of years for the amount to exceed twice the principal.
Solution:
(1) Construct the function t, where t denotes the sum of money received after C years.
Suppose the sum of money is compounded for C years, then
t = u(1 + 4%)A
(2) Find C such that the sum of money after C years > twice the principal, i.e.,
t > 2u
u(1 + 4%)A > 2u
(1 + 4%)A > 2
log[(1 + 4%)A ] > log 2
C log(1 + 4%) > log 2
log 2
C>
(
log 1 + 4%)
Since log(1 + 4%) is positive,
we do not have to change the
inequality sign here.
If , > 6 > 0, then log , > log 6.
C > 17.6729 …
Always check if you are dividing
by a negative number!
Exercise 7.14a
The value of a mobile phone ($w) depreciates at a constant rate of 5% per month. At least how many
months will it take for its value to become less than half of its original value?
Let
n
be
the
V11
5
1 5
EV
7
1
Iggy
c
E
loss
n
I
h
13.51
At least 14 months
s
37
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.14b
In a sale, the selling price of a cushion decreases at a constant rate of 20% per day. At least how
many days will it take for its price to become less than
1
of its original price?
4
Exercise 7.14c
A study is carried out to investigate the growth of algae in a lake. The total area (A m2) covered by
algae on the nth day of the study is given by A = 20(1.2) n . Find the least number of days for the
total area covered by the algae to exceed 500 m2.
38
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Motivation: DSE Paper 2 type Questions
Arrange - = 2012"*$% , 5 = 2013"*$' and x = 2014"*$" in ascending order.
Difficulties: The numbers are too large; your calculator cannot handle it.
Idea: Use log to make the number “smaller”.
Concept:
If log @ > log R, then @ > R > 0.
The proof is better explained when we learn the graph of logarithmic functions.
Example
To compare 100" and 2$** , we consider
log 100" = 2 log 100 = 4
log 2$** = 100 log 2 = 30.1 …
Since log 2$** > log 100" . Therefore, 2$** > 100" .
Exercise
Arrange - = 2012"*$% , 5 = 2013"*$' and x = 2014"*$" in ascending order.
acb
2014 log
2012
72013 log
6653.5
a
c
2013
6650.6
2014W
Exercise
log
a
C
logb
Lon log 2014
6647.76
20132013
20122014
Arrange the numbers 123%(> , 246$'( and 456'"$ in ascending order.
456
log
123
135 log 246
3410914861
246135
952.9
322.7
853.8
45634
123456
39
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Selected Exercises from Textbook:
Exercise 7C:
Level 1: Q9-12.
Level 2: Q17-22.
End of Section Exercises
You may refer to the following formulas to answer all the exercises below.
(a)
The sound intensity level f (in dB) of a sound is defined as:
æ I
è I0
b = 10 logçç
ö
÷÷ ,
ø
where I is the sound intensity (in W/m2) measured, and I0 = 10-12 W/m2.
(b)
The magnitude (M) of an earthquake measured on the Richter scale is given by:
2
E = log h + j,
3
where E is the energy (in J) released in the earthquake, and K is a constant.
(Unless otherwise specified, give your answers in scientific notation and correct to 3 significant
figures if necessary.)
Q1.
Given that the sound intensity of a music concert is 6.3´10-3 W/m2, find its sound intensity
level
Q2.
Given that the sound intensity level of the noise from a busy street is 94 dB, find its sound
intensity.
Q3.
The magnitudes of the earthquakes occurred in city X and city Y were measured 4.5 and 3.2
respectively on the Richter scale. How many times of the energy released in the earthquake
in city Y was the energy released in the earthquake in city X?
Q4.
The energy released in the earthquakes in city A and city B are 4.72 × 10$* J and
6.45 × 10$$ J respectively. Find the difference in magnitudes of the two earthquakes on the
Richter scale.
Q5.
The figure shows the graph of log y against x. The graph
passes through (1,0) and (3,2). It is given that 5 = @R ! ,
where @ and R are constants.
(a) Express log y in terms of @, R and -.
(b) Find the values of @ and R.
40
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Q6.
The number of bacteria (B) in a candy after n hours of exposure is given by
B = 28 000 log(3n + 1) .
(a)
Find the number of bacteria after 6 hours of exposure, correct to the nearest
thousand.
(b)
Given that the number of bacteria after o hours of exposure is 56 000, find the value
of o.
Q7.
The monthly rent of a car park ($y) increases at a constant rate of 8% per year. At least how
many years will it take for the monthly rent to exceed twice its original rent?
Q8.
The sound intensity of loudspeaker A is 3 times that of loudspeaker B. What is the difference
between their corresponding sound intensity levels?
Q9.
The sound intensity level in a concert hall is increased by 6 dB when the singer starts singing.
How many times of the sound intensity in the concert hall before the singer sings is the
sound intensity after the singer sings?
Q10.
In a city, the energy released in the first earthquake with a magnitude of 5.8 on the Richter
scale is 5 times that in the second earthquake.
(a)
Find the magnitude of the second earthquake on the Richter scale.
(b)
If the energy released in the third earthquake is decreased to 30% of that in the
second earthquake, find the magnitude of the third earthquake on the Richter scale.
(Give your answer correct to 1 decimal places.)
Q11.
The figure shows the graph of log y against log x. The graph
passes through (2,4) and (4,7). It is given that log 5 = @ log - +
R, where @ and R are constants.
(a)
Find the values of @ and R.
