4ach05(exponential And Logarithmic Functions)

5

Exponential and Logarithmic Functions

5 Exponential and Logarithmic Functions

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Activity

Activity 5.3 (p.238)

Activity 5.1 (p.216) 1.

1

2.

No

3.

y increases as x increases; the graph lies above the x-axis; it cuts the y-axis at (0, 1).

1.

1

2.

No

3.

y increases as x increases; the graph lies on the right-hand side of the y-axis; it cuts the x-axis at (1,0).

4. x y

4. x y

5.

–1

0 1

1 3

1 3

2 9

0.3 –0.5

0.6 –0.2

1 0

2 0.3

3 0.5

4 0.6

3 27

Yes 5.

Activity 5.2 (p. 222)

They are both defined for x > 0 and cut the x-axis at (1,0); y increases as x increases in both graphs; they both lie on the right-hand side of the y-axis.

1. M

N

1 10

100

logM logN logMN –1

2

1

M N

log 

–3

logM   –2

2

Follow-up Exercise p. 205

100 100 1000 1

2 3

2 –1

4 2

0 4

4 6

1

–2

–1

3

2

1 1000 –2 100

3

1

–5

–4

10 1 100

10

(3 x 3 ) 4 34 ( x 3 ) 4 = 3 2 3 2 3 (2 x ) 2 (x ) 1.

2.

log MN = log M + log N

3.

log 

81 x12 8x 6 81 x12 −6 = 8 81 x 6 = 8 =

M   = log M – log N N 

4.

log M2 = 2 log M

114

5

 a 2 b −3   a −2 b 

−1

[

   

= a 2−( −2 ) b −3−1

(

= a 4 b −4

2.

)

−1

1

]

−1

3

x

2.

=

∵ ∴

b4 a4

2.

∵ ∴

3.

∵ ∴

4.

5.

10 × 10 × 10 = 1000 103 = 1000 =10

1000

11 × 11 = 121 112 = 121

4 × 4 × 4 × 4 = 256 44 = 256

∴

− 256 =−4

∵

1 1 1 1 1 × × × = 3 3 3 3 81

4.

4

6.

∵

5

1 1 = 32 2

3

4

x

1

( x 4 )3 = x 3

1 1 1 1 1 1 × × × × = 2 2 2 2 2 32

5

 4 3× −    3

3

( x)

1 1 = 81 3

1 1   = 32 2 ∴

4 3

5 =  6 125 = 216

4

4

−

3

3

1 1    = 81 3 ∴

=[( −4) 3 ]

2 2  25  2  5     =     36   6  

=− 11

∵

4 3

5 3

= ( −4) −4 1 = ( −4) 4 1 = 256

3.

−216 =−6

− 121

−

−

= ( −4)

(–6) × (–6) × (–6) = –216 (–6)3 = –216 3

5

=x ( −64 )

3

1

(x5 ) 3 1 x3

p. 207 1.

1

=

5

= a −4 b 4 =

Exponential and Logarithmic Functions

5.

x4 = x 3 −1 =x4 −

1

=x 4 1 = 1 x4

p. 210

1.

( x) 3

4

1 = (x 3 )4 4

=x3

115

Certificate Mathematics in Action Full Solutions 4A 1 1 4

a •

1 5

a4

a4

=

a a

6.

=a

−

a

1.

11 20

2.

11 20

3x + 2 – 3x = 216 (32)(3x ) – 3x = 216 3x(32 – 1) = 216 8(3x) = 216 3x = 27 3x = 33 ∴ x=3

3 x +1 − 3 y = 0  (1)  x+y  ( 2) = 16 2 From (1), 3x + 1 = 3y ∴ x+1=y From (2), 2x + y = 24 ∴ x+y=4 Consider the simultaneous equations: x +1 = y  (3)

 x + y = 4  ( 4)

By substituting (3) into (4), we have x + (x +1) = 4 2x = 3

3 2

By substituting x =

116

0.01 = 10–2 2 log 0.01 = −

(b) ∵ ∴

100 000 = 105 log 100 000 = 5

(a) ∵ ∴

10x = 20 x = log 20 = 1.30 (cor. to 3 sig. fig.)

(b) ∵ ∴

10x = 0.8 x = log 0.8 0.0969 =− (cor. to 3 sig. fig.) 10x = 5 x = log 5 = 0.699 (cor. to 3 sig. fig.)

p. 221

52x – 24(5x) = 25 (5x)2 – 24(5x) – 25 = 0 Let y = 5x, the equation becomes y2 – 24y – 25 = 0 (y – 25)(y + 1) = 0 y = 25 or y = –1 ∴ 5x = 25 or 5x = –1 (rejected) 5x = 52 ∴ x= 2

x=

3 5 ,y= . 2 2

(a) ∵ ∴

(c) ∵ ∴

p. 213

3.

The solution is x =

p. 220

4 5

1

=

2.

∴

1 4

1 4 − 4 5

=a

1.

3 5 +1= 2 2

1

(a 4 ) 5 =

y=

3 into (3), we have 2

1.

∵ ∴

92 = 81 log9 81 = 2

2.

∵ ∴

43 = 64 log4 64 = 3

3.

∵ ∴

112 = 121 log11 121 = 2

4.

∵ ∴

25 = 32 log2 32 = 5

p. 225 1.

2 log

2 + log 5 = 2 log

= log (

2

1

22 1 2

+ log 5

)2 + log 5

= log 2 + log 5 = log (2 × 5) = log 10 =1 2.

log2 8 – log2 16 = log2 23 – log2 24 = 3 log2 2 – 4 log2 2 =3–4 1 =−

5

3.

log 16 log 2 4 = log 32 log 2 5 4 log 2 = 5 log 2 4 = 5

7.

1.

log x − log x

5.

log x 2 − log x log x 2 + log x

=

=

6.

=

=

=

3x – 1 = 7 log 3x – 1 = log 7 (x – 1)log 3 = log 7

log 7 log 3 log 7 x= +1 log 3

x–1=

= 2.77 (cor. to 3 sig. fig.) 2.

2 log x

log x 2 − log x

52x = 8 log 52x = log 8 2x log 5 = log 8

log 8 log 5 log 8 x= ÷ 2 log 5

1 2

2x =

log x − log x 2 log x = 1 log x − log x 2 2 log x =  1 1 −  log x 2  2 = 1 2 =4

=

54 6

p. 229

log 5 log 5 = log 25 log 5 2 1 log 5 = 2 2 log 5 1 = 4

2 log x

log 9 = log

= log 54 – log 6 = y −x

1 2

4.

Exponential and Logarithmic Functions

3.

= 0.646 (cor. to 3 sig. fig.) 6x = 3x + 2 log 6x = log 3x + 2 x log 6 = (x + 2) log 3 x log 6 = x log 3 + 2 log 3 x log 6 – x log 3 = 2 log 3 (log 6 – log 3) x = 2 log 3 x=

2 log 3 log 6 −log 3

= 3.17 4.

log (3x – 2) = 2 log (3x – 2) = log 100 3x – 2 = 100 3x = 102 x = 34

5.

log (3x + 1) – log (x – 2) = 1

1 2 1

log x 2 + log x 2 1 2 log x − log x 2 1 2 log x + log x 2 1   2 −  log x 2  1   2 +  log x 2  3 2 5 2 3 5

(cor. to 3 sig. fig.)

 3 x +1  log   = log 10  x −2  3 x +1 = 10 x −2 3x + 1 = 10(x – 2) 3x + 1 = 10x – 20 7x = 21 x=3 6.

Let y = log (x – 1), then the equation [log (x – 1)]2 + 2 log (x – 1) + 1 = 0 becomes y2 + 2y + 1 = 0 (y + 1)2 = 0 y = –1 ∴ log (x – 1) = –1 x–1=

1 10 117

Certificate Mathematics in Action Full Solutions 4A earthquakes measured 6 and 3 respectively. By the definition of the Richter scale, we have 6 = log E1 and 3 = log E2 i.e. E1 = 106 and E2 = 103

11 10

x=

p. 233 1.

