Yimin Math Centre
Year 11 Math Worked Solutions
Grade:
Date:
Score:
nt
re
Student Name:
Table of contents
1
6.1
Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
6.1.1
Base 10 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
6.1.2
The General Definition of a Logarithm . . . . . . . . . . . . . . . . . . . . .
1
6.1.3
Change of Base Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
6.1.4
The Remaining Log Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
6.1.5
Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
M
at
h
Ce
Year 11 Topic 6 — Sequences and Series Part 2 (Worked Solutions)
Yi
m
in
6
This edition was printed on February 7, 2017.
Camera ready copy was prepared with the LATEX2e typesetting system.
Copyright © 2000 - 2017 Yimin Math Centre
Year 11 Worked Solutions
Year 11 Topic 6 Worked Solutions
6
Page 1 of 8
Year 11 Topic 6 — Sequences and Series Part 2 (Worked Solutions)
6.1
6.1.1
Logarithms
Base 10 Logarithms
• If a positive number y is written as a power of 10 in the form y = 10x , the index
part x is called the logarithm of the number y.
– This is written as x = log10 y.
– Thus y = 10x is equivalent to x = log10 y.
• Logarithms based on the power of 10 are often called "common logarithms".
The General Definition of a Logarithm
nt
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6.1.2
– This is written as x = loga y.
Ce
• If a number y is expressed in the form y = ax , then the index part of x is called
the logarithm of y to the base a
at
Exercise 6.1.1 Find the value of the following:
h
– Thus y = ax is equivalent to x = loga y.
M
1. log2 32
Solution:
in
Yi
m
2. log√5 125
log2 25 = 5 log2 2 = 5.
Solution:
√
√
log√5 53 = log√5 ( 5)6 = 6 log√5 5 = 6
3. log4 64
Solution:
4. log2
log4 43 = 3 log4 4 = 3.
1
4
Solution:
log2 2−2 = (−2) log2 2 = −2.
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Year 11 Topic 6 Worked Solutions
6.1.3
Page 2 of 8
Change of Base Law
• The calculator only evaluates common logarithms.
• This law allows us to change a logarithm from one base to another (i.e. base 10).
log
x
• loga x = log10 a
10
Exercise 6.1.2 Use the change of base law to evaluate the following (correct to 3 decimal places):
1. log4 9
log4 9 =
log10 9
= 1.585.
log10 4
log5 60 =
log10 60
= 2.544.
log10 5
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Solution:
Solution:
h
3. log9 56
Ce
2. log5 60
log10 56
= 1.832.
log10 9
at
Solution:
M
log9 56 =
Yi
m
Solution:
in
4. log5 20
log5 20 =
log10 20
= 1.861.
log10 5
log2 10 =
log10 10
= 3.322.
log10 2
log3 40 =
log10 40
= 3.578.
log10 3
5. log2 10
Solution:
6. log3 40
Solution:
7. log2
1
2
Solution:
log2
1
log10 0.5
=
= −1.
2
log10 2
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Year 11 Topic 6 Worked Solutions
6.1.4
Page 3 of 8
The Remaining Log Laws
• loga (xy) = loga x + loga y
• aloga x = x
• loga ( xy ) = loga x − loga y
• loga 1 = 0
• loga xn = n loga x
• loga a = 1
Exercise 6.1.3
Solution:
1
8
Solution:
in
Yi
m
Solution:
M
log6 (4 × 9) = log6 36 = log6 62 = 2.
4. log3 8 − log3 24
log3
8
1
= log3 = log3 3−1 = −1.
24
3
1
2
Solution:
√
= log2 2 = 1.
at
Solution:
5. log2 32 + log2
1
4
1
8
h
log2
3. log6 4 + log6 9