(b)
Express 5 in terms of -.
(c)
Find the percentage change in the value of 5 when - is
increased by 50%.
Q12.
Mrs Wong deposits a sum of money ($P) in a bank at an interest rate of 6% p.a.
compounded half-yearly. Find the least number of years for the amount to exceed three
times the original amount of money.
41
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
7.4
common
Logarithms to an Arbitrary Base
7.4.1 Definition of Logarithms to an Arbitrary Base
e
In the following, we require @ > 0 and @ ≠ 1.
g
4
104
a
X
39
4
Examples:
Examples:
(a) 125 = 5'
(a) log " 32 = 5
⟹ 32 = 2(
⟹ log ( 125 = 3
(b) 16 = 16$
(b) log 2 729 = 3
⟹ 729 = 9'
⟹ log$> 16 = 1
(c) 0.25 = 2&"
(c) log ( 0.008 = −3
⟹ 0.008 = 5&'
⟹ log " 0.25 = −2
$
49
If log , - = 5, then - = @ )
If - = @ ) , then log , - = 5
($"
log x
Definition:
Definition:
(d)
log
$
= 8&'
(d) log ' 7<$8 = −4
$
⟹ log < 7($"8 = −3
$
⟹ <$ = 3&%
In other words, we have the following relation:
! = )!
log " ! = (
Example 7.15 (Refer to Example 7.15 from 4B07)
Find the values of the following logarithms without using a calculator.
(a)
(b)
log % 64
$
log " 7 8
%
(c)
log ( √125
Solution:
(a)
∵ 64 = 4'
∴ log % 64 = 3
(b)
$
∵ % = 2&"
$
∴ log " % = −2
(c)
!
∵ √125 = 5%
∴ log ( √125 =
'
"
42
y
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.15a
Find the values of the following logarithms without using a calculator.
(a)
log 3 49
(b)
$
log ' 7<$8
(c)
log " √32
Exercise 7.15b
Find the values of the following logarithms without using a calculator.
(a)
log 8 512
(b)
æ 1 ö
log 5 ç ÷
è 25 ø
(c)
log 6 3 36
(b)
2./0% ,
(c)
log √" 8
Exercise 7.15c
Find the values of the following.
(a)
log " 2,
43
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
7.4.2 Properties of Logarithms to an Arbitrary Base
Again, we assume @ > 0 and @ ≠ 1.
Properties of loga derived from its definition:
(i)
log , - is only defined for - > 0, i.e., log , 0 and log , (negative) are undefined.
-
log , 0 is undefined, log , (−2) is also undefined.
(ii)
log , @ = 1
(iii)
log , 1 = 0
(iv)
log , @B = R, where R is any real number.
Proof of (ii)-(iv):
(ii): Because @ = @$ , therefore log , @ = 1.
(iii): Because 1 = @* , therefore, log , 1 = 0.
(iv): By writing a very trivial equality, @B = @B , we must have log , @B = R.
Properties of loga:
Let @ > 0 and @ ≠ 1, R > 0 and R ≠ 1. For any E, F > 0,
(v)
log , (EF) = log , E + log , F.
(vi)
log , 7 D 8 = log , E − log , F.
(vii)
log , EA = C log , E, where C is a real number.
(viii)
log , E =
C
./0* C
./0* ,
ß base-change formula, very important in DSE.
The proof of (v)-(vii) are very similar to that of common logarithms (base 10).
Proof of (viii):
Let E > 0, @ > 0 and @ ≠ 1, R > 0 and R ≠ 1.
Furthermore, let - = log , E. Then
E = @!
log B E = log B @ !
log B E = - log B @
∵ 2 = log " $
-=
log B E
log B @
log , E =
log B E
log B @
log " $( = 4 log " $
44
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Take a closer look at the base-change formula
log " $
log ! $ =
log " &
The base we want: 6
The original base: ,
This formula is usually used in some DSE section B MC question.
Example 7.16 (Refer to Example 7.16 from 4B07)
Find the values of the following logarithms correct to 3 significant figures.
(a)
log ' 20
(b)
log *." 3
Solution:
How to use your calculator:
For arbitrary base, you may enter the base using a comma on your calculator.
For example, alculate log " 8 using your calculator.
First method: Using the base-change formula:
log " 8 =
log 8
log 2
then just enter log 8 ÷ log 2 in your calculator.
Second method: Using comma (,):
On your calculator, input log(2,8) to calculate the value of log " 8.
(a)
log ' 20 = 2.73 (3 s. f. )
log *." 3 = −0.683 (3 s. f. )
(b)
Exercise 7.16a
Find the values of the following logarithms correct to 3 significant figures.
(a)
log " 30
(b)
log % 1.5
Exercise 7.16b
Find the values of the following logarithms correct to 3 significant figures.
(a)
log 4 18
(b)
log 1 0.3
3
45
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Example 7.17 (Refer to Example 7.17 from 4B07)
Find the values of the following expressions without using a calculator.
(a)
log " 96 − log " 12
log
a
Ez
(b)
b
2 log > 3 + log > 4
(c)
34 loggy
0
log 8
10963
log 4
1092230
09,9
109,4
IlogI
log "( 125
c
log
4
5,1
I
21
357
9
199,4
5.5135
109636
3
125
345
53
125
F
3
25
logos 253
Exercise 7.17a
3
Find the values of the following expressions without using a calculator.