E1 10 6 = E 2 10 3

∴

Let I1 and I2 be the sound intensities of the concert and that of the football match respectively. By the definition of sound intensity level, we have

 I1  I0

   

 I2 and 80 = 10 log  I  0

   

95 = 10 log  

∴

∴ 2.

 I1  I0

95 – 80 = 10 log  



 I  – 10 log  2  I   0

 I1  I0

 I  − log 2  I   0

15 = 10 log  



= 103 = 1000

 I1 3 = log  I 2  2

   

The strength of an earthquake measured 6 is 1000 times to that measured 3.

Let E1 and E2 be the relative energies released by the earthquakes in Nantou and Turkey respectively. By the definition of the Richter scale, we have 7.3 = log E1 and 6.3 = log E2 i.e. E1 = 107.3 and E2 = 106.3

E1 10 7.3 = E 2 10 6.3

∴

   

= 10 The strength of the earthquake in Nantou is 10 times to that in Turkey.

∴

   

I1 3 = 2 I 2 10 = 31.6 (cor. to 3 sig. fig.) ∴ 2.

The sound intensity of the concert is 31.6 times to that of the football match.

Let I be the original sound intensity produced by the radio, then the sound intensity produced after adjusting the volume is 1.25I. ∵ The original sound intensity level produced by the radio is 30 dB. ∴

 I 30 = 10 log  I  0

   

Exercise Exercise 5A (p.210) Level 1 1

7

1.

x3 = (x3 ) 7 3

= x7

Let β dB be the sound intensity level produced by the radio after adjusting the volume. ∴

1.25 I  I0

β = 10 log  

1.25 I β – 30 = 10 log   I 0 

   

2.

  I  – 10 log   I   0

  1.25 I β – 30 = 10 log      I0

  I  − log   I   0

β – 30 = 10 log 1.25 β = 30 + 10 log 1.25 = 31.0 (cor. to 3 sig. fig.) ∴

   

p.235 1.

Let E1 and E2 be the relative energies released by the

118

8

3

1

= ( x 3 )8 8

=x3

1

    

The sound intensity level produced by the radio after adjusting the volume is 31.0 dB.

( x)

( x) 3.

5

1

=

=

1

( x 2 )5 1 5

x2 =x

−

5 2

5

1 5

4.

x2

1

= =

(x2 ) 1

=x

 = [(m–4 – 3)(n –3 – 4)]–1   

= (m–7n –7)–1 7 7 =m n

2

−

−1

 m −4 n −3  m3n 4 

10. 

1 5

x5

11.

2 5

( m −4 n −2 )

3 2

 7    49    =     25   5   5.

Exponential and Logarithmic Functions

2

1 − 2

( a −1b 3 ) −3 ÷ ( a 4 b 2 )

−

1 2

= a −1×( −3) b 3×( −3) ÷ a

=

3

 2 3  3 4 a b   

2 =  3 8 = 27

3 =  4 27 = 64  27  −   64 

−

1 3

4

=a 3

3 2

13.

=a

1 3 − a 3b 4

1    27  3   = −   64    

 3 = −   4 4 =− 3

9.

1

2

1 3 2 + + 2b2 3

11 13 6 b 6

14.

3

−1

3

= a 3 b 2 •a 2 b 3

−2

 −1 1  ÷a 2 b 2  

    

−

3 4

1 3 − 3 =a b 4

−1

÷a

1 1 − ×( −2) ×( −2) 2 b2

÷ab −1

1 3 −1 − −( −1) 3 =a b 4

=a

−

2 1 3b 4

1

=

−1

 = (a–1+3 – 2)–1   

1

=a 3b

−1

 a −1 a 3   a2 

 1 2× −    2

2

1  3 3   3    =  −     4       

8.

b

2 1 2 3 1 2  ×2 ×2  2b 3 =a 3 b 4 2b3 • a • a   

4

7.

 1 4× −    2

a5 b8

3

3

÷m −1 n −3

= a3b–9 ÷ a–2b–1 = a3 – (–2)b–9 – (–1) = a5b–8

4 4  16  4  2     =     81    3   

2  9  2  3     =     16   4  

 1 −2× −    2

12.

3

6.

n

= m2n ÷ m–1n–3 = m2 – (–1)n1 – (–3) 3 4 =m n

3 2

7 =  5 343 = 125 3

 1 −4× −    2

÷m −1n −3 = m

b4 2

a3

15.

a2a p aq

=

a 2+ p q

(a )

1 2

=

a 2+ p a

q 2

=a

2+ p −

q 2

= (a0)–1 = a0 =1

119

Certificate Mathematics in Action Full Solutions 4A

a2a p



aq

∴

a

1

=1

q 2 +p − 2

x

−

2 3

•

x3

5

=a 0

q ∴ 2 + p − =0 2

   

−1

=

 a −x   ay 



( a − x ) −1 ax = = a x+ y (a y ) −1 a −y

2 3

x4

•

1

(x3 ) 5

=x =x

−

2 3

•

x

1 4 3 5

2 1 3 − + − 3 4 5 −

61 60

1

=

x

18.

61 60

−1

   

=a

3

3

p4

x + y =1 y=1–x When x = 2, y = 1 – 2 = –1. When x = 3, y = 1 – 3 = –2. A possible solution is x = 2, y = –1or x = 3, y = –2. (or any other reasonable answers)

1

p4

÷3 p = 2

(5 p ) 2

∴

∴

−

x

a x +y =a1

∴

=x

=x

q = 2(2 + p) when p = 2, q = 2(2 + 2) = 8. when p = 3, q = 2(2 + 3) = 10. ∴ A possible solution is p = 2, q = 8 or p = 3, q = 10. (or any other reasonable answers)

 a −x 16.   ay 

1

x4

÷( p2 )3

1

( p 5 )2 3

p4

=

÷p

2

2 3

p5 3 2 2 − − 5 3

= p4 Level 2

−

1

(3 x ) 2 3

• 4 x2 =

(x 3 )2 3

x4

1

• (x2 ) 4

x4 =

x x

=x

2 3 3 4

•x

2 3 1 − + 3 4 2

19. 1 2

s5 •

1 s

3

1

3

1

• s 5 = ( s 5 ) 2 • ( s 2 ) −1 • s 5 5

= s2 • s

20.

5

−

1 2

3

• s5

5 1 3 − + 2 5

= x 12

17.

19

= p 60 1 = 19 p 60

= s2

13

=s5

t

−

1 2

÷ 4 t3 •t

−

2 3

=t =t

21.

=t =t =

−

1 2

−

1 2

3

÷t 4 •t

1 3 2 − − +( − ) 2 4 3 −

23 12

1 23

t 12

120

1

÷ (t 3 ) 4 • t −

2 3

−

2 3

5 1 3 4

q •

(q 2 ) 4 9

q

3

1 3 4

=q •

3

1 3 9

(q ) 3

q

= q4 •

q =q

2 3

b2

q2

Exponential and Logarithmic Functions 1 3

b2

• (27 ab ) ÷ a b =

a

a

1 3

• (3 ab) ÷ (a b)

1 3

−

2 3

3

1 3

2 3

2 3

1 2

1 6

1 2

= a b •3 a b ÷a b

1 3

= 9a

2

1 2 1 − + − 3 3 6 1 6

3 1 1 + − 4 2 3

2

= 9a b

b

1 2

2 1 2+ − 3 2

13 6

11

= q 12

22.

1

1 2

1 1

1

1 4

2 3

1

2

= m n ÷ 2 −2 m

1 1 1 1 + +1 1− + 2 2 2 2

b

= m 4 n 3 ÷ 2 −2 m

= a 2b

23.

