NELSON QMATHS 12
MATHEMATICAL METHODS
WORKED SOLUTIONS
Chapter 1 Logarithmic functions
Exercise 1.01 Logarithm laws
Question 1
a
34 = 81
log3 (81) = 4
1
b
16 2 = 4
log16 (4) =
c
1
2
10−4 = 0.0001
log10 (0.0001) = −4
2
d
27 3 = 9
log27 (9) =
2
3
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Question 2
a
log4 (64) = 3
43 = 64
b
1
3
log27 (3) =
1
27 3 = 3
c
log (0.01) = −2
10−2 = 0.01
d
log8 (4) =
2
3
2
83 = 4
Question 3
a
log5 (1) = log5 (5 0 ) = 0 log5 (5) = 0 ×1 = 0
b
1
log2 (0.5) = log2   = log2 ( 2−1 ) = −1 × 1 = −1
2
c
2 log3 (1) = 2 log3 ( 30 ) = 0 × 2 × 1 = 0
d
loga (1), a > 0 = loga (a0) = 0 × 1 = 0
e
log5 (5) = log5 ( 51 ) = 1 × log5 (5) = 1 × 1 = 1
f
[log4 (4)]2 = [log4 (41)]2 = [1 × log4 (4)]2 = [1 × 1]2 = 12 = 1
g
log2 (8) = log2 ( 23 ) = 3 log2 (2) = 3 × 1 = 3
h
log8 (2) = log8 ( 8 3 ) =
1
1
1
1
× log8 (8) = × 1 =
3
3
3
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Question 4
a
log4 (2) + log4 (64)
= log4 (2 × 64)
= log4 (128)
 7
= log4  4 2 
 
=
7
log4 (4)
2
=
7
×1
2
= 3.5
b
log9 (81) – log9 (3)
 12 
= log9 ( 9 ) − log9  9 
 
2
= 2 log9 (9) −
=2−
=1
c
1
log9 (9)
2
1
2
1
or 1.5
2
log2 (10) + 2 log2 (4) − log2 (80)
= log2 (10) + log2 ( 42 ) − log2 (80)
 10 ×16 
= log2 

 80 
= log2 (2)
=1
d
log7 (log5 (5)) = log7 (1) = log7 (7 0 ) = 0 × log7 (7) = 0
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e
1
 49 
 5
5
log7   + 2 log7   – log7   – log7  
4
 11 
 16 
 11 
  1 2 
5
 49 
 5
= log7   + log7     – {log7   + log7   }


 11 
 16 
 11 
 4  
 5 1  49 5  
= log7  × ÷  ×  
 11 16  11 16  
 1 
= log7  
 49 
= log7 ( 7 −2 )
= −2 × log7 ( 7)
= −2
f

1 
log2  5

 1024 
=
1
 1 
log2 

5
 1024 
=
1
 1 
log2  10 
5
2 
=
1
log2 ( 2−10 )
5
= −10 ×
1
log2 ( 2 )
5
= −2
Question 5
a
log6 (7y2) = log6 (7) + log6 (y2)
= log6 (7) + 2 log6 (y)
b
log3 (5xz3) = log3 (5) + log3 (x) + log3 (z3)
= log3 (5) + log3 (x) + 3 log3 (z)
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c
 1 
log4  5  = log4 ( w−5 )
w 
= −5 log4 ( w )
d
 2x 
log8  2 3  = log8 ( 2 ) + log8 ( x ) − log8 ( y 2 ) − log8 ( z 3 )
y z 
 13 
= log8  8  + log8 ( x ) − 2 log8 ( y ) − 3 log8 ( z )
 
=
1
log8 ( 8 ) + log8 ( x ) − 2 log8 ( y ) − 3 log8 ( z )
3
=
1
+ log8 ( x ) − 2 log8 ( y ) − 3 log8 ( z )
3
e
log7 (x2 – 4) = log7 [(x + 2)(x – 2)] = log7 (x + 2) + log7 (x – 2)
f
 y 3 ( y + 3) 
2
log5 
= log5 ( y 3 ) + log5 ( y + 3) − log5 ( y − 2 )
2 
 ( y − 2) 
= 3 log5 ( y ) + log5 ( y + 3) − 2 log5 ( y − 2 )
g
1
 2
2 2 
 x2 + y 2 


x
y
+

log2 
 = log2  
 



xy 
  xy  



=
 x2 + y 2 
1
log2 

2
 xy 
=
1
log 2 ( x 2 + y 2 ) − log 2 ( xy ) 

2 
=
1
log 2 ( x 2 + y 2 ) − log 2 ( x ) − log 2 ( y ) 

2 
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5
h
1
− 
 5 

2
2
log3 
=
log
+
5
z
1
3
(
)



2


 z +1 
1
= log 3 ( 5 ) + log 3 ( z 2 + 1) 2
−
= log3 (5) −
1
log3 ( z 2 + 1)
2
Question 6
a
5 log2 (x) + 3 log2 (y)
= log2 ( x5 ) + log2 ( y 3 )
= log2 ( x 5 y 3 )
b
2 log6 (p) – log6 (11) – log6 (m)
= log6 ( p 2 ) – log6 (11) – log6 (m)
 p2 
= log6 

 11m 
c
2 [log (y) + log (y + 2) – log (y2 – 1)]
  y ( y + 2) 

= 2 log  2
  ( y − 1)  


 y ( y + 2) 
= log  2

 y −1 
d
2
4 [3 log4 (x) – log4 (x + 1) – log4 (x2 – 3)]
= 4 [log4 (x)3 – log4 (x + 1) – log4 (x2 – 3)]4



x3

= 4 log 4 
 ( x + 1) ( x 2 − 3)  






x3

= log 4 
 ( x + 1) ( x 2 − 3) 


4
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e
1
log5 (17) – 3log5 (x) – 4log5 (y)
2
1
= log 5 (17 ) 2 − log 5 ( x ) − log 5 ( y )
3
4
 17 
= log 5  3 4 
x y 
f
1
log7 (x – 3) + 3log7 (x) + 2log7 (3 + x)
3
1
= log 7 ( x – 3) 3 + log 7 ( x ) + log 7 ( 3 + x )
(
3
= log 7 x 3 ( 3 + x ) 3 x − 3
g
2
)
1
[log3 (x) + 3log3 (y) – 5log3 (x – 2)]
2
=
1
5
log 3 ( x ) + log 3 ( y 3 ) – log 3 ( x – 2 ) 

2
=
1
xy 3 
log 3
5
2 
( x − 2 ) 

xy 
= log 3
5
( x − 2 ) 

3

= log 3

h
2
1
2


5
( x − 2 ) 
xy 3
1
1
log2 (x + 1) + log2 (x – 1) + log2 (5)
2
2
= log 2
( x + 1) + log ( x − 1) + log ( 5)
2
(
= log 2 5 ( x + 1)( x − 1)
2
)
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Question 7
(
log a x 2 −3
) = x2 – 3
a
a
b
25log5 8 = 5 (
2 log5 8 )
( log 8 )
2
5
=5
= 5log5 64
= 64
1
c
10 2
log 7
= 10log 7
=
d
7
3
43log4 6 = 4log4 6
= 63
= 216
e
3−2log3 7 = 3log3 7
log3
=3
=
−2
1
49
1
49
Question 8
a
log 2 ( 36 )
=
log 36
log 2
=
1.55630...
0.30102...
= 5.169925...
≈ 5.1699
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b
log100 ( 500 )
=
log 500
log100
=
2.69897
2
= 1.349485...
≈ 1.3495
c
log 0.2 ( 21)
=
log 21
log 0.2
=
1.32221...
−0.69897...
= −1.89166...
≈ −1.8917
d
log 24 ( 3.18 )
=
log 24
log ( 3.18 )
=
0.502427...
1.380211...
= 0.364021...
≈ 0.3640
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Question 9
a
log 3 81
=
log 81
log 3
=
log 34
log 3
=
4 log 3
log 3
=4
b
log 9 27
=
log 27
log 9
=
log 33
log 32
=
3log 3
2 log 3
=
3
2
Question 10
a
log a 30
= log a ( 2 × 3 × 5 )
=
= log a 2 + log a 3 + log a 5
= 0.67 + 1.07 + 1.56
= 3.30
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b
loga (40)
= log a ( 5 × 8 )
= log a ( 5 × 23 )
= log a 5 + 3log a 2
= 1.56 + 3 × 0.67
= 3.57
c
 3
log a  
 10 
 3 
= log a 

 2×5 
= log a 3 − log a 2 − log a 5
= 1.07 − 0.67 − 1.56
= −1.16
d
 1 
log a 

 15 
1
= log a (15 ) 2
−
1
= − log a (15 )
2
1
=
− log a (3 × 5)
2
1
=
− [ log a 3 + log a 5]
2
1
=
− [1.07 + 1.56]
2
−1.32
e
 3
log a  − 
 5
= log a ( −3) − log a 5
log a (−3) is undefined ∴ no solution
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Question 11
a
n
=
If n log
=
x
3 x then 3
3x= 3 × x
= 3 × 3n
= 3n +1
b
log 3 ( 3x 4 )
= log 3 3 + log 3 x 4
= 1 + 4 log 3 x
= 1 + 4n
c
log x ( 243)
= log x ( 35 )
= 5log x 3
 log 3 3 
= 5

 log 3 x 
= 5×
=
d
1
n
5
n
 x2 
log 3  
 27 
= log 3 ( x 2 ) − log 3 27
= 2 log 3 x − log 3 27
= 2n − log 3 ( 33 )
= 2n − 3log 3 3
= 2n − 3
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Question 12
a
=
log y 2 log ( 4 ) − 5log ( x )
= log ( 42 ) − log ( x 5 )
 16 
= log  5 
x 
y=
16
x5
b
=
log 3 ( y ) 4 log 3 ( x ) − 2
= log 3 ( x 4 ) − 2 log 3 3
= log 3 ( x 4 ) − log 3 ( 32 )
 x4 
= log 3  
 9 
y=
x4
9
c
log 4 ( 2 xy ) = 2.5
42.5 = 2 xy
5
( 22 ) 2 = 2 xy
32 = 2 xy
y=
16
x
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d
 2x 
log 3   + 3 =
log 3 (2)
 y 
log 3 ( 2 x ) − log 3 ( y ) + 3 =
log 3 ( 2 )
log 3 ( y ) = log 3 ( 2 x ) + 3log 3 ( 3) − log 3 ( 2 )
= log 3 (2 x) + log 3 ( 33 ) − log 3 ( 2 )
 54 x 
= log 3 

 2 
y = 27 x
Question 13
log m (15) + log m ( x) =
0
log m (15 x ) = log m (1)
15 x = 1
x=
1
15
Question 14
log b (a )
=
log10 ( a )
log10 ( b )
=
1
log10 ( b )
log10 ( a )
=
1
log a ( b )
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Question 15
log a ( x n )
= log a ( x × x × x × .......) , n times
= log a ( x ) + log a ( x ) + log a ( x ) + ...
= n log a ( x )
Question 16
a
log n ( xy 2 )
= log n ( x) + log n ( y 2 )
= log n ( x) + 2 log n ( y )
= 3 + 2×5
= 13
b
log n ( nx 3 y )
=log n ( n ) + log n ( x 3 ) + log n ( y )
=log n ( n ) + 3log n ( x ) + log n ( y )
= 1+ 3× 3 + 5
= 15
c
 x 
log n 
 y 