(a)
a
log % 32 + log % 8
log
32 8
(b)
log ( 40 − 3 log ( 2
b
log
logy Wb
41
0955
In
E
(c)
log "3 9
e
10949
1
4
I
46
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.17b
Find the values of the following expressions without using a calculator.
(a)
log 3 189 - log 3 7
(b)
log 8 16 + 5 log 8 2
log 49 7
(c)
I
É
I
41
a
Exercise 7.17c
Find the value of log , R × log B p × log B p, where @, R and p are numbers greater than 1.
logab
x
log
C X
log
a
II
I
1
47
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
1091
Example 7.18 (Refer to Example 7.18 from 4B07)
Simplify the following expressions, where - > 0 and - ≠ 1.
(a)
$
log " 7!8 + 2 log " - − log " (2-)
(b)
log
./0) !
X
./0% !
Solution:
(a) They all have the same base; we can use the log properties as usual.
1
1
log " H I + 2 log " - − log " (2-) = log " H I + log " - " − log " (2-)
1
= log " H ⋅ - " I − log " (2-)
1 "
⋅= log " |}
2-
L
a
1
= log " H I
2
= −1
(b) They do not have the same base; we should convert them into the same base first.
log " log % - log " 4
=
log " - log " -
log " $ =
=
log " 1
×
log " 4 log " -
=
1
2
log , $
log , ,
Exercise 7.18a
Simplify the following expressions, where - > 0 and - ≠ 1.
(a)
!%
log ' 7 8 − 3 log ' - + log ' (4-)
%
(b)
./0! ! %
./0+ !
4
a
48
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.18b
Simplify the following expressions, where x > 0 and x ¹ 1 .
(a)
log 1 x
æ x ö
log 5 (4 x ) + log 5 ç
÷ - 4 log 5 x
è 100 ø
3
(b)
2
log 2
x
2
7.4.3 Logarithmic Equations
General step to solving logarithm equations:
log
Case 1: Use the properties of log, e.g.
log # / = 3
/ = -$
and other properties as well.
log
X
202
X
2
X
Case 2: Compare the log, e.g.
log # / = log # 2
log X
/=2
2
x
x
32
2
52
We can only do comparison when the bases are the same.
logax
log y
My
logaxalogay
x
y
toga logb
a
b
49
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Example 7.19 (Refer to Example 7.19 from 4B07)
Solve the following logarithmic equations.
(a)
log " (3- − 1) = 3
log > - − log > (- − 1) = 2
(b)
(c)
log 2 4- = log ' 6
Solution:
(a) This belongs to case 1:
log " (3- − 1) = 3
3- − 1 = 2'
3- = 9
-=3
(b) This belongs to both case 1 and 2:
log > - − log > (- − 1) = 2
log > 7
8=2
-−1
= 6"
-−1
- = 36- − 36
35- = 36
-=
36
35
(c) This belongs to case 2, but we must change the bases into the same one:
log 2 4- = log ' 6
log " $ =
log , $
log , ,
log ' 4= log ' 6
log ' 9
log ' 4= log ' 6
2
log ' 4- = 2 log ' 6
"
log ' 4- = log ' 6
4 log " $ = log " $(
4- = 36
-=9
Reminder:
Remember to check if the answers satisfy the original equations by substituting the
answers into the original equations.
50
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.19a
Solve the following logarithmic equations.
(a)
log ( (2- + 1) = 3
(b)
log ' - − log ' (- − 6) = 1
(c)
log % (- + 1) = log " 3
a
log
2
1
3
124
2x
log
x t1
1947
3
x
621
x
logy
log
1 53
2
c
b
3
X
3
log 1
1092
109 us
9
11
It
210923
4923
x
Exercise 7.19b
18
84
Solve the following logarithmic equations.
(a)
log 3 (2 x + 3) = 2
(b)
log 2 x - log 2 (3 x - 4) = -1
(c)
æ1ö
log 2 ( x - 2) = log 1 ç ÷
3
2è ø
log21
2
log
x 2
log
x 2
log
7 2
x
2
X
log
t
logab
logia
logia
logza
log
logax
X
A
0109243T
logs
t
10923
3
5 a
51
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Further Practice P.7.34 Q2
logab
logia
logia
logza
logax
Logax
Solve the following logarithmic equations.
(a)
(b)
log ! 9 = 2
log 9
'
2
b
x2
9
3
x
or
log 1
3
x
log
2
x
x
logsX
XI
x
2
x
I
qt
y
y
Exercise [Using properties that are not taught directly]
Suppose @ > 1 and @ ≠ 0. Show that log # - = − log , -.
-
P 7.57
039140141
logab
logia
logia
logza
logax
log
ex 1
logs
2
1
log
log
log
X
att
2
1 1
2
2x
5
2
X
E
It
y
A
0957
a o
loaf
3
log
a I veg
toga
log
log ( (- + 2) + log # 3 = log ( -
2
1
4
No solution
A
4
reg
52
YETI
log
x2
210g 1
log
x
1
log 1 3
x2
1
4 3
x 46
31
X
6
10
x
01
2
3
y
g
0
Check
5
57
log
log
I
HEFEI
at Lab 32
atb
1
53 3
0.124
9314,7
Ch z
Q4 719 level
I
Ch4
on 5
Q2 6
93
0.24
0,87
7 8
Q1 I 4 6 1 10 111
on6 Q1 I 4 11 1 15118
0h7 Q3 1 4 15 15 17 1 18
121
947m
QI
log x
log5y
T2
logy log 8
3
logx log15g
0
2
log Ey
2
Fy
10
Ey To
lo ox 59
100
59
0
20
log y
log
8
3
log y log 8
y 8 3
8x y
X
4
3
4 5
0
3
log Ees
It
to
100
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Selected Exercises from Textbook:
Further Practice P.7.34
Exercise 7D
Level 1: Q3-6, Q13-16, Q19-20, Q21-22, Q25-26
Level 2: Q30-31, Q34-37, Q39, Q41-42, Q47-48.