1

= 22 m 4  13 12 m n  

 12 13 m n  

−

 1   ÷  m  

1

a2  a 2 a b •   • ab 2 = a 2 b • 1 • ab 2 b b2 =a

−2

2   3 m n ÷ 2 m n     2 3

1 4

1 1 1 1 + −1 + −1   ÷mn = m 3 2 n 2 3  

=m

1 − 6

1 6

m n

−

1 6

17

2

n3

−

4 3

n −2

 1 ÷ 1   2 m

n −2 ÷ ( m

−( −2 )

−

1 2

−

     )

−

1 3

1 3

1

÷m 6

8

n3

8

27. = 4 m 12 n 3

1 − 6

1

= 24.

n

19

= 4m 12

4 −( − ) 3

4 − 3

1 3

28.

1 6

1

xy 4 ÷ 3 64 x 2 y ÷ x

−

1 3

2

1

1

y 5 = xy 4 ÷ ( 4 3 x 2 y ) 3 ÷ x

25.

2

1

 3 ab ÷ 16 ab •  a 4    4

−2

1

1

= (ab ) 2 ÷ (2 4 ab ) 4 • a 1 2

1 2

1 4

1 4

= a b ÷ 2a b • a 1 1 3 − +( − ) 4 2

= 2 −1 a 2 −

5

−

−

1

= xy 4 ÷ 4 x 3 y 3 ÷ x

3 2

= 4 −1 x

2 1 1− −( − ) 3 3

2

3 2

=

1 1 − 4

1 1 2 − − 3 5

y4

29

1 3 − 60 x y 4 2

b2

=

1

= 2 −1 a 4 b 4

x3 29

4 y 60

1

=

b4 5

2a 4 26.

Exercise 5B (p.214) Level 1 2

x

5 2 = 125 3 1.

x 2

5 = (5 ) 3

2 3

x

5 2 = 52 ∴

−

x =2 2 121

1 3

−

1 3 2

y5

2

y5

Certificate Mathematics in Action Full Solutions 4A ∴

x= 4 x 3

8 = 16

2.

x 3

(2 ) = 2 3

9.

4

2 x = 24 ∴

x= 4 x

62

3.

x

62 ∴

+1

10.

= 36

+1

= 62

x +1= 2 2

4x = 4 + x 4 x= 3

23x = 8(4x) 23x = 23(22)x 23x = 23(22x) 23x = 23 + 2x ∴ 3x = 3 + 2x x=3 9(3x + 1) = 272x 32(3x + 1) = (33)2x 3x + 3 = 36x ∴ x + 3 = 6x 3 x= 5

x= 2 x 3

4 = 32 4.

x 3

(2 2 ) = (2 5 ) 2 ∴

2x 3

=2

Level 2

2 5

11. 2 5

1 ) = 19 4 19 4x( ) = 19 4

4x(5 –

2

2x =2 3 x=3

5.

4x + 1 + 4x – 4x – 1 = 19 (4)4x + 4x – (4– 1)4x = 19 4x(4 + 1 – 4– 1) = 19

3x + 2 – 3x = 8 (32)(3x ) – 3x = 8 3x(32 – 1) = 8 8(3x) = 8 3x = 1 3x = 30 ∴ x=0

∴

12. 6x + 2 – 2(6x + 1) – 12(6x) = 2 (62)6x – 2(6)6x – 12(6x) = 2 6x(36 – 12 – 12) = 2 6x(12) = 2 6x = ∴

6.

x

x–1

3(3 ) – 3 = 216 3(3x) – 3x(3– 1) = 216 3x(3 –3–1) = 216 3x(3 –

1 ) = 216 3

8   = 216 3

3x 

∴ 7.

8.

3x = 81 3x = 34 x= 4

2x + 2 + 2x = 10 (22)(2x ) + 2x = 10 2x(22 + 1) = 10 5(2x) = 10 2x = 2 2x = 21 ∴ x =1 42x = 16(2x) (22)2x = 24(2x) 24x = 24 + x

122

4x = 4 x =1

1 6

6x = 6–1 1 x= −

5 13.

14.

22x – 3(2x) + 2 = 0 (2x)2 – 3(2x) + 2 = 0 Let y = 2x, the equation becomes y2 – 3y + 2 = 0 (y – 1)(y – 2) = 0 y = 1 or y = 2 ∴ 2x = 1 or 2x = 2 2x = 20 or 2x = 21 ∴ x = 0 or x = 1 32x – 12(3x) + 27 = 0 (3x)2 – 12(3x) + 27 = 0 Let y = 3x, the equation becomes y2 – 12y + 27 = 0 (y – 3)(y – 9) = 0 y = 3 or y = 9 ∴ 3x = 3 or 3x = 9 3x = 31 or 3x = 32 ∴ x = 1 or x = 2

15. 5(52x) – 26(5x) + 5 = 0 5(5x)2 – 26(5x) + 5 = 0 Let y = 5x, the equation becomes 5y2 – 26y + 5 = 0 (5y – 1)(y – 5) = 0 y=

16.

1 5

or

y=5

1 or 5x = 5 5

∴

5x =

∴

5x = 5–1 or 1 x= −

5x = 51 or x = 1

24x – 20(22x) + 64 = 0 (2 ) – 20(22x) + 64 = 0 Let y = 22x, the equation becomes y2 – 20y + 64 = 0 (y – 4)(y – 16) = 0 y=4 or y = 16 2x 2x ∴ 2 = 4 or 2 = 16 22x = 22 or 22x = 24 ∴ 2x = 2 or 2x = 4 x = 1 or x= 2 2x 2

5 x +2 y =125  (1) 2 x +y =1  ( 2) 25

17. 

From (1), 5x + 2y = 53 ∴ x + 2y = 3 From (2), 252x + y = 250 ∴ 2x + y = 0 y = –2x Consider the simultaneous equations: x + 2 y = 3  (3)

 y = −2 x

 ( 4)

By substituting (4) into (3), we have x + 2(–2x) = 3 –3x = 3 x = –1 By substituting x = –1 into (4), we have

∴

Exponential and Logarithmic Functions

y = –2(–1) = 2 The solution is x = –1, y = 2.

3 x +y = 27  (1) 3 x −y = 4  ( 2) 4

18. 

From (1), 3x + y = 33 ∴ x+y=3 From (2), 43x – y = 41 ∴ 3x – y = 1 Consider the simultaneous equations: x + y = 3  (3)

 3 x − y =1  ( 4)

(3) + (4), (x + y) + (3x – y) = 3 + 1 4x = 4 x=1 By substituting x = 1 into (3), we have 1+y=3 y=2 ∴ The solution is x = 1, y = 2.

3 2 x + y = 729  (1) x −2 y  ( 2) =3 9

19. 

From (1), 32x + y = 36 ∴ 2x + y = 6 From (2), (32)x – 2y = 3 32x – 4y = 31 ∴ 2x – 4y = 1 Consider the simultaneous equations: 2 x + y = 6  (3)

 2 x − 4 y =1  ( 4)

(3) – (4), (2x + y) – (2x – 4y) = 6 – 1 5y = 5 y=1 By substituting y = 1 into (3), we have 2x + 1 = 6 x= ∴

5 2

The solution is x =

5 , y = 1. 2

2 3 x −2 y = 4  (1) x+y = 625  ( 2) 5

20. 

From (1), 23x – 2y = 22 ∴ 3x – 2y = 2 From (2), 5x + y = 54 ∴ x+y=4 y=4–x Consider the simultaneous equations: 3 x − 2 y = 2  (3)

 y = 4 − x

 ( 4)

By substituting (4) into (3), we have

123

Certificate Mathematics in Action Full Solutions 4A 3x – 2(4 – x) = 2 5x = 10 x=2 By substituting x = 2 into (4), we have y=4–2=2 ∴ The solution is x = 2, y = 2.

Exercise 5C (p.226) Level 1 1.