 1
= log n ( x ) − log n  y 2 
 
1
= log n ( x ) − log n ( y )
2
1
=3− ×5
2
= 0.5
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d
log n
( xy )
3
3
1
3 3
= log n ( xy )
=
1
log n ( x ) + log n ( y 3 ) 

3
=
1
log n ( x ) + 3log n ( y ) 
3
1
[ 3 + 3 × 5]
3
=
=6
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Exercise 1.02 Indicial equations
Question 1
a
6x = 1
6 x = 60
x=0
b
9k = 27
(3 ) = 3
2 k
3
2k = 3
k=
c
3
2
4x = 16 2
( 2 )= 2 × 2
2 x
4
1
2
9
22 x = 2 2
2x =
x=
d
9
2
9
4
8–r =
1
2
(2 ) = 2
3 −r
−1
−3r =
−1
r=
1
3
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e
42x = 64
4 2 x = 43
2x = 3
f
x=
3
2
4y =
2
(2 ) = 2
2 y
1
2
2y =
1
2
y=
1
4
x
g
3 2 = 27
x
3 2 = 33
x
=3
2
x=6
h
1
= 3
27 g
( 33g ) = 3
−1
1
2
1
−3 g =
2
g= −
1
6
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Question 2
a
5x = 57x – 2
x 7x − 2
=
−6 x =
−2
x=
b
1
3
71 – u = 7–3
1 − u =−3
−u =−4
u=4
c
52k – 5 = 1
52 k − 5 = 50
2k − 5 =
0
2k = 5
k=
d
5
2
811 – 2m = 27
3 ( − m ) = 33
41 2
4 − 8m =
3
−8m =
−1
m=
1
8
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e
4y
2
= 46 – y
y2 = 6 – y
y2 + y – 6 = 0
(y + 3)(y – 2) = 0
y = –3, 2
f
64x + 1 = 16
1
43( 1) = 4 2
x+
1
3x + 3 =
2
g
3x = −
5
2
x= −
5
6
4z = 16z + 5
4 z = 42( z +5)
=
z 2 z + 10
z = −10
h
95 – 9x =
1
27 x− 2
(3 )
1
= 3( x − 2)
3
2 5–9 x
32(5 9 ) = 3 3(
− x
− x − 2)
−3 x + 6
10 − 18 x =
−15 x =
−4
x=
4
15
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20
Question 3
a
4x – 4x – 1 = 3
4 x − 4 x × 4−1 =
3
Let a = 4 x
a−
a
=
3
4
4a − a =
12
3a = 12
a=4
4x = 4
x =1
b
2m + 4 – 2m = 120
24 × 2m − 2m =
120
Let a = 2m
16a − a =
120
15a = 120
a =8
2m = 8
m=3
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c
3n – 3n – 1 = 54
3n − 3n × 3−1 =
54
Let a = 3n
a−
a
=
54
3
3a − a =
162
2a = 162
a = 81
3n = 34
n=4
d
2x – 2–x =
2x −
3
2
1 3
=
2x 2
Let a = 2 x
a−
1 3
=
a 2
2a 2 − 2 =
3a
2a 2 − 3a − 2 =
0
0
( 2a + 1)( a − 2 ) =
1
or a =
2
a=
−
2
2x =
21
−2−1 or 2 x =
no solution or
x =1
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22
e
2y + 6 – 9 × 2y = 110
2 y × 26 − 9 × 2 y =
110
Let a = 2 y
64a − 9a =
110
55a =110
a=2
2y = 2
y =1
f
27 k − 27 k −1
= 26
3k
33k − 3 ( k − )
= 26
3k
3
1
33k − 33k × 3−3
= 26
3k
Let a = 3k
a3 −
a
a3
27 = 26
1 

a 32 1 − 
 27  = 26
a1
a2 ×
26
26
=
27
a 2 = 27
(3 ) = 3
k 2
3
2k = 3
k=
3
2
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23
Question 4
a
7x = 12
log ( 7 x ) = log (12 )
x log ( 7 ) = log (12 )
x=
log ( 7 )
log (12 )
x = 1.27698...
x ≈ 1.277
b
123x = 38
log (123 x ) = log ( 38 )
3 x log (12 ) = log ( 38 )
3x =
log ( 38 )
log (12 )
3 x = 1.46387...
x = 0.48795...
x ≈ 0.488
c
34x + 1 = 17
log ( 34 x +1 ) = log (17 )
log (17 )
( 4 x + 1) log ( 3) =
log (17 )
4x +1 =
log ( 3)
4 x + 1 =2.5789...
4 x = 1.5789...
x = 0.39472...
x ≈ 0.395
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d
9x + 2 = 4
log ( 9 x + 2 ) = log ( 4 )
log ( 4 )
( x + 2 ) log ( 9 ) =
log ( 4 )
x+2=
log ( 9 )
x+2=
0.63092...
x = −1.36907...
x ≈ −1.369
e
6x – 1 = 4x
log ( 6 x −1 ) = log ( 4 x )
x log ( 4 )
( x − 1) log ( 6 ) =
x log ( 6 ) − log ( 6 ) =
x log ( 4 )
x log ( 6 ) − x log ( 4 ) =
log ( 6 )
x log ( 6 ) − log ( 4 )  =
log ( 6 )
x=
log ( 6 )
log ( 6 ) − log ( 4 )
x = 4.41902...
x ≈ 4.419
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f
53x + 2 – 4x = 0
53 x + 2 = 4 x
log ( 53 x + 2 ) = log ( 4 x )
x log ( 4 )
( 3x + 2 ) log ( 5) =
3 x log ( 5 ) + 2 log ( 5 ) =
x log ( 4 )
3 x log ( 5 ) − x log ( 4 ) =
−2 log ( 5 )
x 3log ( 5 ) − log ( 4 )  =
−2 log ( 5 )
x=
−2 log ( 5 )
3log ( 5 ) − log ( 4 )
x = −0.93517...
x ≈ −0.935
Question 5
a
5x – 1 = 2x + 1
log ( 5 x −1 ) = log ( 2 x +1 )
( x − 1) log ( 5) =
( x + 1) log ( 2 )
x log ( 5 ) − log ( 5 ) = x log ( 2 ) + log ( 2 )
x ( log ( 5 ) − log ( 2 ) ) =log ( 2 ) + log ( 5 )
x=
log ( 2 ) + log ( 5 )
log ( 5 ) − log ( 2 )
x = 2.5129...
x ≈ 2.513
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b
92x + 3 = 75x – 1
log ( 92 x +3 ) = log ( 75 x −1 )
( 2 x + 3) log ( 9 ) =
( 5 x − 1) log ( 7 )
2 x log ( 9 ) + 3log ( 9 ) = 5 x log ( 7 ) − log ( 7 )
x ( 2 log ( 9 ) − 5log ( 7 ) ) =
− log ( 7 ) − 3log ( 9 )
x=
− log ( 7 ) − 3log ( 9 )
2 log ( 9 ) − 5log ( 7 )
x = 1.60026...
x ≈ 1.600
c
122x + 3 = 83x – 1
log (122 x +3 ) = log ( 83 x −1 )
( 2 x + 3) log (12 ) =
( 3x − 1) log (8)
2 x log (12 ) + 3log (12 ) = 3 x log ( 8 ) − log ( 8 )
x ( 2 log (12 ) − 3log ( 8 ) ) =
− log ( 8 ) − 3log (12 )
x=
− log ( 8 ) − 3log (12 )
2 log (12 ) − 3log ( 8 )
x = 7.51602...
x ≈ 7.516
© Cengage Learning Australia Pty Ltd 2018
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d
62x + 5 = 113x – 2
log ( 62 x +5 ) = log (113 x − 2 )
( 3x − 2 ) log (11)
( 2 x + 5) log ( 6 ) =
2 x log ( 6 ) + 5log ( 6 )= 3 x log (11) − 2 log (11)
−2 log (11) − 5log ( 6 )
x ( 2 log ( 6 ) − 3log (11) ) =
x=
−2 log (11) − 5log ( 6 )
2 log ( 6 ) − 3log (11)
x = 3.80995...
x ≈ 3.810
e
7x – 1 – 52x + 1 = 0
7 x −1 = 52 x +1
log ( 7 x −1 ) = log ( 52 x +1 )
( x − 1) log ( 7 ) =
( 2 x + 1) log ( 5)
x log ( 7 ) − log ( 7 =
) 2 x log ( 5) + log ( 5)
log ( 5 ) + log ( 7 )
x ( log ( 7 ) − 2 log ( 5 ) ) =
x=
log ( 5 ) + log ( 7 )
log ( 7 ) − 2 log ( 5 )
x = −2.79296...
x ≈ −2.793
© Cengage Learning Australia Pty Ltd 2018
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f
212x + 1 – 153x – 1 = 0
log ( 212 x +1 ) = log (153 x −1 )
( 3x − 1) log (15)
( 2 x + 1) log ( 21) =
2 x log ( 21) + log ( 21)= 3 x log (15 ) − log (15 )
− log (15 ) − log ( 21)
x ( 2 log ( 21) − 3log (15 ) ) =
x=
− log (15 ) − log ( 21)
2 log ( 21) − 3log (15 )
x = 2.8266...
x ≈ 2.827
Question 6
a
3x = 17
x ≈ 2.579
© Cengage Learning Australia Pty Ltd 2018
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b
2x + 1 = 11
x ≈ 2.459
c
94x + 1 = 85
x ≈ 0.255
d
4 × 5x + 3 = 14
x ≈ −2.222
© Cengage Learning Australia Pty Ltd 2018
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e
7 × 4x – 1 – 5 = 0
7 × 4 x−1 =
5
x ≈ 0.757
f
7 × 62x + 3 – 9 = 0
7 × 62 x+3 =
9
x ≈ −1.430
© Cengage Learning Australia Pty Ltd 2018
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Question 7
a
64 × 16–3x = 163x – 4
43 × 4 2 ( 3 ) =
4 2( 3 4 )
− x
x−
43 × 4−6 x =46 x × 4−8
43 4 6 x
=
46 x 48
411 = 412 x
12 x = 11
x=
b
11
12
243m + 2 × 92m – 3 = 9
3 ( m+ ) × 3 ( m− ) =
32
5
2
2 2
3
35 m × 310 × 34 m × 3−6 =
32
39 m × 34 =
32
39 m = 3−2
9m = −2
m= −
c
2
9
813n + 2
= 34 – n
243− n
34(3n + 2)
= 34− n
−5 n
3
317 n × 38 = 34 × 3− n
318 n = 3−4
18n = −4
n= −
2
9
© Cengage Learning Australia Pty Ltd 2018
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d
162k – 3 × 4–2k + 1 = 32
4 2( 2
k − 3)
× 4−2 k +1 =
2 × 42
1
44 k × 4−6 × 4−2 k × 4 = 4 2 × 42
5
42
42 k = −5
4
15
42 k = 4 2
e
2k =
15
2
=
k
15
3
= 3
4
4
63b × 6–b = 6–2b + 1
63b × 6− b = 6−2b × 6
2b
6=
6−2b × 6
62b
=6
6−2b
64b = 61
4b = 1
b=
f
1
4
3–2x + 3 × 3–2x – 5 = 3–x – 1
3−2 x × 33 × 3−2 x × 3−5 = 3− x × 3−1
3−4 x × 3−2 = 3− x × 3−1
3−3 x = 31
1
−3 x =
x= −
1
3
© Cengage Learning Australia Pty Ltd 2018
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Question 8
22m – 2 – 2m – 2 = 3
22 × 2m × 2−2 − 2m × 2−2 =
3
2m ( 20 − 2−2 ) =
3
2m ×
3
3
=
4
2m = 4
2m = 22
m=2
Question 9
x – 10 x + 21 = 0
2
1
 12 
2
−
×
+ 21 =
x
x
10
0
 