End of Section Exercises:
Q1.
Q2.
Find the values of the following logarithms without using a calculator.
(a)
log 2 2 3
(b)
log 3 3 -5
(c)
log12 121.6
(d)
log 2 4
(e)
log 2 16
(f)
log 3 81
(g)
æ1ö
log 3 ç ÷
è3ø
(h)
æ1ö
log 4 ç ÷
è 16 ø
(i)
log 2 8
(j)
æ 1 ö
log 5 ç
÷
è 125 ø
(k)
log 3 3 9
Find the values of the following logarithms correct to 3 significant figures.
(a)
Q3.
Q4.
log 2 96
(b)
æ1ö
log1.3 ç ÷
è2ø
Find the values of the following expressions without using a calculator.
(a)
æ1ö
log 3 18 + log 3 ç ÷
è2ø
(b)
log 2 160 - log 2 5
(c)
log 4 8 + 5 log 4 2
(d)
2 log 5 15 - log 5 9
(e)
log 7 3
log 7 9
(f)
log 5 8
log 5 4
(g)
log16 64
(h)
æ 1 ö
log 9 ç ÷
è 27 ø
(i)
æ1ö
2 log 6 ç ÷ - log 6 4
è3ø
(j)
log 4 9
log 4 3
(k)
4 log 3 2 - 2 log 3 6 - log 3 4
(l)
æ2ö
log12 ç ÷ + 2 log12 6 - 2 log12 2 (m) log 5 36 ´ log 6 125 (n)
è3ø
log 27 6
log 3 216
Simplify the following expressions, where x > 0 and x ≠ 1.
(a)
æ 1ö
3 log 3 x 2 + log 3 ç 6 ÷
èx ø
(c)
3 log 2 x - log 2 x
log 2 x 3
(f)
5
æ xö
log 5 x - log 5 x + 2 log 5 ç ÷
2
è5ø
(b)
(d)
æ x2
log 4 çç
è 2
ö
÷÷ - log 4 (8 x 2 )
ø
log 2 (3 x)
log 3 (32 x 3 ) - log 3 (2 x)
(e)
log 2 27 + 3 log 2 x
log 3 (4 x)
(g)
log16 x 8
log 4 x 5
53
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Q5.
Solve the following logarithmic equations.
(a)
log 2 ( x + 1) = 4
(b)
log 3 (2 x - 3) = -1
(c)
log 2 ( x - 1) - log 2 6 = -2
(d)
log 4 (5 x + 2) + log 4 3 = 2
(e)
log 9 (3 x + 1) = log 3 4
(f)
log 2 ( x - 1) = log 4 9
(g)
log 7 (4 x - 3) = log 7 ( x - 3) + 1 (h)
log 49 (5 x + 1) - log 7 6 = 0
Q6. Given that log 3 2 = a and log 3 5 = b , express the following in terms of a and/or b.
(a)
log 3 10
(b)
æ 2ö
log 3 ç ÷
è 25 ø
(c)
log 3 15
Q7. Given that log 2 3 = a and log 2 5 = b , express the following in terms of a and b.
(a)
log 2 0.6
(b)
æ 25 ö
log 2 ç ÷
è 9 ø
(c)
log 2 30
Q8. Given that log 3 2 = x and log 3 5 = y , express the following in terms of x and y.
(a)
log 3 60
(b)
æ9ö
log 3 ç ÷
è 10 ø
(c)
log 5 2
54
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Past Paper – Paper 1 [Occurrence: Low] [log is usually tested with F5-6 topics]
HKCEE 1980(3) Q7
HKCEE 1990 Q2(b)
Find - if log ' (- − 3) + log ' (- + 3) = 3
Simplify
HKCEE 1981(1) Q5
./06, % 7#./06B ) 7
,
./0(,B % )
where @, R > 0.
HKCEE 1991 Q7
Solve 4! = 10 − 4!#$ .
Let • and f be the roots of the equation
10- " + 20- + 1 = 0. Without solving the
HKCEE 1982(1) Q2
!&)
If ~4!#) = 4 , solve for - and 5.
4
= 16
equation, find the values of
(a) 4E × 4F .
(b) log • + log f.
HKCEE 1985(B) Q3
Solve 2"! − 3(2! ) − 4 = 0.
HKCEE 1992 Q2(a)
If log - = € and log 5 = •, express log(-5) in
HKCEE 1986(A) Q5(a)
terms of € and •.
$
Evaluate log " 8 + log " $>.
HKCEE 1993 Q5(a)
If 9! = √3, find -.
HKCEE 1986(A) Q5(b)
If 2 log - − log 5 = 0. Show that 5 = - " .
HKCEE 1994 Q7(b)
If log 2 = - and log 3 = 5, express log √12 in
HKCEE 1987(A) Q3(a)
Simplify
./0 , ! B % &./0 ,B %
./0 √,
.
HKCEE 1987(B) Q3
Solve the equation 3"! + 3! − 2 = 0.