∵ ∴

1 000 000 = 106 log 1 000 000 = 6

2.

∵ ∴

625 = 54 log5 625 = 4

3.

∵ ∴

4.

1  − = 2  16 

1 = 10–2 100

 1  =  100 

log 

1

log 2 5 2

− 2

− 2 log 2 5 1 log 2 5 2 = −4 =

10x = 120 x = log 120 = 2.08 (cor. to 3 sig. fig.)

6.

∵ ∴

10x = 88 x = log 88 = 1.94 (cor. to 3 sig. fig.)

7.

log 200 – log 2 = log

200 2

= log 100 = log 102 = 2 log 10 =2 log 4 + log 25 = log(4 × 25) = log 100 = log 102 = 2 log 10 =2

1  1 × 108)  + log3 108 = log3 ( 12 12  

log3 

= log3 9 = log3 32 = 2 log3 3 =2

124

log 2 5 − 2

=

12.

∵ ∴

9.

 1   1  log 2   log 2  2   25  = 5  1 log 2 5 log 2 5 2

log4 

5.

8.

log 64 log 4 3 = log 16 log 4 2 3 log 4 11. = 2 log 4 3 = 2

1 = 4–2 16

∵ ∴

log 9 log 32 = log 27 log 33 2 log 3 10. = 3 log 3 2 = 3

13. log 6 = log (2 × 3) = log 2 + log 3 = x +y 14. log

1

= log

18

= log ( = log

( 2 × 32 ) 2

2

2

1 2

1 2

× 3)

+ log 3

=

1 log 2+ log 3 2

=

1 x +y 2

15. Let loga 8 = x, logb 8 = y. Then ax = 8 and by = 8. ∵ loga 8 logb 8 = 1 ∴ xy = 1 x= Let a = 2, then 2x = 8 2x = 23 x=3

1 y

5

1 3

1  3 2  ×    9  2   

∴

y=

∴

b3 = 8

= log4 

1

1 9  ×  9 4 

= log4 

b = 83 = 512 Let a =

1  4

= log4 

1 , then 2

= log4 4–1 = –log4 4 1 =−

x

1   =8 2 x

1 1   =  2 2 x = −3 1 ∴ y =− 3 b

−

1 3

∴

= ∴

1

−3

log 8 + log 32 log 2 3 + log( 2 5 ) 2 = log 4 log 2 2 5

=8

b=8

20. −3

1 512

A possible solution is a = 2, b = 512 or a =

1 1 ,b= . (or any other reasonable 512 2

answers) 16.

Exponential and Logarithmic Functions

log a =3 log b

log 5 + 2 log 2 −1 log 5 + log 2 2 − log 10 = 2 log 4 − log 0.5 log 4 2 − log 0.5

log a = 3 log b log a = log b3 ∴ a = b3 ∴ A possible solution is a = 8, b = 2 or a = 27, b = 3. (or any other reasonable answers)

Level 2

21.

 2   2   = log 53 + log   25    25   3 2  = log  5 ×  25  

17. 3 log 5 + log 

= log 10 =1

100 10   18. log (100 10 ) – log (10 10 ) = log   10 10    = log 10 =1

3 log 2 + log 2 2 = 2 log 2 5 3 log 2 + log 2 2 = 2 log 2 11 log 2 2 = 2 log 2 11 = 4

 5 ×2 2 log   10  = 42 log 0 .5 log 2 = log 32 log 2 = log 2 5 log 2 = 5 log 2 1 = 5

   

22.

3  2   = log4   + log4 2  18 

19. log4 2 – log4 18 + 2 log4 

3   2

2

125

Certificate Mathematics in Action Full Solutions 4A

log 8 + 3 log 5 − 2 log 8 + log 5 3 − 2 log 10 = 2 log 5 − log 0.25 log 5 2 − log 0.25 log 8 + log 125 − log 100 = log 25 − log 0.25  8 ×125  log    100  = 25 log 0.25 log 10 = log 100 1 = log 10 2 1 = 2 log 10 1 = 2 log x 4 4 log x = 2 3 log x + 4 log x 3( 2 log x) + 4 log x 4 log x = 6 log x + 4 log x 4 log x 23. = (6 + 4) log x 4 = 10 2 = 5

 x3 y 4  log 4 log x 3 y 4 − log xy 4  xy = 1 log 3 x log x 3 log x 2 1 log x 3 2 log x = 1 log x 3 =6 =

25.

26.

2 log x 3 −3 log x 4 2(3 log x ) −3( 4 log x ) = 2 log x y −log xy  x2 y  log   xy     6 log x −12 log x = log x (6 −12 ) log x = log x = −6 27. log 200 = log (4 × 5 × 10) = log 4 + log 5 + log 10 = x +y +1

28. log

320

=

2 log 3 x + log 3 x 3 2 log 3 x + 3 log 3 x = 2 4 log 3 x − 3 log 3 x 4( 2 log 3 x ) − 3 log 3 x

=

( 2 + 3) log 3 x = 8 log 3 x − 3 log 3 x

24.

=

=

5 log 3 x (8 − 3) log 3 x

=

5 5 =1 =

=

= log

1 2 1 2 1 2 1 2

320

1 2

log 320 log (43 × 5) (log 43 + log 5) (3 log 4 + log 5)

1 (3 x + y ) 2

Exercise 5D (p.229) Level 1 1.

2x = 10 log 2x = log 10 x log 2 = 1 x=

1 log 2

= 3.32 (cor. to 3 sig. fig.) 2.

126

  

32x – 1 = 6

5

Exponential and Logarithmic Functions

log 32x – 1 = log 6 (2x – 1) log 3 = log 6

log 6 log 3 log 6 2x = +1 log 3

2x – 1 =

 log 6  +1  ÷2 log 3  

x= 

= 1.32 3.

(cor. to 3 sig. fig.)

41 – 3x = 5 log 41 – 3x = log 5 (1 – 3x) log 4 = log 5

log 5 log 4 log 5 3x = 1 – log 4

1 – 3x =



x= 1 −



log 5   ÷3 log 4  

0.0537 =−

4.

(cor. to 3 sig. fig.)

2(3x + 2) = 5 log [2(3x + 2)] = log 5 log 2 + log 3x + 2 = log 5 (x + 2) log 3 = log 5 – log 2

log 5 −log 2 log 3 log 5 −log 2 x= –2 log 3

x+2=

1.17 =−

5.

(cor. to 3 sig. fig.)

9x + 1 = 8x log 9x + 1 = log 8x (x + 1) log 9 = x log 8 x log 9 + log 9 = x log 8 (log 8 – log 9) x = log 9 x=

log 9 log 8 −log 9

18 .7 (cor. to 3 sig. fig.) =−

127

Certificate Mathematics in Action Full Solutions 4A 6.

42x = 63x – 1 log 42x = log 63x – 1 2x log 4 = (3x – 1) log 6 2x log 4 = 3x log 6 – log 6 (3 log 6 – 2 log 4) x = log 6 x=

8.

9.

x=5

14. log3 (4x + 9) – log3 (3x – 2) = 1

 4x + 9   = log3 3  3x − 2  4 x +9 =3 3x − 2

log3 

log 6 3 log 6 −2 log 4

= 0.688 7.

∴

(cor. to 3 sig. fig.)

