Chapter 2

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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=

=

(b)

=

=

=

(c)

(d) = =

(e)

=

=

=

=

(f)

=

=

=

=

(g)

=

=

=

(h)

=

=

=

4 (a)

(b)

(c)

(d) (rejected)

(e)

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(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 :

(6) (a) Given that

(1) : (2) : ...(4)

Substitute b =1 into (2) :

6 (b)

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(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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(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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(c)

=

=

=

=

=

(d)

=

=

=

=

5 (a)

(b)

6 (a)

=

= Note:

(b)

=

=

= =

(c)

=

= = 1 Note:

(d)

=

=

=

7 (a) = =

=

(b) = = = ( a multiple of 19)

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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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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)

:


3.

4 (a)

(Let u = )

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(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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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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4. (a)

5 (a) = 1 - 3 = -2

(b) = 0 +4 = 4

(c)

= = 8

(d)

=

=

(e) = = 0

(f) = = = 1

6.

7.

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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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(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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(c)

(d)

(e)

(f)

3. (a)

(b)

(c)

4. (a)

(b)

(c)

5. (a)

(b)

(c)

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(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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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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(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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(1) × 3: 3p – 9q = 6 ... (3)

(3) – (2):


and

4 (a)

(b)

5 (a)

(b)

(c)

(d)

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(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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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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3.

4. (a)

(b)

(c)

5. (a)

(b)

(c)

(d)

100 =

(to 3 sig. fig.)

6. (a)

(b)

(c)

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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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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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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)


When ,

When b = 3, a = 4

4. (a)

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(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


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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Substitute into (1): and

(b) (i)

(b) (ii)

10. (a)

(b)


When x = 4, y = 2.

11. (a)

(b)

(c)

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(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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a = 3b – 1 … (2) Substitute (2) into (1):

2(3b – 1) + 3b = 5 9b = 7


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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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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Chapter 2

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