chapter 2
TRANSCRIPT
Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
Exercise 2.1
1. (a) 2 +5 = =
(b) = = = 7
(c) = = =
(d) = =
(e) = = =
(f) = = =
(b)
=
=
= =
(c)
=
=
= = 30 + 12
(d)
=
=
=
=
(e)
=
=
=
=
(f)
=
=
=
=
3. (a)
=
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
=
=
(b)
=
=
=
(c)
(d) = =
(e)
=
=
=
=
(f)
=
=
=
=
(g)
=
=
=
(h)
=
=
=
4 (a)
(b)
(c)
(d) (rejected)
(e)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(f)
5(a) Given that
… (1) … (2)
From (1) :
Substitute into (2) :
(b) Given that
From (2) :
Substitute a = -5 into (1) :
(c) Given that
From (1): a =
Substitute into (2) :
Substitute into :
(6) (a) Given that
(1) : (2) : ...(4)
Substitute b =1 into (2) :
6 (b)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(2) : (1) + (3) :
Substitute b = 2 into (2) :
7 (a) Total length
(b) Volume of the box
8. (a) A =
(b)
9.
10.
If is rational, we can write as ,
where m and n are integers and the HCF of m and n is 1.
Hence, ...(1)
Note that
Let m = 2k where k is an integer. Substitute m = 2k into (1):
Note that
Let where is an integer.
Then which has a common
factor 2. This contradicts the fact that there is no common divisor between m and n greater than 1. Thus, the assumption that is rational is false. In other words, is irrational.
Exercise 2.2
1. (a)
(b)
(c)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(d)
(e)
=
=
= 5
(f)
=
=
=
=
2 (a) Substitute x = 9 into y = 2x :
(b) substitute y = 16 into :
3 (a)
=
=
(b)
=
= = = 2
(c)
4 (a)
=
= =
(b)
=
=
=
=
=
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(c)
=
=
=
=
=
(d)
=
=
=
=
5 (a)
(b)
6 (a)
=
= Note:
(b)
=
=
= =
(c)
=
= = 1 Note:
(d)
=
=
=
7 (a) = =
=
(b) = = = ( a multiple of 19)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
8.
=
=
=
=
=
=
9. LHS =
=
= = = RHS (shown)
Exercise 2.3
1 (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
2 (a)
From (1) : 5
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
From (2) : 3 …….(4)
Substitute (3) into (4) :
…..(5)
Substitute (5) into (3):
(b)
From (1) :
From (2) :
(3) : (4) :
(6) - (5) :
Substitute into (3):
(c)
From (1) :
From (2):
(3) (2) : ….. (5)
:
Substitute into (3) :
3.
4 (a)
(Let u = )
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(b)
5 (a)
(b)
(c)
(d)
(e)
(f)
6.
(no solution) or
The equation is satisfied by only one value of x.
7.
=
= =
=
…..(2)
From (2) :
Substitute into (1) : =
=
8.
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
9.
10. Use substitution: Let . (rejected)
Alternatively, square both sides of the equation :
=0 (rejected)
Exercise 2.4
2. (a)
(b)
(c) (d) (e) (f)
3. (a) when x =2 , 5 - 2x =1 0
is defined at x =2 .
(b) when x = 0.5, 5 - 2x = 4 0 is defined at x= 0.5
(c) when x = 3 , 5 - 2x = -1 0 is not defined at
x = 3 .
(d) when x = 2.5 , 5 - 2x = 0
is not defined at x = 2.5.
(e)
When x = 1, log 1 = 0. As the denominator can not be zero, is not defined at x = 1.
(f) when x = , 5 - 2x = 5 - 0 is defined at
x = .
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
4. (a)
5 (a) = 1 - 3 = -2
(b) = 0 +4 = 4
(c)
= = 8
(d)
=
=
(e) = = 0
(f) = = = 1
6.
7.