4x + 9 = 9x – 6 5x = 15 x=3

log (6x – 4) = 2 log (6x – 4) = log 100 6x – 4 = 100 6x = 104 1 7 x =1 3

15. log (x – 1) + log (2x – 1) = 1 log [(x – 1)(2x – 1)] = log 10 log (2x2 – 3x + 1) = log 10 2x2 – 3x + 1 = 10 2x2 – 3x – 9 = 0 (x – 3)(2x + 3) = 0

log3 (4x – 1) = 3 log3 (4x – 1) = 3 log3 3 log3 (4x – 1) = log3 33 4x – 1 = 27 4x = 28 x=7

x=3 ∴

log (3x – 2) = 0 log (3x – 2) = log 1 3x – 2 = 1 3x = 3 x =1

3 (rejected) 2

x=3

x=

3 2

For y = 2, log2 (2x – 1) = 2 2x – 1 = 4 x=

Level 2 ∴

11. log (8x – 2) – log 3 = 1 log (8x – 2) = log 10 + log 3 log (8x – 2) = log (10 × 3) 8x – 2 = 30 8x = 32 x= 4

3 5 or 2 2

1 1000 999 x=− 1000

x+1=

1 +6 10 61 4x = 10 4x =

For y = 4, log (x + 1) = 4 x + 1 = 10 000 x = 9999 999 ∴ x=− or 9999 1000

61 40

13. log4 (x + 3) + log4 (x – 3) = 2 log4 [(x + 3)(x – 3)] = 2log4 4 log4 (x2 – 9) = log4 42 x2 – 9 = 16 x2 = 25 x=5

x=

5 2

17. Let y = log (x + 1), then the equation [log (x + 1)]2 – log (x + 1) – 12 = 0 becomes y2 – y – 12 = 0 (y + 3)(y – 4) = 0 y = –3 or y = 4 For y = –3, log (x + 1) = –3

12. log (2x – 3) + log 2 = –1 log [(2x – 3) × 2] = –log 10 log (4x – 6) = log 10–1 4x – 6 = 10–1

128

x=−

16. Let y = log2 (2x – 1), then the equation [log2 (2x – 1)]2 – 3 log2 (2x – 1) + 2 = 0 becomes y2 – 3y + 2 = 0 (y – 1)(y – 2) = 0 y = 1 or y = 2 For y = 1, log2 (2x – 1) = 1 2x – 1 = 2

10. log2 (3x + 1) = 4 log2 (3x + 1) = 4 log2 2 log2 (3x + 1) = log2 24 3x + 1 = 16 3x = 15 x=5

x=

or

or

x = –5 (rejected)

18. Let y = log (x – 1), then the equation [log (x – 1)]2 – 3 log (x – 1) + 2 = 0 becomes y2 – 3y + 2 = 0 (y – 1)(y – 2) = 0 y = 1 or y = 2 For y = 1, log (x – 1) = 1 x – 1 = 10

5 x = 11 For y = 2, log (x – 1) = 2 x – 1 = 100 x = 101 ∴ x = 11 or 101 19. x − 2 y = 32

 I2 3 = log  I 2  1

From (2), log x = log y + log 10 log x = log 10y x = 10y ……(3) By substituting (3) into (1), we have 10y – 2y = 32 8y = 32 y=4 By substituting y = 4 into (3), we have x = 10(4) = 40 ∴ The solution is x = 40, y = 4. 20. x − y =1

= 31.6 (cor. to 3 sig. fig.) ∴ 2.

Let I1 and I2 be the sound intensities after the party and during the party respectively. By the definition of sound intensity level, we have

 I1  I0

 I2    I  0

∴

 I1  I0

35 – 67 = 10 log  



 I  – 10 log  2  I   0

 I1  I  − log 2  I  I0   0

–32 = 10 log  



 I1 16 − = log  I 5  2

1 9

   

   

   

I1 16 − = 5 I 2 10

1 into (3), we have 9 1 10 x = 10( )= 9 9 10 1 The solution is x = ,y= . 9 9

By substituting y =

∴

   

and 67 = 10 log  

From (2), log y = log x – log 10 log x = log y + log 10 log x = log 10y x = 10y ……(3) By substituting (3) into (1), we have 10y – y = 1 9y = 1 y=

The sound intensity in restaurant B is 31.6 times to that in restaurant A.

35 = 10 log  

 (1)  ( 2)

 log y = log x −1

   

I2 3 = 2 10 I1  (1)  ( 2)

 log x = log y +1

Exponential and Logarithmic Functions

= 0.000 631 (cor. to 3 sig. fig.) ∴ 3.

The sound intensity after the party is 0.000 631 times to that during the party.

Let I1 and I2 be the sound intensities before adjustment and after adjustment respectively. Then the sound intensity levels before and after

 I1  I0

 I  dB and 10 log  2  I   0

  dB  

Exercise 5E (p.236)

adjustment are 10 log  

Level 1

respectively. ∵ The sound intensity level increases by 5 dB after the adjustment.

1.

Let I1 and I2 be the sound intensities in restaurants A and B respectively. By the definition of sound intensity level, we have

 I1  I0

60 = 10 log  

   



 I2     I0 



 I2  I0

75 – 60 = 10 log  



 I  – 10 log  1  I   0

 I2  I  − log 1  I  I0   0

15 = 10 log 



 I2  I  – 10 log  1  I  I0   0

5 = 10 log  

 I2  I0

5 = 10 log  

and 75 = 10 log   ∴

∴

   

   

 I2 1 = log  I 2  1

 I  − log 1  I   0

   

   

   

I2 1 = 2 I 1 10 = 3.16 (cor. to 3 sig. fig.)

129

Certificate Mathematics in Action Full Solutions 4A ∴ 4.

Let β dB be the sound intensity level of the Hi-Fi after adjustment.

The sound intensity after adjustment is 3.16 times to that before adjustment.

Let E1 and E2 be the relative energies released by the earthquakes measured 8 and 2 respectively. By the definition of the Richter scale, we have 8 = log E1 and 2 = log E2 i.e. E1 = 108 and E2 = 102 ∴

∴

 0.6 I  I0

β – 90 = 10 log  



E1 10 8 = E 2 10 2

5.

β – 90 = 10 log 0.6 β = 90 + 10 log 0.6 = 87.8 (cor. to 3 sig. fig.) ∴ 8.

 I  I0

∴

 I  I0

   

 2I  I0

   

30 = 10 log  

β = 10 log  

 2I  I0

β – 30 = 10 log  



  I  – 10 log   I   0

 2I   I  − log   I  I0   0

β – 30 = 10 log  



β – 30 = 10 log 2 β = 30 + 10 log 2 = 33.0 (cor. to 3 sig. fig.) ∴ 7.

   

   

130

  I  – 10 log   I   0

 1.1I   I  − log   I  I0   0

= 10 log  



∴ 9.

   

   

 I  I0

90 = 10 log  

   

The increase in the sound intensity level is 0.414 dB.

Let E be the relative energy released by an earthquake measured 5.8. By the definition of the Richter scale, we have 5.8 = log E E = 105.8 ∴ The relative energy released by an earthquake that is 20 times to that measured 5.8 = 20 × 105.8 ∴ The magnitude of the earthquake on the Richter scale = log (20 × 105.8) = 7.10 (cor. to 3 sig. fig.)

10. Let E be the relative energy released by an earthquake measured 7.2. By the definition of the Richter scale, we have 7.2 = log E E = 107.2 ∴ The relative energy released by an earthquake that is

1 times to that measured 7.2 8 1 = × 107.2 8

The sound intensity level is 33.0 dB when the master of ceremony speaks.

Let I be the original sound intensity of the Hi-Fi, then the sound intensity is 0.6I after adjustment. ∵ The original sound intensity level is 90 dB. ∴

1.1I  I0

10 log  

  dB respectively.  

= 10 log 1.1 = 0.414 (cor. to 3 sig. fig.)

Let β dB be the sound intensity level when the master of ceremony speaks. ∴

  1.1I  dB and 10 log    I   0



The strength of the earthquake in the Philippines is 20.0 times to that in Indonesia.

Let I be the sound intensity of the gathering when the master of ceremony does not speak, then the sound intensity is 2I when the master of ceremony speaks. ∵ The sound intensity level in the gathering is 30 dB. ∴

   

Let I be the original sound intensity of the noise, then the sound intensity is 1.1I after adjustment. The sound intensity levels before and after adjustment are

Level 2 6.