HKCEE 1988(A) Q6
Given that log 2 = S and log 3 = T, express
terms of - and 5.
HKCEE 1995 Q7
Solve the following equations without using a
calculator:
(a) 3! =
$
√"3
(b) log - + 2 log 4 = log 48
the following in terms of S and T.
(a) log 18,
(b) log 15.
HKCEE 1997 Q2(b)
Simplify
./0 <#./0 %
./0 $>
.
55
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
HKDSE SP Q17
A researcher defined Scale A and Scale B to represent the magnitude of an explosion as shown in the
table:
Scale
Formula
A
E = log % h
B
F = log < h
It is given that E and F are magnitudes of an explosion on Scale A and Scale B respectively, while E
is the relative energy released by the explosion. If the magnitude of an explosion is 6.4 on Scale B,
find the magnitude of the explosion on Scale A.
HKDSE 2014 Q15
The graph in the figure shows the linear relation between log % - and log < 5. The slope and the
$
intercept on the horizontal axis of the graph are − ' and 3 respectively. Express the relation between
- and 5 in the form 5 = ‚- G , where ‚ and o are constants.
HKDSE 2017 Q15
Let @ and R be constants. Denote the graph of 5 = @ + log B - by ƒ. The --intercept of ƒ is 9 and ƒ
passes through the point (243,3). Express - in terms of 5.
56
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Past Paper – Paper 2 [Occurrence: Low]
Exponential Equations [Chapter 6 materials, log can be useful]
HKCEE SP Q2
HKCEE 1986 Q29
If 25! = 125, then - =
If (10! )) = (2H )(5H ), then which of the
A.
(
B.
"
"
following must be true?
.
A. -5 = x
.
(
B. -5 = 2x
C. 5.
D.
'
E.
"
"
'
C. -5 = x "
.
D. - ) = x
.
E. - ) = 2x
HKCEE 1978 Q5
HKCEE 1987 Q7
If 9"! = 27, then - =
If 3"G#$ = 3"G + 6, then o =
$
.
A. − .
$
.
'
B. − ".
C.
"
C.
$
D.
%
D.
$
E.
'
A.
'
B.
"
%
$
.
'
.
'
%
If 10") = 25, then 10&) =
A.
(
.
$
B. − (.
C.
$
"(
.
.
HKCEE 1993 Q34
If 9!#" = 36, then 3! =
A.
"
B.
%
'
'
.
.
C. 2.
$
D. − "(.
E.
"
.
E. 3.
.
HKCEE 1979 Q23
$
%
$
$"(
.
D. √6.
E. 9.
57
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
HKCEE 1995 Q38
HKCEE 1997 Q2
If 5, = 2B = 10I and @, R, p are non-zero,
If 2! ⋅ 8! = 64, then - =
I
I
then , + B =
A.
3
$*
.
B. 1.
C. 7.
E.
./0 "
'
B.
'
C.
>
"
%
(
.
.
.
D. 2.
D. log 7.
$
A.
E. 4.
$
+ ./0 (.
HKCEE 1999 Q4
If 4! = @, then 16! =
A. 4@.
B. @" .
C. @% .
D. 2, .
E. 4, .
Basic Concepts of Logarithms
HKCEE 1983 Q36
HKCEE 1986 Q33
If @ and R are positive numbers, which of the
If log - " + log 5 " = log x " , where -, 5 and x
following is/are true?
are positive numbers, which of the following
(1) log(@ + R) = log @ + log R
,
(2) log 7B8 = log @ − log R
(3)
./0 ,
./0 B
=
must be true?
(1) - " + 5 " = x "
,
(2) log - + log 5 = log x
B
(3) - " 5 " = x "
A. (1) only
B. (2) only
C. (3) only
D. (1) and (2) only
E. (1), (2) and (3)
A. (1) only
B. (2) only
C. (3) only
D. (1) and (2) only
E. (2) and (3) only
58
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
HKCEE 1988 Q35
If log @ > 0 and log R < 0, which of the following is/are true?
E
,
(1) log B > 0
(2) log R" > 0
x
10919
$
(3) log , > 0
x
A. (1) only
B. (2) only
C. (3) only
X
D. (1) and (2) only
E. (2) and (3) only
1
0
9
4
1095
H
logb
log'a
2logb
co
2
log l log a
log a
Logarithm and its application
HKCEE 1974 Q16
HKCEE SP Q12
10./0 B =
If log @ = 0.0490, then log =
$
A. (log R)" .
B. log(log R).
C. log R.
D. R.
E. 10 log R.
HKCEE 1977 Q15
If 3! = 8, then - =
A.
<
B.
./0 <
.
'
A.
$
*.*%2*
.
B. −0.9510.
C. −1.9510.
D. −0.0490.
E. −1.0490.
HKCEE SP Q38
log(0.1) =
A. −2
.
B. −1.
<
C. 0.
'
C. log '.
D. log 5.
E.
,
./0 <
./0 '
.
D. 1.
E. 2.
59
co
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
HKCEE 1978 Q2
HKCEE 1982 Q30
If log @ = 0.5678, then log √@ =
logÖ- ./0 ! Ü =
A. √0.5678.
A. (log -)" .
B. 0.5678 ÷ 2.
B. log(- " ).
C. 0.5678 − 2.
C. - log -.
D. 2 − 0.5678.
D. log(log -).
E. 2.5678.
E. 10&" .
HKCEE 1979 Q14
./0 (
What is ./0 ' equal to?