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
8. (a)
(b) ….(1) ….(2)Substitute (1) into (2): Substitute into (1) :
9. 10.
Exercise 2.5
1. (a) (b) (c) (d)
(e)
(f)
2. (a) x = ln 4 (b) (c) (d) 2x = ln k (e) (f) m + 5 = ln (x – 2)
3. (a) (b) ln 3x = 1 – y y = 1 – ln 3x
(c) ln (y + 1) = x
(d)
y =
(e) 2y = ln (x – 4)
(f) ln (x+ y) – 4x = 0 4 (a) ( to 3 sig. fig. )
(b) (ln x)2 = 3 = 5.65 or 0.177 ( to 3 sig. fig)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(c) ln x= lg 2 x = elg 2
x = 1.35 (to 3 sig. fig)
(d)
(e) ln 2. ln 4x = 3
(f)
(g) ln 4x = lg 3. lg 5 4x = elg3.lg5
(h) lg (x – 1) = ln (e2 – 1)
Exercise 2.6
1. (a)
(e)
(f)
2. (a)
(b)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(c)
(d)
(e)
(f)
3. (a)
(b)
(c)
4. (a)
(b)
(c)
5. (a)
(b)
(c)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(d)
(e)
(f)
6. (a)
(b)
(c)
(d)
7.
Substitute x = 2, y = 3 into the equation:
8 (a) x3 = e6x – 1
3 ln x = (6x – 1)ln e
(b) xe-x = 2.46 ln x + ln e-x = ln 2.46 ln x – x = ln 2.46 ln x x+ 0.9
(c) (xex)2 = 30e-x
2 ln(xex) = ln 30 + ln e-x
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
2 ln x + 2 ln ex = ln 30 – x 2 ln x + 2x = ln 30 – x
2 ln x = ln 30 – 3x
9 (a)
(b)
10 (a)
(b)
(c)
(d)
(e)
(f)
11. (a)
(b)
Exercise 2.7
1. (a)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(b)
(c)
(d)
(e)
(f)
(g)
(h)
2. ln(3x – y) =2 ln 6 – ln 9
ln(3x – y) = ln 4 3x – y = 4 … (1)
(1) – (2): x = 3Substitute x = 3 into (2): y = 5
x = 3 and y = 5
3.
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(1) × 3: 3p – 9q = 6 ... (3)
(3) – (2):
Substitute into (1):
and
4 (a)
(b)
5 (a)
(b)
(c)
(d)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(e)
6 (a)
(b)
(c)
(d)
7.
Note: Since a is the base, a > 0 and a 1. The logarithms are defined for both values of x.
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
8.
For the logarithms to be defined, x = 40.
Exercise 2.8
1. (a)
(b)
(c)
(d)
(e) ex = 7 x = ln 7 = 1.95 (to 3 sig. fig)
(f) e3x = 14 3x = ln 14
= 0.880 (to 3 sig. fig)
(g) 4e2x = 21 ln 4 + 2x = ln 21
= 0.829 (to 3 sig. fig)
(h) e4x – 125 = 0 e4x = 125 4x = ln 125 x = 1.21 (to 3 sig. fig) (i)
2.
(a) When x = 3,
(b) When y = 12,
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
3.
4. (a)
(b)
(c)
5. (a)
(b)
(c)
(d)
100 =
(to 3 sig. fig.)
6. (a)
(b)
(c)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
7. (a)
(b)
8.
Substitute (1) into :
When x = 0.631, y = 1.631.
Exercise 2.9
1. (a) When x = 0, Its initial temperature is .
(b) When x = 15,
(c) When T = 30,
2. (a) When t = 0,
There are 100 fruits flies at the beginning of the experiment.
(b) When t = 4,
The population is 741.
(c) When N = 400,
3. (a) When t = 10 000,
(b) When R = 50,
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
It will take 1620 years.
4. (a) When T = 1.5 mm,
The intensity is 0.811 when the thickness is 1.5 mm.
