4ach05(exponential And Logarithmic Functions)

5

Exponential and Logarithmic Functions

5 Exponential and Logarithmic Functions • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

4.

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.

4. x y

5.

log M2 = 2 log M

–1 1 3

0 1

1 3

2 9

3 27

x y

0.3 –0.5

0.6 –0.2

1 0

2 0.3

3 0.5

4 0.6

Yes

Activity 5.2 (p. 222) 5.

1. M

N

100 1 10 100 100 1000 1 10 10 1 100 1 1000

logM logN logMN

2  M  logM log   N –3 –2

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

2

1

2 3

2 –1

4 2

0 4

4 6

1

–2

–1

3

2

(3 x 3 ) 4 3 4 ( x 3 ) 4 = (2 x 2 ) 3 23 ( x 2 ) 3

–2

3

1

–5

–4

=

100 2.

log MN = log M + log N

3.

M  log   = log M – log N N

Follow-up Exercise p. 205

1.

81x12 8x 6 81x12−6 = 8 81x 6 = 8

114

5  a 2 b −3  −2  a b

−1

  = a 2−( −2 ) b −3−1 

[

(

= a 4 b −4

2.

=a b −4

=

)

]

1

−1

−1

3

2.

b4 a4

2.

5

=x −

( −64)

3

1

(x5 ) 3 1 x3

∵ 10 × 10 × 10 = 1000 103 = 1000 ∴

1

= =

4

p. 207 1.

x

5

Exponential and Logarithmic Functions

4 3

3

5 3

= [(−4) 3 ] = ( −4)

1000 = 10

− 216 = − 6

∵ 11 × 11 = 121 112 = 121 ∴

4.

3

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

5 =  6 125 = 216

∵ 4 × 4 × 4 × 4 = 256 44 = 256 ∴

5.

4.

− 121 = − 11

∵

− 4 256 = − 4

( x) 4

1 1 1 1 1 × × × = 3 3 3 3 81

3

x

∴

4

1 1 = 81 3

1

( x 4 )3 = x x4 = x

5.

3

∵

1 1 1 1 1 1 × × × × = 2 2 2 2 2 32 5

5

1

=x 4 1 = 1

1 1   = 2 32   ∴

−1

= x4 −

6.

3

3

4

1 1   = 3 81  

4 3

 4 3× −   3

3

3.

−

= ( −4) −4 1 = (−4) 4 1 = 256

3.

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

−

x4 1

1 1 = 32 2

1 4

a •

1 5

a4

a4

=

1

(a 4 ) 5 1

=

p. 210

1.

( x) 3

4

1

6.

a4 4

a5 1 4 − 5

= (x 3 )4

= a4

4 x3

=a

=

=

−

11 20

1 11

a 20

115

Certificate Mathematics in Action Full Solutions 4A p. 213 1.

(c) ∵ ∴

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

2.

p. 221

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

∴ 5x = 25 or 5x = 52 ∴ x=2 3.

3 x +1 − 3 y = 0  (1)  x+ y  ( 2) 2 = 16

By substituting (3) into (4), we have x + (x +1) = 4 2x = 3 3 x= 2 3 By substituting x = into (3), we have 2 3 5 y= +1= 2 2 3 5 ∴ The solution is x = , y = . 2 2

∵

2.

∵

3.

∵

4.

∵ 25 = 32 ∴ log2 32 = 5

112 = 121 ∴ log11 121 = 2

1.

1

2 log 2 + log 5 = 2 log 2 2 + log 5 1

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

3.

log2 8 – log2 16 = log2 23 – log2 24 = 3 log2 2 – 4 log2 2 =3–4 = −1 log16 log 2 4 = log 32 log 2 5 4 log 2 = 5 log 2 4 = 5 1

(a) ∵ 0.01 = 10–2 ∴ log 0.01 = −2 (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.)

116

43 = 64 ∴ log4 64 = 3

p. 225

p. 220

2.

92 = 81 ∴ log9 81 = 2

1.

5x = –1 (rejected)

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)

1.

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

4.

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

5 2 log x log x − log x

5.

=

log x 2 + log x

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 2 log 3 x= log 6 − log 3 = 3.17 (cor. to 3 sig. fig.)

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  3x + 1  log  = log 10  x−2  3x + 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 1 x–1= 10 11 x= 10

1

=

=

=

=

=

7.

1 2

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

log x 2 − log x

6.

3.

2 log x

Exponential and Logarithmic Functions

log x 2 − log x 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

54 6 = log 54 – log 6 = y−x

log 9 = log

p. 233 p. 229 1.

2.

3x – 1 = 7 log 3x – 1 = log 7 (x – 1)log 3 = log 7 log 7 x–1= log 3 log 7 x= +1 log 3 = 2.77 (cor. to 3 sig. fig.) 52x = 8 log 52x = log 8 2x log 5 = log 8 log 8 2x = log 5 log 8 x= ÷2 log 5 = 0.646 (cor. to 3 sig. fig.)

1.