A.
HKCEE 1985 Q8
A.
(
'
B. log(5 − 3)
C. log 5 − log 3
(
D. log 7 8
'
E. log ' 5
HKCEE 1980 Q4
If C = 10, , then log C =
log carb cat
.
loglatb tlog
2 log(@ − R).
log(@" − R" ) =
B.
./0 ,
./0 B
D. log(@ + R) + log(@ − R).
O
E. (log @ + log R)(log @ − log R).
log % 2√2 =
B. 10 .
B.
'
C. C, .
C.
$
A
D. @A .
E. @.
2 2
HKCEE 1989 Q42
'
A. 10 .
<
%
%
.
.
.
!
D. 2) .
!
E. 2. .
is
10942
3 Thy
2 I
L
HKCEE 1981 Q8
HKCEE 1990 Q5
If log - + log 4 = log(- + 4), what is the
If 2 = 10J , 3 = 10K , express log > in terms of
value of -?
€ and •.
$
A. 0
A. −€ − •
B. 1
B.
C.
%
'
D. 4
E. - may be any positive number
b
C. 2 log @ − 2 log R.
A.
,
a
C.
$
JK
$
J#K
D. ۥ
E. € + •
60
2
2
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
HKCEE 1991 Q34
HKCEE 1996 Q38
If log - : log 5 = B: C, then - =
Let - > 5 > 0. If log(- + 5) = @ and log(- −
A.
L)
A
.
5) = R, then log ]- " − 5 " =
B. (B − C)5.
C. B − C + 5.
O
/
0
D. 5 .
E.
L ./0 )
A
logy
log x
log
.
x
HKCEE 1992 Q5
$
log
,#B
B.
,B
O
"
.
.
"
II
D. √@R.
E. √@ + √R.
HKCEE 1997 Q5
B. 10 + √@.
A. 2 − @.
C. 5@.
B. 100 − @.
,
C.
.
"
If log(€ + •) = log € + log •, then
A. € = • = 1.
K
B. € = K&$.
K
K#$
D. € =
K#$
E. € =
K&$
K
K
Suppose log 2 = @ and log 3 = R. Express
log 15 in terms of @ and R.
A. −@ + R + 1
C. @ + 2R
D. (@ + R)R
.
B
B. 2@ + ".
"
@ + R.
'
#
D. @" + R% .
E.
tb
.
.
B
#
,
B. −@ + 10R
A. 2@ + '.
@" R % .
$**
.
If log 2 = @ and log 9 = R, then log 12 =
'
t
HKCEE 1998 Q40
E.
HKCEE 1994 Q34
C.
af
a
E. 100 − log @.
HKCEE 1993 Q8
"
I
log
D. 2 − log @.
,
E. 1 + ".
C. € =
I log x y
I log lay
If log(- + @) = 2, then - =
A. 10√@.
D.
Egg's
log
C. √@ + R.
y't
If log R = 1 + " log @, then R =
A.
$*,
B
HKCEE 1999 Q39
$
If log 5 = 1 + log -, then
"
A. 5 = √10-.
B. 5 = 100 + - " .
C. 5 = (10 + -)" .
D. 5 = 10- " .
E. 5 = 100- " .
61
x
y
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
HKCEE 2000 Q38
HKCEE 2004 Q39
If log(- − @) = 3, then - =
If 5 = 10, and 7 = 10B , then log (* =
A. 3'#, .
'
B. @ .
C. 1000@.
D. 1000 + @.
E. 30 + @.
3
A. R − @ − 1.
B. R − @ + 1.
C.
D.
B
,
.
B
,#$
.
HKCEE 2001 Q37
If log - " = (log -)" , then - =
A. 1.
HKCEE 2005 Q39
B. 10.
If @ and R are positive integers, then
C. 100.
logÖ@B R, Ü =
D. 1 or 10.
A. @R log(@R).
E. 1 or 100.
B. @R(log @)(log R).
C. (@ + R) log(@ + R).
HKCEE 2002 Q40
D. R log @ + @ log R.
If log - " = log 3- + 1, then - =
A. 2.
B. 5.
C. 30.
HKCEE 2006 Q38
D. 0 or 30.
Let @ and R be positive numbers. If log
,
$*
=
2 log R, then @ =
HKCEE 2003 Q40
A. 10R" .
If 10,#B = p, then R =
B. 20R.
A. log p − @.
C. R" + 10.
B. @ − log p.
D. 2R + 10
C.
I
$*
− @.
D. p − 10, .
62
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
HKCEE 2007 Q39
Which of the following is the greatest?
A. 500'***
B. 2000"(**
C. 2500"***
D. 3000(**
HKCEE 2009 Q38
Which of the following is the best estimate of
HKCEE 2011 Q39
1234'"'( ?
Which of the following is the least?
A. 10%***
A. 1234$<$$
B.
B. 2345$3$$
C. 10(***
C. 3456$($$
D. 10$****
D. 7890$%$$
E. 10"****
HKDSE
HKDSE PP Q36
Let R > 1. If @ = log$" R, then , =
$
A. log B $".
logub
B. log B 12.
z
$
I
C. log$" .
B
D.
$
./01 $"
Ib
at
logub
a
$
.