(b) When I = 0.5,
The thickness of the material needed is 4.98 mm.
5. (a) When n = 5,
The population at the beginning of 2005 is 34300.
(b)
1990 + 28.7 = 2018.7The population would be expected to have first increased to 90000 in year 2018.
6. (a) When t = 5,
The amount would be $6083 at the end of 5 years.
(b)
At the end of 12 years the amount would first exceed $8000.
7. (a) When t = 4,
The average score after 4 months was 65.3.
(b) When s = 55,
The average score was 55 after 27 months.
8. (a) When t = 4,
In this case, overestimate is better than underestimate. So we take y = 30. i.e. 30 students are infected after 4 days.
(b) When y = 1500 × 0.4 = 600,
After 9 days the school will cancel classes.
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
9. When T = 28.8,
2.5 hours before 7.00 am is 4.30 am. The estimated time of death of the man is 4.30 am.
Miscellaneous Exercise 2
1. (a)
(b)
2. (a)
(b)
3.
Given
From (1): a = 15b – 41 ... (3)
Substitute (3) into (2):
When ,
When b = 3, a = 4
4. (a)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(b)
(c)
(d)
(e)
(f)
5. (a)
Let h cm be the height of triangle.
(b)
AB = 10 cm
Perimeter of triangle
6. Substitute (3,4) into :
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h cm
cm
A
B C
Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
Substitute (9, 220) into :
(2) (1):
Substitute n = 2 into (1): a = 3
Substitute (-1, k) into :
7. (a)
(b)
(c)
(d)
8. (a)
(b)
(c)
9. (a)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
Substitute (1) into (2):
Substitute into (1): and
(b) (i)
(b) (ii)
10. (a)
(b)
Substitute (1) into (2):
When x = 4, y = 2.
11. (a)
(b)
(c)
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
(d)
(e)
(f)
(g)
(h)
(i)
(j)
12.
When x2y = 32, 22a+3b = 25
2a + 3b = 5 … (1)
When ,
a – 3b = -1
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
a = 3b – 1 … (2) Substitute (2) into (1):
2(3b – 1) + 3b = 5 9b = 7
Substitute into (2):
Thus, .
13. (a) log2 xy = log2 x + log2 y
log2 xy = log2 x + 2q
(b)
(c) logx 4y = logx 4 + logx y
(d) x = p2 and y = q4
x2y = (pq)4
14.
15. From (1) and (2),
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Panpac Additional Math : Worked Solutions Chapter 2 Surds, Indices and Logarithms
16(a)
(b)(i) T = 80(0.95)6
= 58.80 C
(ii)
(c) ex+2 = e6-3x
x + 2 = 6 – 3x x = 1 y = e3 = 20.1 Coords of P = (1, 20.1)
17(a)(i) m = 32e-0.4
= 21.5 (ii) t = 0, m = 32 Half of its value = 16 16 = 32e-0.02t
t = 34.7
18(a) 22+p – 21+ p = 9(2) – 2 4(2p) – 2(2p) = 16 2p(2) = 16 2p = 23
p = 3
(b) 22x+3 – 2x+3 = 9(2x) – 2 (2x)2(8) – 8(2x) = 9(2x) – 2 Let 2x = y.
8y2 – 17y + 2 = 0 (8y – 1)(y – 2) = 0
y = 2 or y =
x = -3 or 1
19(a) lg (3x – 24-x) = 2 + lg 2 – lg 2x
(b) (lg 5)2 + lg 2 [lg 5 + (lg 2 + lg 5)] = (lg 5)2 + 2lg 2 lg 5 + (lg 2)2
= (lg 5 + lg 2)2
= (lg 10)2
= 1
20(a)
(b) lg 3 lg 3x = lg 4 lg 4x (lg 3)2 + lg 3 lg x = (lg 4)2 + lg 4 lg x lg x(lg 3 – lg 4) = (lg 4)2 – (lg 3)2
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