 
1
Let a = x 2
a 2 − 10a + 21 =
0
0
( a − 7 )( a − 3) =
=
a 7=
or a 3
1
2
1
2
=
x
7=
or x
3
=
=
x 49
or x 9
© Cengage Learning Australia Pty Ltd 2018
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Question 10
52a – b =
1
⇔ 52 a – b = 5−4
625
102b – 6a = 0.01 ⇔ 102b – 6 a =10−2
∴ 2a − b =−4 and 2b − 6a =−2
4a − 2b =
−8
[1]
−6a + 2b =
−2 [ 2]
[1] + [ 2]
−2a =
−10
a=5
20 − 2b =
−8
−2b =
−28
b = 14
© Cengage Learning Australia Pty Ltd 2018
35
Exercise 1.03 Logarithmic graphs
Question 1
a
b=5
=
y log 4 ( x) + 5
b
b = −3
=
y log 1 ( x) − 3
4
c
b=2
=
y log 7 ( x) + 2
d
b = −7
=
y log 0.6 ( x) − 7
Question 2
a
c=5
=
y log 5 ( x + 5)
b
c = −3
=
y log 0.4 ( x − 3)
c
c=2
=
y log 1 ( x + 2)
8
d
c = −7
=
y log 3 ( x − 7)
© Cengage Learning Australia Pty Ltd 2018
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Question 3
a
y = log2 (x) + 3
a = 2, so a > 1
b = 3, log 2 ( x) is translated 3 units up.
Zero = a − b
= 2−3
=
1
8
= 0.125
The graph passes through (0.125, 0).
The graph passes through (1, b).
(1, b) = (1, 3)
© Cengage Learning Australia Pty Ltd 2018
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b
y = log0.5 (x) – 2
a = 0.5, so 0 < a < 1
b = −2, log 0.5 ( x) is translated 2 units down.
Zero = a − b
= 0.5 ( )
− −2
= 0.25
The graph passes through (0.25, 0).
The graph passes through (1, b).
(1, b) = (1, −2)
© Cengage Learning Australia Pty Ltd 2018
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c
y = log5 (x) – 1
a = 5, so a > 1
b = −1, log 5 ( x) is translated 1 units down.
Zero = a − b
=5 ( )
− −1
=5
The graph passes through (5, 0).
The graph passes through (1, b).
(1, b) = (1, −1)
© Cengage Learning Australia Pty Ltd 2018
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d
y = log0.8 (x) + 3
a = 0.8, so 0 < a < 1
b = 3, log 0.8 ( x) is translated 3 units up.
Zero = a − b
= 0.8−3
= 1.953125...
≈ 1.95
The graph passes through (1.95, 0).
The graph passes through (1, b).
(1, b) = (1, 3)
© Cengage Learning Australia Pty Ltd 2018
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e
y = log3 (x) + 2
a = 3, so a > 1
b = 2, log 3 ( x) is translated 2 units up.
Zero = a − b
= 3−2
=
1
9
≈ 0.111
The graph passes through (0.111, 0).
The graph passes through (1, b).
(1, b) = (1, 2)
© Cengage Learning Australia Pty Ltd 2018
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f
y = log0.25 (x) – 1
a = 0.25, so 0 < a < 1
b = −1, log 0.25 ( x) is translated 1 units down.
Zero = a − b
= 0.25 ( )
− −1
= 0.25
The graph passes through (0.25, 0).
The graph passes through (1, b).
(1, b) = (1, −1)
© Cengage Learning Australia Pty Ltd 2018
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Question 4
a
y = log3 (x + 2)
a = 3, so a > 1
c = 2, log 3 ( x) is translated 2 units to the left.
Zero = 1 − c
= 1− 2
= −1
The graph passes through (−1, 0).
31 = 3 , so
log 3 ( 3) = 1
x+2=
3
x =1
The graph passes through (1, 1).
Vertical asymptote is at x = −2.
© Cengage Learning Australia Pty Ltd 2018
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b
y = log0.5 (x – 2)
a = 0.5, so 0 < a < 1
c = −2, log 0.5 ( x) is translated 2 units to the right.
Zero = 1 − c
= 1 − ( −2 )
=3
The graph passes through (3, 0).
( 0.5)
−1
= 2 , so
log 0.5 ( 2 ) = −1
x−2=
2
x=4
The graph passes through (4, −1).
Vertical asymptote is at x = 2.
© Cengage Learning Australia Pty Ltd 2018
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c
y = log0.25 (x + 3)
a = 0.25, so 0 < a < 1
c = 3, log 0.25 ( x) is translated 3 units to the left.
Zero = 1 − c
= 1− 3
= −2
The graph passes through (−2, 0).
( 0.25)
−1
= 4 , so
log 0.25 ( 4 ) = −1
x+3=
4
x =1
The graph passes through (1, −1).
Vertical asymptote is at x = −3.
© Cengage Learning Australia Pty Ltd 2018
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d
y = log5 (x + 1)
a = 5, so a > 1
c = 1, log 5 ( x) is translated 1 unit to the left.
Zero = 1 − c
= 1−1
=0
The graph passes through (0, 0).
51 = 5 , so
log 5 ( 5 ) = 1
x +1 =
5
x=4
The graph passes through (4, 1).
Vertical asymptote is at x = −1.
© Cengage Learning Australia Pty Ltd 2018
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e
y = log4 (x – 4)
a = 4, so a > 1
c = −4, log 4 ( x) is translated 4 units to the right.
Zero = 1 − c
= 1 − ( −4 )
=5
The graph passes through (5, 0).
41 = 4 , so
log 4 ( 4 ) = 1
x−4=
4
x =8
The graph passes through (8, 1).
Vertical asymptote is at x = 4.
© Cengage Learning Australia Pty Ltd 2018
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f
y = log0.6 (x + 2)
a = 0.6, so 0 < a < 1
c = 2, log 0.6 ( x) is translated 2 units to the left.
Zero = 1 − c
= 1− 2
= −1
The graph passes through (−1, 0).
Choose y = 3 to find another point.
( 0.6 )
−3
= 4.6296...
, so
≈ 4.63
log 0.6 ( 4.63) = −3
x+2=
4.63
x = 2.63
The graph passes through (2.63, −3).
Vertical asymptote is at x = −2.
© Cengage Learning Australia Pty Ltd 2018
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Question 5
a
y = log2 (x + 2) – 1
a = 2, so a > 1
b = −1, log 2( x ) . is translated 1 unit down.
c = 2, log 2 ( x) is translated 2 units to the left.
x-intercept, let y = 0
=
0 log 2 ( x + 2 ) − 1
=
1 log 2 ( x + 2)
1
2=
( x + 2)
x=0
∴graph passes through (0, 0).
Let x = 6
=
y log 2 ( 8 ) − 1
= 3 −1
=2
∴graph passes through (6, 2).
Vertical asymptote is at x = −2.
© Cengage Learning Australia Pty Ltd 2018
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b
y = log0.5 (x – 2) + 2
a = 0.5, so 0 < a < 1
b = 2, log 0.5 ( x) is translated 2 units up.
c = −2, log 0.5 ( x) is translated 2 units to the right.
∴graph passes through (3, 2).
x-intercept, let y = 0
=
0 log 0.5 ( x − 2 ) + 2
=
−2 log 0.5 ( x − 2)
( 0.5) = ( x − 2)
−2
x−2=
4
x=6
∴graph passes through (6, 0).
Let x = 10
y log 0.5 ( 8 ) + 2
=
= log 0.5 ( 0.5 ) + 2
−3
=−3 + 2
= −1
∴graph passes through (10, −1).
Vertical asymptote is at x = 2.
© Cengage Learning Australia Pty Ltd 2018
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c
y = log3 (x – 1) + 2
a = 3, so a > 1
b = 2, log 3 ( x) is translated 2 units up.
c = −1, log 3 ( x) is translated 1 unit to the right.
x-intercept, let y = 0
=
0 log 3 ( x − 1) + 2
=
−2 log 3 ( x − 1)
2
3−=
( x − 1)
1
x −1 =
9
x=
10
9
x ≈ 1.11
∴graph passes through (1.11, 0).
Let x = 10
=
y log 3 ( 9 ) + 2
= log 3 32 + 2
= 2+2
=4
∴graph passes through (10, 4).
Vertical asymptote is at x = 1.
© Cengage Learning Australia Pty Ltd 2018
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d
y = log5 (x + 2) + 1
a = 5, so a > 1
b = 1, log 5 ( x) is translated 1 unit up.
c = 2, log 5 ( x) is translated 2 units to the left.
x-intercept, let y = 0
=
0 log 5 ( x + 2 ) + 1
=
−1 log 5 ( x + 2)
1
5−=
( x + 2)
1
x+2=
5
x= −
9
5
x = −1.8
∴graph passes through (−1.8, 0).
Let x = 3
=
y log 5 ( 5 ) + 1
= 1+1
=2
∴graph passes through (3, 2).
Vertical asymptote is at x = −2.
© Cengage Learning Australia Pty Ltd 2018
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Question 6
y = log2 (8x)
=
y log 2 ( 8 ) + log 2 ( x )
=
y log 2 ( 23 ) + log 2 ( x )
=
y 3log 2 ( 2 ) + log 2 ( x )
y= 3 + log 2 ( x )
∴A vertical translation of 3 units up maps y = log 2 ( x ) to y = log2 (8x).
Question 7
y = log3 (x)
log 3 ( x ) =
log ( x )
log ( 3)
log 9 ( x ) =
log ( x )
log ( 9 )
=
=
log ( x )
log ( 32 )
log ( x )
2 log ( 3)
log ( x )
= 2 log 9 ( x )
log ( 3)
∴ log 3 ( x ) =
2 log 9 ( x )
∴A vertical dilation with dilation factor of 2.
© Cengage Learning Australia Pty Ltd 2018
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Exercise 1.04 Logarithmic equations
Question 1
a
log5 (7x + 3) = log5 (5x + 9)
(7x + 3) = (5x + 9)
2x = 6
x=3
Check: Let x = 3
log 5 ( 21 +=
3) log 5 (15 + 9 )
log 5 ( 24 ) = log 5 ( 24 )
∴x = 3 is a solution.
b
log8 (x) + log8 (x + 6) = log8 (5x + 12)
log8 ( x ( x + 6=
) ) log8 ( 5 x + 12 )
x 2 + 6 x = 5 x + 12
0
x 2 + x − 12 =
0
( x + 4 )( x − 3) =
3
x=
−4 or x =
Check: Let x = −4
log8 ( −4 ) + log8 ( −4 + =
6 ) log8 ( −20 + 12 )
x = –4 gives a negative number in the logarithms, so x = –4 is not a solution.
Check: Let x = 3
log8 ( 3) + log8 ( 3 +=
6 ) log8 (15 + 12 )
log8 ( 3 × 9 ) =
log8 ( 27 )
log8 ( 27 ) = log8 ( 27 )
∴x = 3 is a solution.
© Cengage Learning Australia Pty Ltd 2018
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c
log (x – 2) – log (2x – 3) = log (2)
 ( x − 2) 
log 
 = log ( 2 )
 ( 2 x − 3) 
( x − 2=) 2 ( 2 x − 3)
x − 2 = 4x − 6
3x = 4
x=
Check: Let x =
4
3
4
3
 4
 