É
HKDSE SP Q32
109612
The graph in the figure shows the linear relation between - and log ( 5. If 5 = @R ! , then @ =
A. 1.
logg ab
loggy
B. 2.
C. 5.
O
D. 25.
loggy
loggatloggb
loggy
loggatxlogsb
J
A
slope
B
B
4
slope
IB
Intercept
on
vertical
axis
T
slope
x
logga
log
9
a
9
2
52
63
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
HKDSE 2012 Q32
The graph in the figure shows the linear relation between - and log ' 5. If 5 = BC ! , then C =
A.
B.
$
<$
.
$
.
2
O
D. 81.
C. 9.
loggy
log 1mn
log y
log
x logan
slope
log
HKDSE 2013 Q32
m t
4
slopes
h
2
h
32
h
9
The figure above shows the graph of 5 = @R ! , where @ and R are constants. Which of the following
graphs may represent the relation between - and log 3 5?
A.
B.
64
2
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
C.
D.
HKDSE 2013 Q34
If - − log 5 = - " − log 5 " − 10 = 2, then 5 =
A. 100.
B. 2 or −4.
C.
D.
$
$**
or 10 000.
$
$****
or 100.
HKDSE 2014 Q33
Which of the following is the greatest?
A. 124"%$
B. 241"$%
C. 412$%"
D. 421$"%
n
99
y abt
IX
Iggy
y kx
65
I
logx
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
y
HKDSE 2015 Q32
y 9x
ki
The graph in the figure shows the linear relation between log ' - and log ' 5. Which of the following
É
must be true?
loggy log
A. - " 5 ' = 729
GO
I
x
gg
B. - ' 5 " = 729
C. - " + 5 ' = 729
'
"
D. - + 5 = 729
x
g g
g
x y
f
3
HKDSE 2016 Q32243
4
log y
flog
xtlogst
2
slope n
93
9
t
729
The graph in the figure shows the linear relation between - and log 2 5. If 5 = @R ! , then R =
A. −2.
B.
C.
$
<$
$
"
loggy
.
.
O
loggba
HKDSE 2017 Q34
Cab
sloggatthsffe
4 L
slope
D. 3.
log
171410
pyo
2
b 3
log 2 5 = - − 3
If [
, then 5 =
2(log 2 5)" = 4 − $
A. −1 or ".
$
B. 1 or '.
3
C. 2 or .
"
$
D. 3 or 2.
66
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
HKDSE 2018 Q33
In the figure, the straight line ˆ shows the relation between log % - and log % 5. It is given that ˆ
passes through the points (1,2) and (9,6). If 5 = o- , , then o =
A.
$
B.
'
"
"
.
.
C. 2.
D. 8.
HKDSE 2019 Q31
It is given that log 2 5 is a linear function of log ' -. The intercepts on the vertical axis and on the
horizontal axis of the graph of the linear function are 7 and 8 respectively. Which of the following
must be true?
A. - % 5 3 = 3(>
B. - 3 5 % = 3(>
C. - 3 5 < = 3(>
D. - < 5 3 = 3(>
HKDSE 2019 Q32
If
'
' ./0 !&"
+7=
"
" ./0 !#$
$
, then log =
!
A. −3 or 2
B. −2 or 3
$
$
$
$
"
'
C. − ' or "
D. − or
HKDSE 2020 Q32
If the roots of the equation (log M -)" − 10 log M - + 24 = log M - are • and f, then •f =
A. ‰ $* .
B. ‰ $$ .
C. log M 10.
D. log M 11.
67
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
HKDSE 2021 Q33
Let @, R and p be positive constants. On the same coordinate rectangular coordinate system, the
graph 5 = @ + log B - and the graph of 5 = log I - cuts the --axis at the point t and Š respectively.
Denote the origin by ‹. Find ‹Š: ‹t.
A. 1: R,
B. 1: p ,
C. R, : 1
D. p , : 1
HKDSE 2021 Q34
The graph in the figure shows the linear relation between log ( - and log ( 5. Which of the following
must be true?
A. -5 " = 625
B. - " 5 = 625
C.
D.
)%
!
)
!%
= 625
= 625
HKDSE 2022 Q31
Which of the following is the least?
A. (−345)3><
B. 453&3<>
$
C. 7%'(8
D. 7
"
8
<>3
<3>
(%'
HKDSE 2022 Q32
It is given that log , 5 is a linear function of -, where 0 < @ < 1. The intercepts on the vertical axis
and on the horizontal axis of the graph of the linear function are 6 and 3 respectively. If 5 = BC ! ,
which of the following is/are true?
I.
B<1
II.
C<1
III.
BC' = 1
A. I only
B. II only
C. I and III only
D. II and III only
68
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
7.5
Graphs of Logarithmic Functions and their Features
First case: the graph of 5 = log , - where @ > 1
The figure below shows the graph of 5 = log " - and 5 = log ' -.
Observation:
When 3 > 0, log 2 2 > log 3 2
When 3 < 0, log 2 2 < log 3 2
(1) The graph must pass through (K, L), because when - = 1, 5 = log , 1 = 0.
-
When 0 < - < 1, 5 = log , - is negative, i.e., below the --axis.
-
When - > 1, 5 = log , - is positive, i.e., above the --axis.
(2) The graph will NOT intersect the l-axis and will always lie on the RHS of the l-axis,
because log , 0 and log , (negative) are undefined.
Second case: the graph of 5 = log , - where 0 < @ < 1
The figure below shows the graph of 5 = log # - and 5 = log # -.
%
!