   

The sound intensity level of the Hi-Fi is 87.8 dB after adjustment.

10 log  

E1 10 7.3 = E 2 10 6.0 = 101.3 = 20.0 (cor. to 3 sig. fig.)

∴

  I  – 10 log   I   0

 0. 6 I   I  − log   I  I0   0



= 10 The strength of an earthquake measured 8 is 106 times to that measured 2.

Let E1 and E2 be the relative energies released by the earthquakes in the Philippines and Indonesia respectively. By the definition of the Richter scale, we have: For Philippines, For Indonesia, 7.3 = log E1 6.0 = log E2 E1 = 107.3 E2 = 106.0 ∴

   

β – 90 = 10 log  

6

∴

 0.6 I  I0

β = 10 log  

∴

The magnitude of the earthquake on the Richter scale = log (

1 × 107.2) 8

= 6.30

(cor. to 3 sig. fig.)

5

Exponential and Logarithmic Functions

3

2

−3

a b c

12

−3

= (a b c ) 2

12

1 3

2

3.

= a 3 b −1c 4

(a)

2

a 3 c4 = b Revision Exercise 5 (p.244) Level 1

6

( x ) = (x ) 7

5

1.

(a)

1 5

1

(b)

x

4

4

=

(x ) 1

−

 216     125 

2.

3

x2

=

−

7 6

1 7

x6

 3  6   =     5       6 2  =     5    

2 3

−1

( x −5 ) 3 3 4 x −15 • x = 3 • (x4 ) 3 x3 x 1

4

= x −18 • x 3

−1

    

(c)

=x

−1

=x =

−1

 36  =   25  25 = 36

−18+

−

4 3

50 3

1 50

x3 (d) 1

 27  −   343 

x3

=x

4 − 3

1 − 3

1 2

= x3

2    216  3  =    125    

(a)

3

(x )

1 3 − 2

1 3

4

2 3

1 6

(x )

(b)

x3 =x

=

=

1

=

x

3

2

1

7 5

=x

3

7

x

2

1    27 3   = −   343    

−1

1

1 1

=a 2 −1

(b)

1  3 3   3    =  −      7       −1

 3 = −   7 7 =− 3

1

a 2 b • ( ab 2 ) −1 ÷( 4 ab ) 2 = a 2 b • a −1b 2

=a =a

−

−1 1−

b

1 1 2b 2

1 2

×−1

1

÷[( ab ) 4 ]2 1

÷( ab ) 2

1 1 ÷a 2 b 2

1 1 1 1 − − − 2 2b 2 2

= a −1 1 = a

131

Certificate Mathematics in Action Full Solutions 4A

2 x = 3 16 4.

(b) ∵ ∴

1 4 3

(a)

= (2 ) =2 ∴

102x = 20 2x = log 20 x = 0.651

(cor. to 3 sig. fig.)

4 3

7.

4 x= 3

(a) log 400 – log 8 + log 2 = log

= log 50 + log 2 = log (50 × 2) = log 100 =2

4 x = 64 1

(b) [( 2 ) ] 2 = 2 2

x

(b) log 125 + log 80 = log (125 × 80) = log 10 000 = log 104 =4

6

1 2

(2 2 x ) = 2 6 2 x = 26 ∴

(c) log6 108 + log6 2 = log6 (108 × 2) = log6 216 = log6 63 =3

x=6

3 5 3 5(5x) – 2(5x) = 5 3 3(5x) = 5 1 5x = 5

(c) 5x + 1 – 2(5x) =

1  1    = log9 9 ×  81   81   1 = log9   9

(d) log9 9 + log9 

= log9 9–1 1 =−

5x = 5–1 1 x= − (d)

5.

2x + 2 – 2x = 24 22(2x) – 2x = 24 (22 – 1)2x = 24 3 × 2x = 24 2x = 8 2x = 23 x=3

(a) ∵ ∴

  4 = − 

0.000 000 1 = 10–7 7 log 0.000 000 1 = −

(c) ∵ ∴

64 = 82 log8 64 = 2

∴ (a) ∵ ∴

132

log 5

1 = 10–4 10 000

1 5

log 256 log 4 4 = log 16 log 4 2 4 log 4 = 2 log 4 =2

log 625

(b) ∵ ∴

(d) ∵

6.

(e)

 1 log  10 000 

400 + log 2 8

=5

−

1 2

(f)

8.

(a)

10x = 45 x = log 45 = 1.65 (cor. to 3 sig. fig.)

log 5 4 1

log 5 2 4 log 5 = 1 log 5 2 4 = 1 2 =8

5x = 10 log 5x = log 10 x log 5 = 1 x=

 1  1  log 5   5 =− 2  

=

1 log 5

= 1.43 (cor. to 3 sig. fig.) (b)

2x – 1 = 6 log 2x – 1 = log 6 (x – 1) log 2 = log 6

5 log log log x= log

x–1=

6 2 6 +1 2

 I1  I0

45 = 10 log  

= 3.58 (cor. to 3 sig. fig.) (c)

Exponential and Logarithmic Functions

32x + 3 = 33 log 32x + 3 = log 33 (2x + 3) log 3 = log 33

 I2     I0 

and 53 = 10 log   ∴

 I2  I0

53 – 45 = 10 log  

log 33 log 3 log 33 2x = –3 log 3



2x + 3 =

x = 0.0913

 I2 4 = log  I 5  1

(cor. to 3 sig. fig.)

11. Let I1 and I2 be the sound intensities of the cassette player before and after adjustment. ∵ The sound intensity level increases by 8 dB. ∴

 I2  I0

 I  – 10 log  1  I   0



 I2  I  − log 1  I  I0   0

10 log  

10 log  



 I2  I1

log  

(f) log2 (6x – 4) = 5 log2 (6x – 4) = log2 25 6x – 4 = 32 6x = 36 x=6

  = 8  

 4 =  5 

I2 = ∴

1 (5 x +2 y ) 2

10. Let I1 and I2 be the sound intensities of groups A and B respectively. By the definition of sound intensity level, we have

10

4 5

I1

The percentage increase of the sound intensity: =

1 2

1 = log (25 × 32) 2 1 = (log 25 + log 32) 2 1 = [5 log 2 + 2 log 3] 2 =

 =8  

I2 4 = 5 10 I1

(a) log 144 = log (24 × 32) = log 24 + log 32 = 4 log 2 + 2 log 3 = 4 x +2 y

288

   

The sound intensity of group B is 6.31 times to that of group A.

x = 2.2

= log

   

= 6.31 (cor. to 3 sig. fig.)

x  −1 = –1 2  x  1  log  −1 = log   2    10  1 x –1= 10 2 x = 1.1 2

288

   

I2 4 = 5 10 I1

log 

(b) log

 I2  I  − log  1  I  I0   0



∴

9.

 I  – 10 log  1  I   0

8 = 10 log 

(d) log (3x + 1) = 2 log (3x + 1) = log 100 3x + 1 = 100 3x = 99 x = 33 (e)

   

I 2 − I1 ×100% I1

= 10

4 5

I1 − I 1 × 100% I1

= 531 % (cor. to 3 sig. fig.) 12. Let E1 and E2 be the relative energies released by the earthquakes measured 7.5 and 6.5 respectively. By the definition of the Richter scale, we have 7.5 = log E1 and 6.5 = log E2 i.e. E1 = 107.5 and E2 = 106.5 ∴

E1 10 7.5 = E 2 10 6.5 = 10

133

Certificate Mathematics in Action Full Solutions 4A ∴

The strength of an earthquake measured 7.5 is 10 times to that measured 6.5.