  3 − 2 
  = log 2
log  
( )
4
  2 × − 3 
 3 


 2
 −3 


log  −1  = log ( 2 )
 3 






log(2) = log(2)
∴x =
4
is a solution.
3
© Cengage Learning Australia Pty Ltd 2018
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d
log3 [(x – 2)(x + 3)] = log3 (14)
14
( x − 2 )( x + 3) =
14
x2 + x − 6 =
0
x 2 + x − 20 =
0
( x + 5)( x − 4 ) =
4
x=
−5or x =
Check: Let x = −5
log 3 ( −7 × −2 ) = log 3 (14 )
log 3 (14 ) = log 3 (14 )
Check: Let x = 4
log 3 ( 2 × 7 ) =
log 3 (14 )
log 3 (14 ) = log 3 (14 )
∴x = −5 and x = 4 are solutions.
© Cengage Learning Australia Pty Ltd 2018
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e
log6 (2x +1) = log6 (x + 2) – log6 (3)
 x+2
log 6 ( 2 x + 1) =
log 6 

 3 
x+2
2x +1 =
3
6x + 3 = x + 2
5 x = −1
x= −
1
5
Check: Let x = −
1
5
 1

−5+2
 2 
log 6  − + 1 =
log 6 

 5 
 3 


3
9 1
log=
log 6  × 
6 
5
 5 3
3
3
log 6   = log 6  
5
5
∴x = −
1
is a solution.
5
© Cengage Learning Australia Pty Ltd 2018
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f
log7 (x + 4) + log7 (x – 2) = log7 (4x)
log 7 ( ( x + 4 )( x − 2 ) ) =
log 7 ( 4 x )
x2 + 2x − 8 =
4x
x2 − 2 x − 8 =
0
0
( x − 4 )( x − 2 ) =
=
x 4=
or x 2
Check: Let x = 4
log 7 ( 8 ) + log 7 ( 2 ) =log 7 ( 4 × 4 )
log 7 ( 8 × 2 ) =
log 7 (16 )
log 7 (16 ) = log 7 (16 )
∴x = 4 is a solution
Check: Let x = 2
log 7 ( 6 ) + log 7 ( 0 ) =log 7 ( 4 × 2 )
log 7 ( 6 × 0 ) =
log 7 ( 8 )
log 7 ( 0 ) ≠ log 7 ( 8 )
log 7 (0) is undefined, ∴x = 2 is not a solution.
© Cengage Learning Australia Pty Ltd 2018
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g
log (3x + 5) = log (7x – 12)
3 x + 5 = 7 x − 12
−4 x =
−17
x=
17
4
Check: Let x =
17
4
 17

 17

log  3 × + 5=
 log  7 × − 12 
4
4




 71 
 71 
log   = log  
 4
 4
∴x =
h
17
is a solution.
4
log2 (3x + 2) = log2 (x − 2)
3x + 2 = x − 2
2 x = −4
x = −2
Check: Let x = –2
LHS = log2 (3 × –2 + 2) = log2 (–4)
RHS = log2 (–2 – 2) = log2 (–4)
So LHS = RHS, but both undefined.
So no solution.
© Cengage Learning Australia Pty Ltd 2018
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i
log6 (2x) + log6 (4) = log6 (x + 12) – log6 (2)
 x + 12 
log 6 ( 2 x × 4 ) =
log 6 

 2 
8x =
x + 12
2
16 x= x + 12
15 x = 12
=
x
Check: Let x =
12 4
=
15 5
4
5
4

 5 + 12 
 4 
log 6  2 × × 4  =
log 6 

 5 
 2 


 32 
 64 1 
log=
6
 log 6  × 
 5 
 5 2
 32 
 32 
log 6   = log 6  
 5 
 5 
∴x =
4
is a solution.
5
© Cengage Learning Australia Pty Ltd 2018
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j
log8 (2x + 1) + log8 (2x − 1) = 3 log8 (3)
log8 ( 2 x + 1)( 2 x − 1) =
log8 ( 33 )
log8 ( 4 x 2 − 1) =
log8 ( 27 )
4 x 2 − 1 =27
4 x 2 = 28
x2 = 7
x= ± 7
Check: Let x = − 7
(
)
(
)
log8 2 × − 7 + 1 + log8 2 × − 7 − 1 = 3log8 ( 3)
log8 ( −4.29 ) + log8 ( −6.29 ) =
3log8 ( 3)
Log of negative number is undefined, ∴x = − 7 is not a solution.
Check: Let x =
(
7
(
)
)
log8 2 × 7 + 1 + log8 2 × 7 − 1 =
3log8 ( 3)
(
)(
)
log8  2 × 7 + 1 2 × 7 − 1  =
log8 ( 33 )


log8 ( 4 × 7 − 1) =
log8 ( 27 )
log8 ( 27 ) = log8 ( 27 )
∴x =
7 is a solution.
© Cengage Learning Australia Pty Ltd 2018
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Question 2
a
log2 (5x + 7) = 5
log 2 ( 5 x + 7 ) =
5log 2 ( 2 )
log 2 ( 5 x + 7 ) =
log 2 ( 25 )
5x + 7 =
32
5 x = 25
x=5
Check: Let x = 5
log 2 ( 5 × 5 + 7 ) =
5
log 2 ( 32 ) = 5
log 2 ( 25 ) = 5
5log 2 ( 2 ) = 5
5=5
∴x = 5 is a solution.
© Cengage Learning Australia Pty Ltd 2018
62
b
log (4x – 1) = 3
log ( 4 x − 1) =
3log (10 )
log ( 4 x − 1) =
log (103 )
4x −1 =
1000
4 x = 1001
x = 250.25
Check: Let x = 250.25
log ( 4 × 250.25 − 1) =
3
log (1000 ) = 3
log (103 ) = 5
3log (10 ) = 3
3=3
∴x = 250.25 is a solution.
© Cengage Learning Australia Pty Ltd 2018
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c
log3 (5x – 11) = 2
log 3 ( 5 x − 11) =
2 log 3 ( 3)
log 3 ( 5 x − 11) =
log 3 ( 32 )
5 x − 11 =
9
5 x = 20
x=4
Check: Let x = 4
log 3 ( 5 × 4 − 11) =
2
log 3 ( 9 ) = 2
log 3 ( 32 ) = 2
2 log 3 ( 3) = 2
2=2
∴x = 4 is a solution.
© Cengage Learning Australia Pty Ltd 2018
64
d
log5 (4x + 11) = 2
log 5 ( 4 x + 11) =
2 log 5 ( 5 )
log 5 ( 4 x + 11) =
log 5 ( 52 )
4 x + 11 =
25
4 x = 14
x=
7
2
Check: Let x =
7
2
 7

log 5  4 × + 11 =
2
 2

log 5 ( 25 ) = 2
log 5 ( 52 ) = 2
2 log 5 ( 5 ) = 2
2=2
∴x =
7
is a solution.
2
© Cengage Learning Australia Pty Ltd 2018
65
e
log3 (9x + 2) = 4
log 3 ( 9 x + 2 ) =
4 log 3 ( 3)
log 3 ( 9 x + 2 ) =
log 3 ( 34 )
9x + 2 =
81
9 x = 79
=
x
Check: Let x =
79
7
= 8
9
9
79
.
9
 79

log 3  9 × + 2  =
4
9


log 3 ( 79 + 2 ) =
4
log 3 ( 81) = 4
log 3 ( 34 ) = 4
4=4
∴x=
79
is a solution.
9
© Cengage Learning Australia Pty Ltd 2018
66
f
log3 (4x – 9) = 3
log 3 ( 4 x − 9 ) =
3log 3 ( 3)
log 3 ( 4 x − 9 ) =
log 3 ( 33 )
4x − 9 =
27
4 x = 36
x=9
Check: Let x = 9
log 3 ( 4 × 9 − 9 ) =
3
log 3 ( 27 ) = 3
log 3 ( 33 ) = 3
3log 3 ( 3) = 3
3=3
∴x = 9 is a solution.
© Cengage Learning Australia Pty Ltd 2018
67
g
log3 (3x + 11) = 4
log 3 ( 3 x + 11) =
4 log 3 ( 3)
log 3 ( 3 x + 11) =
log 3 ( 34 )
3 x + 11 =
81
3 x = 70
=
x
Check: Let x =
70
1
= 23
3
3
70
3
 70

log 3  3 × + 11 =
4
3


log 3 ( 81) = 4
log 3 ( 34 ) = 4
4 log 3 ( 3) = 4
4=4
∴x =
70
is a solution.
3
© Cengage Learning Australia Pty Ltd 2018
68
h
log7 (6x – 5) = 2
log 7 ( 6 x − 5 ) =
2 log 7 ( 7 )
log 7 ( 6 x − 5 ) =
log 7 ( 7 2 )
6x − 5 =
49
6 x = 54
x=9
Check: Let x = 9
log 7 ( 6 × 9 − 5 ) =
2
log 7 ( 49 ) = 2
log 7 ( 7 2 ) = 2
2 log 7 ( 7 ) = 2
2=2
∴x = 9 is a solution.
© Cengage Learning Australia Pty Ltd 2018
69
i
log3 (x2 – 6x) = 3
log 3 ( x 2 − 6 x ) =
3log 3 ( 3)
log 3 ( x 2 − 6 x ) =
log 3 ( 33 )
x2 − 6x =
27
x 2 − 6 x − 27 =
0
0
( x − 9 )( x + 3) =
x = 9 or x = −3
Check: Let x = 9
log 3 ( 92 − 6 × 9 ) =
3
log 3 ( 27 ) = 3
log 3 ( 33 ) = 3
3log 3 ( 3) = 3
3=3
∴x = 9 is a solution.
Check: Let x = −3
(
)
log 3 ( −3) − 6 × ( −3) =3
2
log 3 ( 27 ) = 3
log 3 ( 33 ) = 3
3log 3 ( 3) = 3
3=3
∴x = −3 is a solution.
© Cengage Learning Australia Pty Ltd 2018
70
Question 3
a
log2 (x + 5) – log2 (2x –1) = 5
 x+5 
log 2 
 = 5log 2 2
 2x −1 
 x+5 
5
log 2 
 = log 2 ( 2 )
 2x −1 
x+5
= 32
2x −1
x +=
5 32(2 x − 1)
x + 5= 64 x − 32
63 x = 37
x=
37
63
Check: Let x =
37
63
 37