Observation:
When 3 > 0, log ! 2 > log ! 2
"
#
"
#
When 3 < 0, log ! 2 < log ! 2
(3) The graph must pass through (K, L), because when - = 1, 5 = log , 1 = 0.
-
When 0 < - < 1, 5 = log , - is positive, i.e., above the --axis.
-
When - > 1, 5 = log , - is negative, i.e., below the --axis.
(4) The graph will NOT intersect the l-axis and will always lie on the RHS of the l-axis,
because log , 0 and log , (negative) are undefined.
69
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Range of a
@ > 1
0 < @ < 1
Graph of y = log a x
Common features
1.
The graph cuts the x-axis at (1, 0).
2.
The graph never cuts the y-axis. It lies on the right-hand side of
the y-axis.
3.
The graph has neither a maximum point, a minimum point nor
an axis of symmetry.
Differences
1. (a) 5 < 0 for 0 < x < 1
4. (a) 5 > 0 for 0 < x < 1
(b) 5 > 0 for x > 1
(b) 5 < 0 for x > 1
2. The value of y increases as x
5. The value of y decreases as x
increases.
increases.
3. As x increases, the rate of
6. As x increases, the rate of
increase of y becomes smaller.
decrease of y becomes smaller.
Assume @ > 1. In fact, there is a relation between log , - and log # -.
-
Using the base-change formula, we have
log , 1
log , 7 8
@
log , =
log , @&$
log $ - =
,
= − log , In other words, their graphs are symmetric with respect to the --axis. For example,
70
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
A summary on the looks of the graphs:
When ; > <
The larger the value of ,, the flatter the graph
of 3 = log " 2
When = < ; < <
The smaller the value of ,, the flatter the graph
of 3 = log " 2
Example 7.20
The figure shows the graph of 5 = log ' -.
(a) Using the graph, find the values of
(i)
log ' 1.5
(ii)
log # 4.2
!
(b) Solve the following equations graphically.
(i)
log ' - = 1.6
(ii)
log # - = 0.6
!
Solution:
The same method of looking at the graph of a function has been taught in chapter 3, 4 and 6 already.
From the graph, we have
(a) (i)
(ii)
(b) (i)
(iii)
log ' 1.5 ≈ 0.4
log # 4.2 = − log ' (4.2). When - = 4.2, 5 = log ' 4.2 ≈ 1.3. Hence log # 4.2 ≈ −1.3
!
!
When 5 = 1.6, - ≈ 5.8.
log # - = − log ' - ⟹ − log ' - = 0.6 ⟹ log ' - = −0.6. Hence, - ≈ 0.5.
!
71
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Exercise 7.20a
The figure shows the graph of 5 = log # -.
%
(a) Using the graph, find the values of
(i)
log # 2.3
%
(ii)
log " 3.5
(b) Solve the following equations graphically.
(i)
log # - = −2.2 (ii)
%
log " - = −1.3
Exercise 7.20b
Using the figure shown in Example 7.20,
(a)
find the values of the following,
(i)
log 3 1.9
(ii)
log 1 3.4
3
(b)
solve the following equations graphically.
(i)
log 3 x = -0.2 (ii)
log 1 x = 1.1
3
72
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
The graphs of exponential functions and logarithmic functions are reflection of each other with
respect to the 45-degree line, i.e.,
The reason is that they are “inverse” to each other. The idea of inverse function is out of the DSE
syllabus. Interested students may think of what other pair of functions are inverse of each other.
Selected Exercises from Textbook:
Exercise 7E:
Level 1: Q1-Q3
Level 2: Q6-7
End of Section Exercises:
The figure shows the graph of 5 = 2! and 5 = -. The graph of
5 = 2! is reflected about the line 5 = - to obtain
the graph of 5 = •(-).
(a)
x
It is given that the graph of y = 2 and y = f (x)
cuts the y-axis at A and B respectively.
(b)
(i)
Write down the coordinates of A and B.
(ii)
Mark B on the given graph.
(i)
Write down the algebraic representation of the function f(x).
(ii)
Sketch the graph of y = f (x) on the given graph.
73
St. Catharine’s School for Girls | HKDSE – Logarithmic Functions | Notes
Past Paper – Paper 2
HKDSE 2014 Q32
The figure shows the graph of 5 = R ! and the graph of 5 = p ! on the same rectangular coordinate
system, where R and p are positive constants. If a horizontal line ˆ cuts the 5-axis, the graph of 5 =
R ! and the graph of 5 = p ! at ‚, Ž and • respectively, which of the following are true?
(1) R < p
(2) Rp > 1
(3)
NO
NP
= log B p
A. (1) and (2) only
B. (1) and (3) only
C. (2) and (3) only
D. (1), (2) and (3)
HKDSE 2018 Q32
The figure shows the graph of 5 = log , - and the graph of5 = log B - on the same rectangular
coordinate system, where @ and R are positive constants. If a vertical line cuts the graph of
5 = log , -, the graph of 5 = log B - and the --axis at ‚, Ž and • respectively, which of the following
is/are true?
I.
@>1
II.
@>R
III.
NO
X
NP
= log ,
larger
lower
the base
the
graph
o
B. II only
I
C. I and IIi only
e
D. II and III only
109
1
logac
196
1
5
Cio
gie
I
I
2
B
,
A. I only
Afc
Acb
togax last
fab
Ky
logax
Local
74
logoff
logabi.gg
I
log a a
base
O
110
X
E
O
f
log x
s
baseof
fix
t
logg X
Exp
log
for