6. log4
10
= log5 5 = 1.
2
Ce
2. log2 14 − log2
log5
nt
re
1. log5 10 − log5 2
1
log2 (32 × ) = log2 16 = log2 24 = 4.
2
16 − log4 64
Solution:
log4 4 − log4 64 = log4
4
1
= log4
= log4 4−2 = −2.
64
16
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Year 11 Topic 6 Worked Solutions
Page 4 of 8
Exercise 6.1.4 Solve for x in the following:
1. loga x = loga 3 + loga 5
Solution:
loga x = loga 3 + loga 5 ⇒ loga x = loga 15 ⇒ ∴ x = 15.
2. loga x + loga 4 = log2 0
Ce
nt
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Solution: log x + log 4 = log 20 ⇒ log 4x = log 20 ⇒ 4x = 20 ⇒ ∴ x = 5.
a
a
a
a
a
3. loga 6 − loga x = loga 12
6
6
1
= loga x, ⇒
= 12 ⇒ ∴ x = .
x
x
2
h
loga 6 − loga x = log, 12, ⇒ loga
M
at
Solution:
x+1
x+1
= loga x, ⇒ x =
5
5
1
⇒ 5x = x + 1 ⇒ 4x = 1, ⇒ ∴ x = .
4
loga (x + 1) − loga 5 = loga x ⇒ loga
Yi
m
Solution:
in
4. loga (x + 1) − loga 5 = loga x
5. logx
1
3
= − 13
Solution:
1
x− 3 =
√
3
1
1
1
⇒ √
=
3
3
3
x
x = 3 ⇒ x = 33 ⇒ ∴ x = 27.
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Year 11 Topic 6 Worked Solutions
6.1.5
Page 5 of 8
Miscellaneous Problems
Exercise 6.1.5 Simplify the following:
log x10
1. loga x
a
loga x10
10
loga
x
=
= 10.
loga x
log
x
a
Solution:
loga 2x3 −loga x2
loga 2x
loga 2x3 − loga x2
loga (2x3 ÷ x2 )
=
loga 2x
loga 2x
loga 2x
=
= 1.
loga 2x
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2.
Ce
Solution:
3
h
at
Solution:
3
loga ( xy ) + loga ( xy )
loga x3 − loga y + loga y − loga x
√
=
1
loga x
2 loga x
M
3.
loga ( xy )+loga ( xy )
√
loga x
loga x3 − loga x
1
2 loga x
2
loga
x
= 1
= 4.
loga x
2
Yi
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in
=
4.
loga 4x+loga ( x2 )
loga 2
Solution:
loga 4x + loga ( x2 )
loga 4 + loga x + loga 2 − loga x
=
loga 2
loga 2
loga 4 + loga 2
=
loga 2
3
loga
2
=
= 3.
loga
2
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Year 11 Topic 6 Worked Solutions
Page 6 of 8
Exercise 6.1.6 Solve the following equations:
1. log3 (x + 2) + 2 log3 9 = 3
Solution:
log3 (x + 2) + 2 log3 9 = 3 ⇒ log3 (x + 2) + log3 92 = 3,
log3 ((x + 2) × 81) = 3, ⇒ 33 = 81(x + 2)
1
2
(x + 2) =
⇒ ∴ x = −1 .
3
3
2. loga (x + 2) = 13 loga 8 + loga 3
1
loga 8 + loga 3 ⇒ loga (x + 2) = loga 2 + loga 3
3
loga (x + 2) = loga 6 ⇒ x + 2 = 6 ⇒ ∴ x = 4.
nt
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loga (x + 2) =
Ce
Solution:
3. 3 loga 5x + 12 loga 25 = loga 5000
1
loga 25 = loga 5000 ⇒ loga (5x)3 + loga 5 = loga 5000
2
loga (5 × 5x3 ) = loga 5000 ⇒ 125x3 = 5000
h
3 loga 5x +
M
x3 = 8 ⇒ ∴ x = 2
at
Solution:
2
log2 x + log2 (x − 4) − log2 (x + 2) = 3 ⇒ log2
x(x − 2)
(x
+2)
log2
= 3 ⇒ log2 x(x − 2) = 3
(x
+2)
Yi
m
Solution:
in
4. log2 x + log2 (x2 − 4) − log2 (x + 2) = 3
x(x2 − 4)
(x + 2)
=3
23 = x(x − 2) ⇒ 8 = x2 − 2x ⇒ x2 − 2x − 8 = 0
(x − 4)(x + 2) ∴ x = 4 or x = −2.
log 9
5. log2 3 = log2 2x
2
Solution:
log2 9
2
log2
3
= log2 2x ⇒
= log2 2x
log2 3
log2 3
⇒ 2 = log2 2x, ⇒ 22 = 2x
⇒ 2x = 4 ⇒ ∴ x = 2.
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Year 11 Topic 6 Worked Solutions
Page 7 of 8
Exercise 6.1.7 Solve each pair of simultaneous equations by converting both sides to a common
base:
1. 22x−y = 32
24x+y = 128
Solution:

22x−y = 32
24x+y = 128

22x−y = 25 ⇒ 2x − y = 5 . . . (1)
⇒
24x+y = 27 ⇒ 4x + y = 7 . . . (2)

x = 2
(1) + (2) ⇒ 6x = 12, ⇒ ∴ x = 2 ⇒ y = 2x − 5 ⇒ y = −1, ∴
y = −1.

5x+y = 1
5
53x+2y = 1
Ce

5x+y = 5−1 ⇒ x + y = −1 . . . (1)
⇒
53x+2y = 50 ⇒ 3x + 2y = 0 . . . (2)

x = 2
(1) × 2 − (3) x = 2, y = −1 − (2) = −3 ∴
y = −3.
3. 3x+y = 81
81x−y = 3

3x+y = 81
81x−y = 3

3x+y = 34 ⇒ x + y = 4 . . . (1)
⇒
34(x−y) = 31 ⇒ 4(x − y) = 1 . . . (2)
Yi
m
in
Solution:
M
at
h
Solution:
nt
re
2. 5x+y = 15
53x+2y = 1
From (2) x − y =
1
. . .(3)
4

x =
17
15
17
(1) + (3) 2x =
⇒ x=
⇒ y =4−x=
⇒∴
y =
4
8
8
17
8
15
8
4. 7x+y = 49
49x−y = 7
Solution:

7x+y = 49
49x−y = 7

x + y = 2 . . . (1)
⇒
2(x − y) = 1
(1) × 2 ⇒ 2x + 2y = 4 . . . (3) ⇒ (3) + (2) ⇒ 4x = 5

x = 5
5
3
4
x = , ⇒ y = 2 − x ⇒ y = ⇒∴
y = 3 .
4
4
4
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Year 11 Topic 6 Worked Solutions
Page 8 of 8
Exercise 6.1.8 Practical Exam Questions
1. If log64 a = 13 , find the value of a.
Solution:
log64 a =
1
1
⇒ 64 3 = a ∴ a = 4.
3
2. Simplify (log6 4 + log6 9)2 .
Solution:
(log6 4 + log6 9)2 = [log6 (4 × 9)]2
= (log6 36)2
nt
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= (log6 62 )2 = 4.
Ce
3. Solve log(x − 2) + log(x − 1) = log(x + 2).
Solution: log(x − 2) + log(x − 1) = log(x + 2) ⇒ log[(x − 2)(x − 1)] = log(x + 2)
⇒ (x − 2)(x − 1) = (x + 2) ⇒ x2 − 3x + 2 = x + 2 ⇒ x2 − 4x = 0
M
at
h
x(x − 4) = 0 ∴ x = 4, but x = 0 gives a log of a negative number.
4. Simplify log4 9 + log4 8 − 2 log4 6.
9×8
62
= log4 2
in
log4 9 + log4 8 − 2 log4 6 = log4
Yi
m
Solution:
1
1
= log4 4 2 = .
2
5. If f (x) = log( 1+x
), prove that f (u) = 2f (x), where u =
1−x
Solution:
f (u) = log
1+u
1−u
=
2x
.
1+x2
2x
1+x2
log
2x
1 − 1+x
2
2
1+x +
1+
!
2x
= log
1 + x2 − 2x
1+x 2
= log
1−x
1+x
= 2 log
= 2f (x).
1−x
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