 63 + 5 
log 2 
=5
37
 2 × −1 
63 

log 2 ( 32 ) = 5
log 2 ( 25 ) = 5
5log 2 ( 2 ) = 5
5=5
∴x =
37
is a solution.
63
© Cengage Learning Australia Pty Ltd 2018
71
b
2 log3 (x) − 1 = log3 (2x − 3)
2 log 3 ( x ) − log 3 3= log 3 ( 2 x − 3)
log 3 ( x 2 ) − log 3 3= log 3 ( 2 x − 3)
 x2 
log=
 log 3 ( 2 x − 3)
3
 3
x2
= 2x − 3
3
2
x=
6x − 9
x2 − 6x + 9 =
0
0
( x − 3) =
2
x=3
Check: Let x = 3
2 log 3 ( 3=
) − 1 log3 ( 6 − 3)
2 log 3 ( 3) − 1 =log 3 ( 3)
2 ×1 − 1 =1
1=1
∴ x = 3 is a solution.
© Cengage Learning Australia Pty Ltd 2018
72
c
log4 (4x – 2) – log4 (3x +1) = 5
 4x − 2 
log 4 
 = 5log 4 ( 4 )
 3x + 1 
 4x − 2 
5
log 4 
 = log 4 ( 4 )
 3x + 1 
4x − 2
= 1024
3x + 1
x − 2 1024 ( 3 x + 1)
4=
−3068 x =
1026
1026
513
−
=
−
x=
3068
1534
Check: Let x = −
513
1534
513
513




− 2  − log 4  3 × −
+ 1 = 5
log 4  4 × −
1534
1534 



5 
 5120 

log 4  −
5
 − log 4  −
=
 1534 
 1534 
Log of a negative number is undefined.
∴x = −
513
is not a solution.
1534
∴No solution possible.
© Cengage Learning Australia Pty Ltd 2018
73
d
log2 (x + 1) – log2 (x – 4) = 3
 ( x + 1) 
log 2 
 = 3log 2 2
 ( x − 4) 
 ( x + 1) 
3
log 2 
 = log 2 ( 2 )
 ( x − 4) 
( x + 1) = 8
( x − 4)
x + 1= 8( x − 4)
x + 1 = 8 x − 32
7 x = 33
x=
Check: Let x =
33
7
33
7
 33 
 33

log 2  + 1 − log 2  − 4  =
3
 7

 7

 40 
5
log 2   − log 2   =
3
 7 
7
log 2 ( 8 ) = 3
log 2 ( 23 ) = 3
3log 2 ( 2 ) = 3
3=3
∴x =
33
is a solution.
7
© Cengage Learning Australia Pty Ltd 2018
74
e
log2 (x) + log2 (2 − 2x) = −1
− log 2 ( 2 )
log 2 ( x ) + log 2 ( 2 − 2 x ) =
log 2 ( x ( 2 − 2 x ) ) =
log 2 ( 2−1 )
1
2x − 2x2 =
2
4x2 − 4 x + 1 =
0
0
( 2 x − 1) =
2
x=
Check: Let x =
1
2
1
2
1
1

log 2   + log 2  2 − 2 ×  =− l
2
2

log 2 ( 2−1 ) + log 2 (1) =
−1
−1 + 0 =−1
−1 =−1
∴x =
1
is a solution.
2
© Cengage Learning Australia Pty Ltd 2018
75
f
log3 (x) + log3 (2x + 1) = 1
log 3 ( x ) + log 3 ( 2 x + 1) =
log 3 ( 3)
log 3 ( x ( 2 x + 1) ) =
log 3 ( 3)
2x2 + x =
3
2x2 + x − 3 =
0
0
( 2 x + 3)( x − 1) =
3
− or x =
x=
1
2
Check: Let x = −
3
2
3 
 3

log 3  −  + log 3  2 × − + 1 =1
2 
 2

Log of a negative number is undefined.
∴x = −
3
is not a solution.
2
Check: Let x = 1
log 3 (1) + log 3 ( 2 ×1 + 1) =1
0 +1 =
1
1=1
∴x = 1 is a solution.
© Cengage Learning Australia Pty Ltd 2018
76
g
log6 (x) + log6 (x + 5) = 2
log 6 ( x ( x + 5 ) ) =
2 log 6 ( 6 )
log 6 ( x 2 + 5 x ) =
log 6 ( 62 )
x2 + 5x =
36
x 2 + 5 x − 36 =
0
0
( x + 9 )( x − 4 ) =
x=
−9 or x =
4
Check: Let x = −9
log 6 ( −9 ) + log 6 ( −9 + 5 ) =2
Log of a negative number is undefined.
∴x = −9 is not a solution.
Check: Let x = 4
log 6 ( 4 ) + log 6 ( 4 + 5 ) =
2
log 6 ( 36 ) = 2
log 6 ( 62 ) = 2
2 log 6 ( 6 ) = 2
2=2
∴x = 4 is a solution.
© Cengage Learning Australia Pty Ltd 2018
77
h
log2 (2x + 12) − log2 (3x) = 2
 2 x + 12 
log 2 
 = 2 log 2 ( 2 )
 3x 
log 2 ( ) = log 2 ( 22 )
2 x + 12
=4
3x
2 x + 12 =
12 x
10 x = 12
x=
Check: Let x =
6
5
6
5
 6

 6
log 2  2 × + 12  − log 2  3 ×  =
2
 5

 5
 72 
 18 
log 2   − log 2   =
2
 5 
5
 72 18 
log 2  ÷  =
2
 5 5
log 2 ( 4 ) = 2
log 2 ( 22 ) = 2
2 log 2 ( 2 ) = 2
2=2
∴x =
6
is a solution.
5
© Cengage Learning Australia Pty Ltd 2018
78
i
log4 (2x – 3) – log4 (x + 2) = 1
 2x − 3 
log 4 
 =1
 x+2 
 2x − 3  1
=

 4= 4
 x+2 
2 x − 3= 4 ( x + 2 )
−2 x =
11
x= −
Check: Let x = −
11
2
11
2
11 

 11 
log 4  2 × − − 3  − log 4  − + 2  =
1
2


 2

 7
log 4 ( −8 ) − log 4  −  =
1
 2
Log of a negative number is undefined.
∴x = −
11
is not a solution.
2
∴No solution possible.
© Cengage Learning Australia Pty Ltd 2018
79
j
log3 (x) – log3 (x − 1) = 2
 x 
log 3 
 = 2 log 3 ( 3)
 x −1 
 x 
2
log 3 
 = log 3 ( 3 )
 x −1 
x
=9
x −1
=
x 9 ( x − 1)
−8 x =
−9
x=
9
8
Check: Let x =
9
8
9
9 
log 3   − log 3  − 1 =
2
8
8 
9
1
log 3   − log 3   =
2
8
8
9 1
log 3  ÷  =
2
8 8
log 3 ( 9 ) = 2
log 3 ( 32 ) = 2
2 log 3 ( 3) = 2
2=2
∴x =
9
is a solution.
8
© Cengage Learning Australia Pty Ltd 2018
80
Question 4
a
log6 (x) + log6 (x – 9) = 2
log 6 ( x ( x − 9 ) ) =
2 log 6 ( 6 )
log 6 ( x 2 − 9 x ) =
log 6 ( 62 )
x2 − 9x =
36
x 2 − 9 x − 36 =
0
0
( x − 12 )( x + 3) =
x=
−3or x =
12
Check: Let x = −3
log 6 ( −3) + log 6 ( −3 − 9 ) =2
Log of a negative number is undefined.
∴x = −3 is not a solution.
Check: Let x = 12
log 6 (12 ) + log 6 (12 − 9 ) =
2
log 6 ( 36 ) = 2
log 6 ( 62 ) = 2
2 log 6 ( 6 ) = 2
2=2
∴x = 12 is a solution.
© Cengage Learning Australia Pty Ltd 2018
81
b
log4 (x) + log4 (x – 12) = 3
log 4 ( x ( x − 12 ) ) =
3log 4 ( 4 )
log 4 ( x 2 − 12 x ) =
log 4 ( 43 )
x 2 − 12 x =
64
x 2 − 12 x − 64 =
0
0
( x − 16 )( x + 4 ) =
x =−4 =16
Check: Let x = −4
log 4 ( −4 ) + log 4 ( −4 − 12 ) =3
Log of a negative number is undefined.
∴x = −4 is not a solution.
Check: Let x = 16
log 4 (16 ) + log 4 (16 − 12 ) =
3
log 4 ( 4 2 ) + log 4 ( 4 ) =
3
2 log 4 ( 4 ) + log 4 ( 4 ) =
3
2 +1 =
3
3=3
∴x = 16 is a solution.
© Cengage Learning Australia Pty Ltd 2018
82
c
log6 (x) + log6 (x − 1) = 1
log 6 ( x ( x − 1) ) =
log 6 ( 6 )
log 6 ( x 2 − x ) =
log 6 ( 6 )
x2 − x =
6
x2 − x − 6 =
0
0
( x + 2 )( x − 3) =
x=
−2 or x =
3
Check: Let x = −2
log 6 ( −2 ) + log 6 ( −2 − 1) =
1
Log of a negative number is undefined.
∴x = −2 is not a solution.
Check: Let x = 3
log 6 ( 3) + log 6 ( 3 − 1) =
1
log 6 ( 6 ) = 1
1=1
∴x = 3 is a solution.
© Cengage Learning Australia Pty Ltd 2018
83
d
log6 (x2 – 2x) = log6 (5x – 12)
x 2 − 2 x = 5 x − 12
x 2 − 7 x + 12 =
0
0
( x − 3)( x − 4 ) =
=
=
x 3or
x 4
Check: Let x = 3
log 6 ( 32 − 2 ×
=
3) log 6 ( 5 × 3 − 12 )
log 6 ( 3) = log 6 ( 3)
∴x = 3 is a solution.
Check: Let x = 4
log 6 ( 42 − 2 ×=
4 ) log 6 ( 5 × 4 − 12 )
log 6 ( 8 ) = log 6 ( 8 )
∴x = 4 is a solution.
© Cengage Learning Australia Pty Ltd 2018
84
e
log7 (6x) – log7 (4 – x) = log7 (3)
 6x 
log 7 
 = log 7 ( 3)
 4− x
6x
=3
4− x
=
6x 3( 4 − x)
x 12 − 3 x
6=
9 x = 12
x=
4
3
Check: Let x =
4
3
4
 4