13. Let E1 and E2 be the relative energies released by the earthquakes in Indonesia and Pakistan respectively. By the definition of the Richter scale, we have: For Indonesia, For Pakistan, 7.5 = log E1 4.6 = log E2 E1 = 107.5 E2 = 104.6 ∴

∴

p

−

3 5

•

6

p3

5

p2

1 −

=p

3 5

•

( p3 ) 6 1

( p2 )5 1 −

=p

3 5

p2

•

2

p5

(b)

E1 10 7.5 = E 2 10 4.6

−

=p

= 102.9 = 794 (cor. to 3 sig. fig.) The strength of the earthquake in Indonesia is 794 times to that in Pakistan.

3 5

•p

1 10

1 − 2

=p

1

=

1

14. (a) log (4x – 2) – log 2x = y

p2

 4x − 2  =y  2x  2 x −1 = 10y x

log 

4

x

2x – 1 = x10y x (2 – 10y) = 1

3 4

1

1

1 1 = =1 0 2 −10 2 −1 1 1 When y = 1, x = =− 8 2 −10 1 A possible solution is x = 1, y = 0 or x = − y = 1. (or any other reasonable answers)

1

−

1 2

2

3

3

1

2

3

= x4 y4 • y2 • x4 ÷ x3 y

(b) When y = 0, x =

∴

−

÷ x3 y

−

= ( xy 3 ) 4 • y 2 • x 4 ÷ x 3 y

1 2 −10 y

x=

2

y

xy 3 •

1 3 + 4 4

=x

y

3 1 + 4 2

2 3

÷x y

2 3

5 4

= xy ÷ x y 1 , 8

=x

1−

2 3

5

y4

1

16. (a)

−

−

1 2

1 2

−

1 2

1 2

1 −( − ) 2

7

= x3 y4

Level 2

(4 a ) 2 3

a

•a

1 − 3

=

1 4

(a ) 2 a

•a

1 3

1

= 15. (a)

a2 1

•a

−

3 4

a ab ÷ (3 a 2 b ) −1 •   b

1 − 3

= ab ÷ [( a 2 b) ]−1 •

1 3

1

3

−

−

2

1

−

1 2

−

1 2

−

1 2

1

1

= ab 4 ÷ ( a 3 b 3 ) −1 • a 2 b 2

1 3

3

−

2

= ab 4 ÷ a 3 b

1

=a 6 1 = 1 a6

=a

2 1−( − ) 3 5 3

=a b =a

17. (a)

13 12

5 1 − 3 2 7

(b)

134

a b

a3 = a6 •a

1 3

3 4

−

b

b

−

1 3

3 1 −( − ) 4 3 1 − 2

•a b

−

1

1

•a 2b2 −

1 2

•a b 1 2

13 1 + 12 2

19

= a 6 b 12 5x + 1 + 5x – 5x – 1 = 29 (5)5x + 5x – (5– 1)5x = 29

1 2

5 5x(5 + 1 – 5– 1) = 29

y = 2(

1 5x(6 – ) = 29 5 29 5x( ) = 29 5 5x = 5 x =1

∴

∴

(d)

18. (a)

 3 x + y =1  ( 4)

By substituting (3) into (4), we have 3x + x = 1 4x = 1 x=

 ( 4)

x=

3 5

By substituting x =

3 into (4), we have 5

1 4

1 1 into (3), we have y = 4 4 1 1 The solution is x = ,y= . 4 4

By substituting x = ∴

19. (a)

−

1 1 − log 625 – log 4 = log 2 – log 4 625 2

= log

1 625

= log

1 25 4

= log

1 100

– log 4

= log 10–2 2 =−

From (1), 4x + 2y = 43 ∴ x + 2y = 3 From (2), 162x – y = 160 ∴ 2x – y = 0 y = 2x Consider the simultaneous equations: x + 2 y = 3  (3) By substituting (4) into (3), we have x + 2(2x) = 3 5x = 3

3 6 ,y= . 5 5

From (1), 3x + y = (32)2x – y 3x + y = 34x – 2y ∴ x + y = 4x – 2y y=x From (2), 93x + y = 91 ∴ 3x + y = 1 Consider the simultaneous equations:  (3) y = x

4 x +2 y = 64  (1)  2 x−y =1  ( 2) 16

 y = 2 x

The solution is x =

3 x +y = 9 2 x −y  (1) 3 x +y  ( 2) =9 9

22x – 5(2x) + 4 = 0 (2x)2 – 5(2x) + 4 = 0 Let y = 2x, the equation becomes y2 – 5y + 4 = 0 (y – 1)(y – 4) = 0 y = 1 or y = 4 x ∴ 2 = 1 or 2x = 4 2x = 20 or 2x = 22 ∴ x = 0 or x = 2 42x – 10(4x) + 16 = 0 (4x)2 – 10(4x) + 16 = 0 Let y = 4x, the equation becomes y2 – 10y + 16 = 0 (y – 2)(y – 8) = 0 y = 2 or y = 8 x ∴ 4 = 2 or 4x = 8 22x = 21 or 22x = 23 2x = 1 or 2x = 3 1 3 ∴ x= or x = 2 2

3 6 )= 5 5

(b) 

(b) 3x + 3 – 2(3x + 2) + 3x = 30 (33)3x – 2(32)3x + 3x = 30 3x(27 – 18 + 1) = 30 3x(10) = 30 3x = 3 ∴ x =1 (c)

Exponential and Logarithmic Functions

(b)

32 log log 32 − log 64 64 = log 128 log 128 1 log 2 = log 128 log 2 −1 log 2 7 − log 2 = 7 log 2 1 =− 7 =

135

Certificate Mathematics in Action Full Solutions 4A

(c)

3 log 3 − log 243 log 33 − log 243 = log 9 log 9 27 log 243 = log 9 1 log 9 = log 9

(c)

log xy − log x y log 3 xy − log( x • 3 y )

1

1

1

1

1

1

log x 2 y 2 − log xy 2

     = 1 1  3 3 x y  log 1   xy 3   

1.5 log 16 − 2 log 2 1.5 log 2 4 − 2 log 2 = log 36 − 2 log 3 log 36 − log 3 2 1.5( 4 log 2) − 2 log 2 = log 36 − log 9 (6 − 2) log 2 = 36 log 9 4 log 2 = log 4 4 log 2 = log 2 2 4 log 2 = 2 log 2 =2

=

log x

− −

1 2 2

log x 3 1 − log x = 2 2 − log x 3 3 = 4  a 2b    log ab  log a 2 b − log ab  = log ab 2 − log a 2 b 2  ab 2  log  2 2  a b 

log a 4 + log b 8 log a 4 b 8 = log ab 2 log ab 2 (d)

4 log ab 2 log ab 2 =4 =

log a 1 log a log a = log a −1 log a = − log a = −1 =

21. (a) log (3x – 2) + log 2 = 1 log [(3x – 2) × 2] = log 10 (3x – 2) × 2 = 10 3x – 2 = 5 3x = 7 7 x= 3 (b) log 2x + 6 log 2 = log 96

136

1

 12 12 x y log 1  xy 2 

(d)

4 log x 2 + 2 log x 4 8 log x +8 log x = 3 log x 3 log x (8 +8) log x = 3 log x (b) 16 log x = 3 log x 16 = 3

1

log x 3 y 3 − log xy 3

log 9 −1 log 9 − log 9 = log 9 = −1

20. (a)

1

log( xy ) 3 − log xy 3 =

=

log( ab 2 ) 4 = log ab 2

=

1

log( xy ) 2 − log xy 2

5 log 2x + log 26 = log 96 log (2x × 26) = log 96 2x × 26 = 96 128x = 96 3 x= 4

=

∴

 4x − 2   = log3 3  x +1  4x −2 =3 x +1

(d) log5 (3x + 3) – log5 (x – 1) = 2 log5

log

……(4)

By substituting (4) into (3), we have

 2 +2y  2   y = 100  3  2y2 + 2y3 = 300 y + y2 – 150 = 0 (y – 5)(y2 + 6y + 30) = 0 3

y = 5 or y =

−6 ± 6 2 −4(1)( 30 ) 2(1)

……(3)

xy = log 10 2 xy = 10 2

xy = 20 ……(4) By substituting (3) into (4), we have

3x + 3 = 5(x – 1) 3x + 3 = 5x – 5 2x = 8 x= 4

From (1), log x + log y2 = log 100 log xy2 = log 100 xy2 = 100 ……(3) From (2), 3x = 2 + 2y

10 +5 y 6

From (2),

5

 (1)  ( 2)

 ( 2)

From (1), 6x – 5y = 10 6x = 10 + 5y x=

(e) Let y = log3 (2x + 3), then the equation [log3 (2x + 3)]2 – 4 log3 (2x + 3) + 3 = 0 becomes y2 – 4y + 3 = 0 (y – 1)(y – 3) = 0 y = 1 or y = 3 For y = 1, log3 (2x + 3) = 1 2x + 3 = 3 2x = 0 x=0 For y = 3, log3 (2x + 3) = 3 2x + 3 = 27 2x = 24 x = 12 ∴ x = 0 or 12

 (1)

 log xy −log 2 =1

1  3x + 3  log5   = log5 (5 2 ) 2  x −1  3 x +3 =5 x −1

2 +2 y 3

The solution is x = 4, y = 5.