log 7  6 ×  − log 7  4 −  =
log 7 ( 3)
3
 3

8
log 7 ( 8 ) − log 7   =
log 7 ( 3)
3
 8
log 7  8 ÷  =
log 7 ( 3)
 3
log 7 ( 3) = log 7 ( 3)
∴x =
f
4
is a solution
3
log (x) + log (x + 3) = log (5x)
log  x ( x + 3)  =
log ( 5 x )
5x
x 2 + 3x =
0
x2 − 2 x =
0
x ( x − 2) =
x = 0, 2
But log (0) is not defined.
∴ x = 2.
© Cengage Learning Australia Pty Ltd 2018
85
g
log5 (x – 1) + log5 (x + 3) = log5 (x2 – 3x + 12)
log 5 ( x − 1)( x +=
3)  log 5 ( x 2 − 3x + 12 )
log 5 ( x 2 + 2 x =
− 3) log 5 ( x 2 − 3x + 12 )
x 2 + 2 x − 3 = x 2 − 3x + 12
5 x = 15
x=3
Check: Let x = 3
+ 3) log 5 ( 32 − 3 × 3 + 12 )
log 5 ( 3 − 1) + log 5 ( 3=
log 5 ( 2 ) + log 5 ( 6 ) =
log 5 (12 )
log 5 ( 2 × 6 ) =
log 5 (12 )
log 5 (12 ) = log 5 (12 )
∴x = 3 is a solution
© Cengage Learning Australia Pty Ltd 2018
86
h
[log3 (x)]2 = 3 − 2 log3 (x)
Let a = log 3 ( x )
a 2 = 3 − 2a
a 2 + 2a − 3 =
0
0
( a + 3)( a − 1) =
a=
−3or a =
1
log 3 ( x ) =
−3
log 3 ( x ) =
1
−3
=
x 3=
x 31
1
=
x 3
27
=
x
Check x =
1
27
2

 1 
 1 
log 3  27   = 3 − 2 log 3  27 
 
 

log 3 ( 3−3 )  = 3 − 2 log 3 ( 3−3 )


2
 −3log 3 ( 3)  = 3 − 2 × −3log 3 ( 3)
2
9= 3 + 6
9=9
∴x =
1
is a solution.
27
Check: Let x = 3
log 3 ( 3)  = 3 − 2 log 3 ( 3)
2
1= 3 − 2
1=1
∴x = 3 is a solution.
© Cengage Learning Australia Pty Ltd 2018
87
i
log5 (2x) + log5 (x + 2) = log5 (6)
log 5 ( 2 x ( x + 2 ) ) =
log 5 ( 6 )
log 5 ( 2 x 2 + 2 x ) =
log 5 ( 6 )
2x2 + 2x =
6
x2 + x − 6 =
0
0
( x + 5)( x − 1) =
x=
−5or x =
1
Check: Let x = −5
log 5 ( 2 × −5 ) + log 5 ( −5 + 2 ) = log 5 ( 6 )
log 5 ( −10 ) + log 5 ( −3) ≠ log 5 ( 6 )
∴x = −5 is not a solution.
Check: Let x = 1
log 5 ( 2 ×1) + log 5 (1 + 2 ) =
log 5 ( 6 )
log 5 ( 2 ) + log 5 ( 3) =
log 5 ( 6 )
log 5 ( 6 ) = log 5 ( 6 )
∴x = 1 is a solution.
© Cengage Learning Australia Pty Ltd 2018
88
j
log8 (2) + log8 (4x2) = 2
log8 ( 2 × 4 x 2 ) =
2 log8 ( 8 )
log8 ( 8 x 2 ) = log8 ( 82 )
8 x 2 = 64
0
x2 − 8 =
0
( x − 2 2 )( x + 2 2 ) =
x = ±2 2
Check: Let x = −2 2
( (
log8 ( 2 ) + log8 4 × −2 2
) ) =2
2
log8 ( 2 ) + log8 ( 32 ) =
2
log8 ( 64 ) = 2
log8 ( 82 ) = 2
2 log8 ( 8 ) = 2
2=2
∴x = – 2 2 is a solution.
Check: Let x = 2 2
(
(
log8 ( 2 ) + log8 4 × 2 2
2
) )=
2
log8 ( 2 ) + log8 ( 32 ) =
2
log8 ( 64 ) = 2
log8 ( 82 ) = 2
2 log8 ( 8 ) = 2
2=2
∴x = 2 2 is a solution.
© Cengage Learning Australia Pty Ltd 2018
89
Question 5
2log2 ( x−5) + 2(x – 5) – 12 = 0
0
( x − 5) + 2 ( x − 5) − 12 =
3 x − 27 =
0
3 x = 27
x=9
Check: Let x = 9
2log2 (9−5) + 2 ( 9 – 5 ) –12 =
0
2log2 ( 4) + 2 × 4 − 12 =
0
2
( ) −4=
0
log 2 22
2 × 2log2 ( 2) − 4 =
0
2 × 21 − 4 =
0
0=0
Question 6
log10 (5x + x – 31) = x[1 – log10 (2)]
log10 ( 5 x =
+ x − 31) x log10 (10 ) − log10 ( 2 ) 
log10 ( 5 x + x − 31) =
x log10 ( 5 )
log10 ( 5 x + x − 31) =
log10 ( 5 x )
5 x + x − 31 =
5x
x = 31
© Cengage Learning Australia Pty Ltd 2018
90
Exercise 1.05 Applications of logarithmic functions
Question 1
 x
M(x) = log  
 x0 
 758 x0 
M ( x ) = log 

 x0 
= log ( 758 )
= 2.87966...
≈ 2.9
© Cengage Learning Australia Pty Ltd 2018
91
Question 2
 I 
L = 10 log  
 I0 
Drill:
 I 
50 = 10log  −12 
 10 
5 = log (1012 I )
105 = 1012 I
I = 10−7 W/m 2
Compressor:
 I 
62 = 10log  −12 
 10 
6.2 = log (1012 I )
106.2 = 1012 I
I = 10−5.8 W/m 2
10−5.8
= 101.2
10−7
= 15.8489...
≈ 15.85
∴Compressor noise is 101.2 or approximately 15.85 W/ m 2 more intense than drill noise.
© Cengage Learning Australia Pty Ltd 2018
92
Question 3
 r
FV = PV 1 + 
 n
nt
FV = 2PV
12 t
 0.1 
=
2 PV PV 1 +

 12 
12 t
 0.1 
2= 1 +

 12 
log ( 2 ) = log (1.0083...)
12 t
log ( 2 ) = 12t log (1.0083...)
12t =
log (1.0083...)
log(2)
12t = 83.5237...
t = 6.9603...
t ≈ 7 years
© Cengage Learning Australia Pty Ltd 2018
93
Question 4
 x
M(x) = log  
 x0 
 x
8.3 = log  
 x0 
108.3 =
x
x0
x = 108.3 x0
 4 ×108.3 x0 
M ( x ) = log 

x0


= log ( 4 ×108.3 )
= 8.90205...
≈ 8.9
Chilean earthquake measured approximately 8.9 on the Richter scale.
© Cengage Learning Australia Pty Ltd 2018
94
Question 5
 x
M(x) = log  
 x0 
Ecuador:
x 
8.3 = log  1 
 x0 
108.3 =
x1
x0
x1 = 108.3 x0
New Zealand:
x 
7.1 = log  2 
 x0 
107.1 =
x2
x0
x2 = 107.1 x0
x1 108.3 x0
=
x2 107.1 x0
= 101.2
= 15.8489...
≈ 15.85
The earthquake in Ecuador is approximately 16 times greater.
© Cengage Learning Australia Pty Ltd 2018
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Question 6
Amp 1: 25 W/m2
 25 
L = 10log  −12 
 10 
= 133.9794...
≈ 133.98
Amp 2: 500 W/m2
 500 
L = 10log  −12 
 10 
= 147.7815...
≈ 147.78
Difference = 147.78 − 133.98 = 13.8 dB
Question 7
L = 160 dB
I=?
 I 
160 = 10log  −12 
 10 
160 = 10log (1012 I )
16 = log (1012 I )
1016 = 1012 I
I = 104 W/m 2
© Cengage Learning Australia Pty Ltd 2018
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Question 8
a
¢=
 f 
1200
log  2 
log(2)
 f1 
=
f1 27.5,
=
f 2 29.135
¢ =
1200
 29.135 
log 

log ( 2 )
 27.5 
= 99.986...
≈ 100
∴There are 100 ¢ between A and B flat.
b
¢=
 f 
1200
log  2 
log(2)
 f1 
=
f1 261.63,
=
f 2 440
¢ =
1200
 440 
log 

log ( 2 )
 261.63 
= 899.97...
≈ 900
∴There are 900 ¢ between middle C and A4.
c
¢=
 f 
1200
log  2 
log(2)
 f1 
=
=
f1 261.63,
f 2 587.33
¢ =
1200
 587.33 
log 

log ( 2 )
 261.63 
= 1399.972...
≈ 1400
∴There are 1400 ¢ between middle C and C5.
© Cengage Learning Australia Pty Ltd 2018
97
Question 9
pH = −log[H+]
pH = 1.8, [H+] = ?
1.8 = − log  H + 
(
1.8 = log  H + 
101.8 =  H + 
−1
)
−1
1
 H +  = 1.8
10
= 0.0158...
≈ 0.016
Concentration of hydrogen ions is 0.016 mole/litre.
Question 10
a
pH = −log[H+]
pH = 0, [H+] = ?
0 = − log  H + 
(
0 = log  H + 
100 =  H + 
−1
)
−1
1
 H +  = 0
10
=1
Hydrogen ion concentration for battery acid is 1 mole/litre.
© Cengage Learning Australia Pty Ltd 2018
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b
pH = 3, [H+] = ?
3 = − log  H + 
(
3 = log  H + 
103 =  H + 
−1
)
−1
1
 H +  = 3
10
= 10−3
= 0.001
Hydrogen ion concentration for vinegar is 10−3 or 0.001 mole/litre.
c
pH = 8, [H+] = ?
8 = − log  H + 
(
8 = log  H + 
108 =  H + 
−1
)
−1
1
 H +  = 8
10
= 10−8
Hydrogen ion concentration for baking soda is 10−8 mole/litre.
d
pH = 11, [H+] = ?
11 = − log  H + 
(
11 = log  H + 
1011 =  H + 
−1
)
−1
1
 H +  = 11
10
10−11
Hydrogen ion concentration for household ammonia is 10−11 mole/litre.
© Cengage Learning Australia Pty Ltd 2018
99
Question 11
a
Mercury: m1 =
−2, I1 =
6.31
SN1006: m2 =
−7, I 2 =
?
I 
m1 – m2 = –2.5 log10  1 
 I2 
 6.31 
−2 − ( −7 ) =−2.5log10 

 I2 
 6.31 
5 = −2.5log10 

 I2 
 6.31 
−2 =
log10 

 I2 
10−2 =
6.31
I2
I2 =
6.31
10−2
I 2 = 631
© Cengage Learning Australia Pty Ltd 2018
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b
Mercury: m1 =
−2, I1 =
6.31
Proxima Centauri:=
m2 11,
=
I2 ?
 6.31 
−2 − (11) =−2.5log10 

 I2 
 6.31 
−13 =
−2.5log10 

 I2 
 6.31 
5.2 = log10 

 I2 
105.2 =
6.31
I2
I2 =
6.31
105.2
I 2 = 0.000 039 81...
≈ 3.98 × 10−5
Brightness of Proxima Centauri is approximately 0.000 0398 or 3.98 ×10−5 .
© Cengage Learning Australia Pty Ltd 2018
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c
Sun: m1 = −27 , Moon: m2 = −13
I 
−27 − ( −13) =−2.5log10  1 
 I2 
I 
−14 =
−2.5log10  1 
 I2 
I 
5.6 = log10  1 
 I2 
105.6 =
I1
I2
I1
5.6
=
10=
398107
I2
The Sun is approximately 398 107 (or about 400 000) times brighter than the Moon.
© Cengage Learning Australia Pty Ltd 2018
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Question 12
a
When t = 5750 , N (t ) =
N0
2
N(t) = N0 × 2.72−kt
N0
= N 0 × 2.72− k ×5750
2
0.5 = 2.72−5750 k
log ( 0.5 ) = log ( 2.72−5750 k )
log ( 0.5 ) = −5750k log ( 2.72 )
log ( 0.5 )
−5750k =
log ( 2.72 )
= −0.692709...
k = 0.000120471...
k ≈ 0.0001205
b
Skeleton loses 70% of carbon-14 atoms ⇒ skeleton has 30% left.
∴ N (t ) =
0.3 N 0
0.3 N=
N 0 × 2.72−0.0001205t
0
0.3 = 2.72−0.0001205t
log ( 0.3) = log ( 2.72−0.0001205t )
log ( 0.3) = −0.0001205t log ( 2.72 )
log ( 0.3)
−0.0001205t =
log ( 2.72 )
= −1.2032125...
t = 9985.16612...
t ≈ 9986
The skeleton is approximately 9986 years old.
© Cengage Learning Australia Pty Ltd 2018
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Question 13
a
Q(t) = Q0 × 2.5−nt
Initial amount occurs when t = 0.
Q(t) = Q0 × 2.5−n×0
Q(t) = Q0 × 2.50
Q(t) = Q0
b
When t = 1000 , Q(t ) =
Q0
2
Q0
= Q0 × 2.5− n × 1000
2
0.5 = 2.5−1000 n
log ( 0.5 ) = log ( 2.5−1000 n )
log ( 0.5 ) = −1000n log ( 2.5 )
log ( 0.5 )
−1000n =
log ( 2.5 )
= −0.7564707...
n = 0.00075647...
n ≈ 0.00076
c
Q0 = 45
= 45 × 2.5−0.00076 × 10
Q(10)
= 45 × 2.5−0.0076
= 44.6877...
≈ 44.7
In 10 years there would be 44.7 kg. ∴The radioactive substance would decay.
45 − 44.7 = 0.3 kg in 10 years.
© Cengage Learning Australia Pty Ltd 2018
104
Question 14
a
P = 6.9(1.011)t
2011 to 2045 = 34 years, ∴t = 34.
P = 6.9(1.011)34
P = 10.00894...
P ≈ 10.0
∴Population in 2045 will be approximately 10.0 billion.
b
In 2011 t = 0
P = 6.9(1.011)0
P = 6.9
Hence, let P = 6.9 × 2 = 13.8
13.8 = 6.9(1.011)t
2 = (1.011)t
log(2) = log(1.011)t
log(2) = t log(1.011)
t=
log(2)
log(1.011)
t = 63.3593...
t ≈ 64 years
(Need to round up to allow population to double.)
2011 + 64 = 2075
∴The population will double in 2075.
© Cengage Learning Australia Pty Ltd 2018
105
Question 15
N = 100 (2)
t
15
Let N = 300
t
300 = 100 ( 2 )15
t
3 = ( 2 )15
t
log(3) = log ( 2 )15
log(3) =
t
log ( 2 )
15
log(3)
t
=
15 log ( 2 )
t
= 1.58496...
15
t = 23.7744...
t ≈ 24
There will be 300 bacteria after 24 hours.
© Cengage Learning Australia Pty Ltd 2018
106
Chapter review
Question 1
a
25 = 32
log2 (32) = 5
2
b
64 3 = 16
2
3
log 64 (16 ) =
c
6−3 =
1
216
 1 
log 6 
 = −3
 216 
Question 2
a
log3 (81) = 4
34 = 81
b
log25 (5) =
1
2
1
2
25 = 5
c
log (0.001) = −3
10−3 = 0.001
d
log8 (32) =
5
3
5
8 3 = 32
© Cengage Learning Australia Pty Ltd 2018
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Question 3
a
log7 (1)
= log 7 (7 0 )
= 0 log 7 (7)
= 0 ×1
=0
b
log2 (0.25)
1
= log 2  
4
= log 2 ( 2−2 )
= −2 log 2 (2)
=−2 × 1
= −2
c
2 log6 (6)
= 2 ×1
=2
Question 4
a
log3 (27) + log3 (9) – log3 (81)
= log 3 ( 33 ) + log 3 ( 32 ) − log 3 ( 34 )
= 3log 3 ( 3) + 2 log 3 ( 3) − 4 log 3 ( 3)
= log 3 ( 3)
=1
© Cengage Learning Australia Pty Ltd 2018
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b
2 log4 (7) – 2 log4 (28)
= log 4 ( 7 2 ) − log 4 ( 282 )
 49 
= log 4 

 784 
1
= log 4  
 16 
= log 4 ( 4−2 )
= −2 log 4 (4)
=−2 ×1
= −2
c
 1 
log3  4

 81 
1
 1 4
= log 3  4 
3 
1
= log 3 ( 3−4 ) 4
= log 3 ( 3−1 )
= −1log 3 (3)
=−1×1
= −1
d
49log7 5
= 7 2log7 5
=7
( )
log 7 52
= 52
= 25
© Cengage Learning Australia Pty Ltd 2018
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Question 5
log5 (9x3)
a
= log 5 ( 9 ) + log 5 ( x3 )
= log 5 ( 9 ) + 3log 5 ( x )
b
 3
log7  4 
z 
= log 7 ( 3) − log 7 ( z 4 )
= log 7 ( 3) − 4log 7 ( z )
 m 2 (m − 4) 
log7 
2 
 (m + 3) 
c
2
= log 7  m 2 ( m − 4 )  − log 7 ( m + 3) 


= log 7 ( m 2 ) + log 7 ( m − 4 ) − 2 log 7 ( m + 3)
= 2 log 7 ( m ) + log 7 ( m − 4 ) − 2 log 7 ( m + 3)
d
 x3 − y 
log6 

 2 xy 


1
 3

2


x
−
y


= log 6 
  2 xy  


=
 x3 − y  
1
log
 6

2
 2 xy  
=
1
log 6 ( x3 − y ) − log 6 ( 2 xy ) 

2
=
1
log 6 ( x3 − y ) − ( log 6 ( 2 ) + log 6 ( x ) + log 6 ( y ) ) 

2
=
1
log 6 ( x3 − y ) − log 6 ( 2 ) − log 6 ( x ) − log 6 ( y ) 

2
© Cengage Learning Australia Pty Ltd 2018
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Question 6
a
2 log5 (x) + 5 log5 (y)
= log 5 ( x 2 ) + log 5 ( y 5 )
= log 5 ( x 2 y 5 )
b
2 [log (m) + log (3 – m) – log (2m3 + 7)]
  m (3 − m ) 
= 2 log 

3
  2m + 7  
 m (3 − m) 
= log 

3
 2m + 7 
c
2
1
log5 (17) – 3log5 (x) – 4log5 (y)
2
 12 
= log 5 17  − log 5 ( x3 ) − log 5 ( y 4 )


 17 
= log 5  3 4 
x y 
Question 7
log7 (25)
=
log 25
log 7
=
1.39794...
0.84509...
= 1.65417...
≈ 1.6542
© Cengage Learning Australia Pty Ltd 2018
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Question 8
a
log m ( 60 )
= log m ( 3 × 4 × 5 )
= log m ( 3 × 22 × 5 )
=log m ( 3) + log m ( 22 ) + log m 5
=log m ( 3) + 2 log m ( 2 ) + log m 5
= 1.15 + 2 × 0.73 + 1.68
= 4.29
b
8
log m  
 15 
 23 
= log m 

 3× 5 
= log m ( 23 ) − log m ( 3) − log m ( 5 )
= 3log m ( 2 ) − log m ( 3) − log m 5
=
3 × 0.73 − 1.15 − 1.68
= −0.64
c
 2
log m 

 9
 12 
2
= log m  
 3 
 
 
 12 
= log m  2  − log m ( 3)
 