(b) log( 6 x −5 y ) =1

4x – 2 = 3(x + 1) 4x – 2 = 3x + 3 x=5

x=

2 + 2(5) =4 3

x=

log3 

 3 x − 2 y = 2

− 6 ± −84 2

(rejected) By substituting y = 5 into (4), we have

(c) log3 (4x – 2) – log3 (x + 1) = 1

22. (a) log x + 2 log y = 2

Exponential and Logarithmic Functions

 10 + 5 y    y = 20 6   10y + 5y2 = 120 y + 2y – 24 = 0 (y – 4)(y + 6) = 0 y = 4 or y = –6 By substituting y = 4 into (3), we have 2

x=

10 +5( 4) =5 6

By substituting y = –6 into (3), we have x= ∴

10 +5( −6) 10 =− 3 6 10 ,y=– 3

The solution is x = 5, y = 4 or x = −

6. 23. Let I be the original sound intensity, then the increased sound intensity is 1.8I. The change of the corresponding sound intensity level



 1.8 I   I  −10 log   I  I0   0

= 10 log  

 



 1.8 I   I  − log   I  I0   0

= 10 log  



  dB   

  dB  

= (10 log 1.8) dB = 2.55 dB (cor. to 3 sig. fig.) ∴

The corresponding sound intensity level is increased by 2.55 dB.

24. (a) Let E be the relative energy released by an earthquake measured 2. By the definition of the Richter scale, we have 2 = log E E = 102 ∴ The relative energy released by the earthquake

137

Certificate Mathematics in Action Full Solutions 4A in City A is 1.5 times to that measured 2 = 1.5 × 102 The magnitude of the earthquake in City A on the Richter scale = log (1.5 × 102) = 2.18 (cor. to 3 sig. fig.)

∴

3 x +y = 27  (1)  2 x −y =1  ( 2) 2 From (1), 3x + y = 33 ∴ x+y=3 From (2), 22x – y = 20 ∴ 2x – y = 0

(b) Let E be the relative energy released by an earthquake measured 8. By the definition of the Richter scale, we have 8 = log E E = 108 ∴ The relative energy released by the earthquake in City B is half of that measured 8 = 0.5 × 108 ∴ The magnitude of the earthquake in City B on the Richter scale = log (0.5 × 108) = 7.70 (cor. to 3 sig. fig.)

Consider the simultaneous equations: x + y = 3  (3)

 2 x − y = 0  ( 4)

(3) + (4), (x + y) + (2x – y) = 3 + 0 3x = 3 x=1 By substituting x =1 into (3), we have 1+y=3 y=2 ∴ The solution is x = 1, y = 2.

Multiple Choice Questions (p.246) 1.

Answer: B

( x) 4

x

3

3

=

3

3

1 4

(x ) (x )

=

x x

=x =x 2.

1 2

5. Answer: C 32x = 23y log 32x = log 23y log (32)x = log (23)y x log 32 = y log 23 x log 9 = y log 8

3 2

x log 8 = y log 9

3 4

log 8 x : y = log 9

3 3 − 2 4

6.

3 4

Answer: C log3 (3x + 12) – log3 (2x + 1) = 1

 3 x +12   = log3 3  2 x +1  3 x +12 =3 2 x +1

log3 

Answer: C

 18    100 

log 0.18 = log 

= log 18 – log 100 = log (2 × 32) – log 100 = log 2 + log 32 – log 100 = log 2 + 2 log 3 – log 100 = a +2b −2 3.

7.

Answer: D (log x)2 – 2 log x2 + 3 = 0 (log x)2 – 4 log x + 3 = 0 Let y = log x, then the equation (log x)2 – 4 log x + 3 = 0 becomes y2 – 4y + 3 = 0 (y – 1)(y – 3) = 0 y = 1 or y = 3 For y = 1, log x = 1 x = 10 For y = 3, log x = 3 x = 1000 ∴ x = 10 or 1000

8.

Answer: A

Answer: C 9(32x) – 10(3x) + 1 = 0 9(3x)2 – 10(3x) + 1 = 0 Let y = 3x, the equation becomes 9y2 – 10y + 1 = 0 (9y – 1)(y – 1) = 0 y=

4.

3x + 12 = 3(2x + 1) 3x + 12 = 6x + 3 3x = 9 x=3

or

y=1

1 or 3x = 1 9

∴

3x =

∴

3x = 3–2 or 2 x= −

Answer: A

138

1 9

3x = 30 or x = 0

Since

log 1 x is undefined for x ≤ 0, the graph of 2

5

y = log 1 x does not cut the y-axis.

(xyz)60 = x60y60z60 = (x6)10(y10) 6(z15)4 = t10t 6t4 = t20 ∴ logxyz t20 = 60 20 logxyz t = 60 logxyz t = 3 ∴ d=3

2

∴

C and D cannot be the answer.

When x =

9.

1 y = log 1 , = 1. 1 2 2 2

∴

The graph passes through the point (

∴

The answer is A.

1 , 1). 2

Answer: B When x = 0, y = a0 = 1. ∴ The graph of y = ax passes through the point (0, 1). ∴ III cannot be the answer.

10. Answer: D Since the graph cuts the y-axis, it cannot be a logarithmic function. ∴ A and C cannot be the answer. Read from the graph, the function passes through the point (1, 3). The point (1, 3) satisfies the function y = 3x but does not satisfy the function y = 10x. ∴ The answer is D. 11. Answer: B f(5 +2x)• f(5 – 2x) = 52(5 +2x) + 1•52(5 – 2x) + 1 = 510 +4x + 1+ 10 – 4x + 1 22 =5 12. Answer: B x4 log x = 10 log x4 log x = 1 4 log x log x = 1

Exponential and Logarithmic Functions

3.

S = log144 3 2 + log144 6 3 = log144 = log144 = = = = = =

1 6 1 6 1 6 1 6 1 6

2 2

1 3

+ log144

1 2× 6

1

36

+ log144

3

1 6

1 log144 4 + log144 3 6 (log144 4 + log144 3) log144 12 log144 ×

144

1 2

1 2

1 12

1 4 1 1 log x = or log x = − (rejected) 2 2

(log x)2 =

1

x=

10 2

=

10

HKMO (p. 248) 1.

4a = 10 log 4a = 1 a log 4 = 1

∴

and and and

25b = 10 log 25b = 1 b log 25 = 1

1 1 = log 4 and = log 25 a b 1 1 + = log 4 + log 25 a b = log (4 × 25) = log 100 =2

2.

∵ ∴ ∵ ∴ ∵ ∴

logx t = 6 x6 = t logy t = 10 y10 = t logz t = 15 z15 = t

139

Certificate Mathematics in Action Full Solutions 4A

140