1
log m ( 2 ) − log m ( 3)
2
=
1
=× 0.73 − 1.15
2
= −0.785
© Cengage Learning Australia Pty Ltd 2018
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Question 9
a
4x – 1 = 64
4 x −1 = 43
x − 1 =3
x=4
b
32x – 4 = 92 – x
32 x − 4 = 32( 2− x )
2x − 4 = 4 − 2x
4x = 8
x=2
c
9x =
3
( 3 ) = ( 3)
2 x
d
2x =
1
2
x=
1
4
95 – 9x =
(3 )
1
2
1
27 x− 2
2 5−9 x
= ( 3−3 )
x−2
2 (5 − 9x ) =
−3 ( x − 2 )
10 − 18 x =
−3 x + 6
−15 x =
−4
x=
4
15
© Cengage Learning Australia Pty Ltd 2018
113
Question 10
a
52x – 24 × 5x = 25
Let a = 5 x
25
( 5 ) − 24 ( 5 ) =
x 2
x
a 2 − 24a =
25
a 2 − 24a − 25 =
0
0
( a − 25)( a + 1) =
a = 25or a = −1
5 x = 25or 5 x = −1
5 x = 52 or no solution
x=2
b
22x – 3 × 2x + 2 = 0
Let a = 2 x
a 2 − 3a + 2 =
0
0
( a − 1)( a − 2 ) =
a 1or
a 2
=
=
2 x 1or
2x 2
=
=
0
2 x 2=
or 2 x 21
=
or x 1
=
x 0=
© Cengage Learning Australia Pty Ltd 2018
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Question 11
a
52x – 3 = 84
log ( 52 x −3 ) = log ( 84 )
log ( 84 )
( 2 x − 3) log ( 5) =
log ( 84 )
2x − 3 =
log ( 5 )
2x − 3 =
2.75302...
2 x = 5.75302...
x = 2.87651...
x ≈ 2.877
b
7 2 x – 3 = 53 x+ 1
log ( 7 2 x −3 ) = log ( 53 x +1 )
( 2 x − 3) log ( 7 ) =
( 3x + 1) log ( 5)
( 2 x − 3) = log ( 5)
( 3x + 1) log ( 7 )
( 2 x − 3) = 0.827087...
( 3x + 1)
2 x − 3 0.827087... ( 3 x + 1)
=
2 x − 3 2.481262...x + 0.827087...
=
0.481262...x = −3.827087...
x = −7.95219...
x ≈ −7.952
© Cengage Learning Australia Pty Ltd 2018
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Question 12
a
113x + 1 = 111
x ≈ 0.321
b
5 × 72x – 1 – 28 = 0
5 × 72x – 1 = 28
x ≈ 0.943
© Cengage Learning Australia Pty Ltd 2018
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Question 13
a
y = log3 (x) + 2
a = 3, so a > 1
b = 2, log 3 ( x) is translated 2 units up.
Zero = a − b
= 3−2
=
1
9
≈ 0.111
The graph passes through (0.111, 0).
The graph passes through (1, b).
(1, b) = (1, 2)
© Cengage Learning Australia Pty Ltd 2018
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b
y = log0.2 (x) – 2
a = 0.2, so 0 < a < 1
b = −2, log 0.2 ( x) is translated 2 units down.
Zero = a − b
= 0.2−( −2)
= 0.04
The graph passes through (0.04, 0).
The graph passes through (1, b).
(1, b) = (1, −2)
© Cengage Learning Australia Pty Ltd 2018
118
Question 14
a
y = log2 (x + 3)
a = 2, so a > 1
c = 3, log 2 ( x) is translated 3 units to the left.
Zero = 1 − c
= 1− 3
= −2
The graph passes through (−2, 0).
22 = 4 , so
log 2 ( 4 ) = 2
x+3=
2
x =1
The graph passes through (1, 2).
Vertical asymptote is at x = −3.
© Cengage Learning Australia Pty Ltd 2018
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b
y = log0.6 (x – 2)
a = 0.6, so 0 < a < 1
c = −2, log 0.5 ( x) is translated 2 units to the right.
Zero = 1 − c
= 1 − ( −2 )
=3
The graph passes through (3, 0).
( 0.6 )
−3.15
= 5 , so
log 0.6 ( 5 ) = −3.15
x−2=
5
x=7
The graph passes through (7, −3.15).
Vertical asymptote is at x = 2.
© Cengage Learning Australia Pty Ltd 2018
120
Question 15
a
log2 (5x) = log2 (2x + 9)
x 2x + 9
5=
3x = 9
x=3
Check: Let x = 3
log 2 ( 5 =
× 3) log 2 ( 2 × 3 + 9 )
log 2 (15 ) = log 2 (15 )
∴x = 3 is a solution.
b
log5 (4m − 5) = log5 (2m − 1)
4m − 5= 2m − 1
2m = 4
m=2
Check: Let m = 2
log 5 ( 4 × 2 −=
5 ) log 5 ( 2 × 2 − 1)
log 5 ( 3) = log 5 ( 3)
∴m = 2 is a solution.
© Cengage Learning Australia Pty Ltd 2018
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c
log3 (16 + 2k) = log3 (k2 – 4k)
16 + 2k = k 2 − 4k
k 2 − 6k − 16 =
0
0
( k − 8)( k + 2 ) =
k = 8or k = −2
Check: Let k = −2
(
log 3 (16 + 2 ×=
−2 ) log 3 ( −2 ) − 4 × ( −2 )
2
)
log 3 (16 −=
4 ) log 3 ( 4 + 8 )
log 3 (12) = log 3 (12)
∴k = −2 is a solution.
Check: Let k = 8
(
log 3 (16 +=
2 × 8 ) log 3 ( 8 ) − 4 × ( 8 )
2
)
log 3 (16 + 16
=
) log3 ( 64 − 32 )
log 3 (32) = log 3 (32)
∴k = 8 is a solution.
© Cengage Learning Australia Pty Ltd 2018
122
d
log 7 ( 2 p ) + log 7 ( p + 2 ) =
log 7 ( 6 )
log 7 ( 2 p ) + log 7 ( p + 2 ) =
log 7 ( 6 )
log 7 ( 2 p ( p + 2 ) ) =
log 7 ( 6 )
2 p2 + 4 p =
6
2 p2 + 4 p − 6 =
0
p2 + 2 p − 3 =
0
0
( p + 3)( p − 1) =
p=
−3or p =
1
Check: Let p = −3
log 7 ( 2 × −3) + log 7 ( −3 + 2 ) = log 7 ( 6 )
log 7 ( −6 ) + log 7 ( −1) =
log 7 ( 6 )
Log of negative undefined ∴ p = −3 is not a solution.
Check: Let p = 1
log 7 ( 2 ×1) + log 7 (1 + 2 ) =
log 7 ( 6 )
log 7 ( 2 ) + log 7 ( 3) =
log 7 ( 6 )
log 7 ( 2 × 3) =
log 7 ( 6 )
log 7 ( 6 ) = log 7 ( 6 )
∴p = 1 is a solution.
© Cengage Learning Australia Pty Ltd 2018
123
Question 16
a
log2 (5w + 6) = 4
5w + 6 = 24 = 16
5w = 10
w=2
Check: Let w = 2
log 2 ( 5 × 2 + 6 ) =
4
log 2 (16 ) = 4
log 2 ( 24 ) = 4
4 log 2 (2) = 4
4=4
∴w = 2 is a solution.
© Cengage Learning Australia Pty Ltd 2018
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b
log3 (v) + log3 (v – 6) = 3
log 3 ( v ) + log 3 ( v − 6 ) =
3log 3 ( 3)
log 3 ( v ( v − 6 ) ) =
log 3 ( 33 )
v 2 − 6v =
27
v 2 − 6v − 27 =
0
0
( v + 3)( v − 9 ) =
−3or v =
v=
9
Check: Let v = −3
log 3 ( −3) + log 3 ( −3 – 6 ) =
3
log 3 ( −3) + log 3 ( −9 ) =
3
Log of negative undefined ∴ v = −3 is not a solution.
Check: Let v = 9
log 3 ( 9 ) + log 3 ( 9 – 6 ) =
3
log 3 ( 32 ) + log 3 ( 3) =
3
2 ×1 + 1 =3
3=3
∴v = 9 is a solution.
© Cengage Learning Australia Pty Ltd 2018
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c
log2 (m +1) – log2 (m – 4) = 3
log 2 ( m + 1) − log 2 ( m − 4 ) =
3log 2 ( 2 )
 m +1 
3
log 2 
 = log 2 ( 2 )
m−4
m +1
=8
m−4
m +=
1 8 ( m − 4)
m + 1= 8m − 32
7 m = 33
=
m
Check: Let m =
33
5
= 4
7
7
33
7
 33 
 33

log 2  + 1 − log 2  − 4  =
3
 7

 7

 40 5 
log 2  ÷  =
3
 7 7
log 2 ( 8 ) = 3
log 2 ( 23 ) = 3
3log 2 (2) = 3
3=3
∴m =
33
is a solution.
7
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d
log12 (x) + log12 (x − 1) = 1
log12 ( x ) + log12 ( x − 1) =
log12 (12 )
log12 ( x ( x − 1) ) =
log12 (12 )
x2 − x =
12
x 2 − x − 12 =
0
0
( x − 4 )( x + 3) =
x = 4 or x = −3
Check: Let x = −3
log12 ( −3) + log12 ( −3 –1) = 1
log12 ( −3) + log12 ( −4 ) = 3
Log of negative undefined ∴ x = −3 is not a solution.
Check: Let x = 4
log12 ( 4 ) + log12 ( 4 –1) =
1
log12 ( 4 × 3) = 1
log12 = 1
1=1
∴x = 4 is a solution.
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Question 17
1.5 × 3.5
3 t TNT ⇔ M = 3.5 ∴=
101.5 M 10
=
105.25
1.5 × 9.1
x t TNT ⇔ M = 9.1 ∴=
101.5 M 10
=
1013.65
x
1013.65
= =
108.4
5.25
3000 10
=
x 3000 ×108.4 t
= 753 565 929.5 t
= 753.565 929 5 Mt
≈ 754 Mt
Question 18
 I 
L = 10 log  
 I0 
 100 
L = 10log  −12 
 10 
=
L 10 log (1012 ×102 )
L = 10 log (1014 )
L= 10 ×14 log (10 )
L = 140
The loudness of the rock concert measures 140 dB.
∴You will need ear plugs.
© Cengage Learning Australia Pty Ltd 2018
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Question 19
 r
FV = PV 1 + 
 n
nt
PV = 6500, r = 0.08, n = 12, t = 7
12×7
 0.08 
=
PV 6500 1 +

12 

PV = 6500 (1.00666...)
84
PV = 11358.24333...
PV ≈ 11358.24
 r
FV = PV 1 + 
 n
nt
PV = 6500, r = 0.08, n = 365, t = 7
 0.08 
=
PV 6500 1 +

365 

365×7
PV = 6500 (1.000219...)
2555
PV = 11378.67296...
PV ≈ 11378.67
Extra money = $11 378.67 − $11 358.24 = $20.43 ≈ $20.40
© Cengage Learning Australia Pty Ltd 2018
129
Question 20
8b = 42a + 3
(2 ) = (2 )
2 2 a +3
3 b
23b = 2 4 a + 6
3=
b 4a + 6 [1]
log2 (b) = log2 (a) + 3
log
=
log 2 ( a ) + 3log 2 ( 2 )
2 (b)
log
=
log 2 ( a ) + log 2 ( 23 )
2 (b)
log 2 ( b ) = log 2 ( 8a )
b = 8a
[2]
Substitute [2] into [1]
3 × 8a = 4a + 6
24=
a 4a + 6
20a = 6
a=
3
10
b =×
8
3 12
2
2
= =
10 5
5
© Cengage Learning Australia Pty Ltd 2018
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Question 21
y = log3 (9x + 20) – 1
20 

= log 3 9  x +  − 1
9 

20 

= log 3 ( 9 ) + log 3  x +  − 1
9 

20 

= log 3 ( 32 ) + log 3  x +  − 1
9 

20 

= 2 log 3 ( 3) + log 3  x +  − 1
9 

20 

=
2 + log 3  x +  − 1
9 

20 

= log 3  x +  + 1
9 

∴Vertical translation of 1 unit up and horizontal translation of
20
units to the left.
9
Question 22
m=
1
+ 2 log 9 ( x )
2
2m = 1 + 4 log 9 ( x )
2m – 1 = 4 log 9 ( x )
2m − 1
= log 9 ( x )
4
9
2m − 1
4
x=9
=x
2m − 1
4
© Cengage Learning Australia Pty Ltd 2018
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Question 23
Let log x ( 81) = z
[1]
∴ xz =
81
m = log3 (3x2)
∴ 3m =
3x 2
3m −1 = x 2
1
m −1 2
1
2 2
(3 ) = ( x )
 m −1 


3 2  = x [ 2]
Substitute [2] into [1].
z
  m2−1  
3   = 81


3
 m −1 
z

 2 
= 34
z ( m − 1)
=4
2
8
z ( m − 1) =
z=
8
m −1
∴ log x ( 81) =
8
m −1
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