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  95 = 10 log    I0   I2 and 80 = 10 log   I0

   

 I1 ∴ 95 – 80 = 10 log   I0   I1 15 = 10 log   I 0  I1 3 = log  2  I2

 I  – 10 log  2  I   0  I  − log 2  I   0

   

   

  

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

117

Certificate Mathematics in Action Full Solutions 4A ∴ The sound intensity of the concert is 31.6 times to that of the football match. 2.

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    I0  Let β dB be the sound intensity level produced by the radio after adjusting the volume.  1.25 I   ∴ β = 10 log    I0   1.25 I β – 30 = 10 log   I0

  I  – 10 log   I   0

  1.25 I   I  − log β – 30 = 10 log  I I  0   0 

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

1.

7

x3 = ( x3 ) 7 3

= x7

2.

( x) 3

1

8

= ( x 3 )8

   

= 1

   

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

( x) 3.

=

5

4.

x2

=

118

−

5 2

1

=

1

(x2 ) 5 1 2

x5 =x

−

2 5

3

3

5.

2 2  49  2  7     =     25   5  

7 =  5 343 = 125

∴ 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.

5

=x 1

2.

1

( x 2 )5 1 x2

p.235 Let E1 and E2 be the relative energies released by the 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 6 E1 10 = ∴ E 2 10 3 = 103 = 1000

1

=

5

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

1.

8 x3

3

3

3

6.

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

2 =  3 8 = 27

3

5 3

3

7.

 27  −   64 

−

3

1 1

 3 = −   4 4 =− 3

9.

a a  2  a −1

3

−4

=a

−1

1 3

=     

14.

−1

15.

−1

11. ( m n )

÷m n −1

3 4

1 1 − × ( −2) × ( −2) 2 b2

÷ ab −1

3 −1 − − ( −1) 4

=m

−3

−

1 2

2 1 3b 4

−

(a q )

a

2+ p −

q 2

1 2

=

a 2+ p a

q 2

b4 2 a3

=a

2+ p −

q 2

=1 = a0

q =0 2 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)

 1 −4× −   2

n

 a−x 16.  y a  1 −2× −   2

÷m n −1

= a −1×( −3) b 3×( −3) ÷ a

 a− x   ay 

−3

 1 4× −   2

b

 1 2× −   2

= =

11 13 a6b6

=a

1

2

4 1 3 2 + + a 3 2b 2 3

−1

Level 2

2 1 2 3 1 2  2 3 ×2 ×2  3 4 2b 3 = a 3 b 4 • a 2b 3 a b • a    

=

   

a x + y = a1 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)

(3 x ) 2

4 3 a 3b 2

−1

 ( a − x ) −1 a x  = y −1 = − y = a x + y (a ) a 

∴ ∴

= a3b–9 ÷ a–2b–1 = a3 – (–2)b–9 – (–1) = a5b–8 a5 = 8 b

13.

−

÷a

∴ 2+ p−

= m2n ÷ m–1n–3 = m2 – (–1)n1 – (–3) 3 4 =m n −1 3 −3 4 2 12. ( a b ) ÷ (a b )

aq

∴

−1

1 2

a 2+ p

=

aq

  = [(m–4 – 3)(n –3 – 4)]–1  = (m–7n –7)–1 7 7 =m n −

a 2a p

a 2a p

  = (a–1+3 – 2)–1  = (a0)–1 = a0 =1

−2

3 4

1

−1

 m −4 n −3 10.  3 4  mn

−

= a3 b

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

8.

1

= a 3b

= a 3b

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

1 3

−2

 −1 1    ÷a 2b 2     

1 3 − a 3b 4

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

3 =  4 27 = 64

Exponential and Logarithmic Functions

x

3 4

• x = 4

2

( x 3 )2 x

3 4

2

=

1 2 • a 2b 3

x3 3

•x

1

• (x2 ) 4 1 2

x4 2 3 1 − + 4 2

= x3 17.

5

= x 12

119

Certificate Mathematics in Action Full Solutions 4A 1

x

2 3

−

1

x4

•

5

=x

x3

−

2 3

•

=x

2 3

1

=x

23.

x

1 1 1 1 + −1 + −1  13 12  12 13   m n  m n  ÷ mn = m 3 2 n 2 3      

61 60

−

p

(5 p )

2

3 4

p

÷ 3 p2 =

3

p =p

÷p

2 5

−2

2 3

1

s5 •

1 s

−

=

25.

1

= s •s

3

1

b2

1 − 2

•s

3

a

2

1

• (27 ab) 3 ÷ a 3 b =

5

−

2 3

=t =t

21.

=t =t

−

1

= 9a

−

1

÷ (t 3 ) 4 • t

1 2

3

÷t 4 •t

−

9

q3

q2 1

3 1 1 + − 2 3

11

120

2

−

4 −( − ) 3

19 1 −

1

q3

= q 12

1

−

8

= 4m 12 6 n 3

q2

1

22.

2 3

1

(q 3 ) 9

= q4

1 4

= 22 m 4

=q •

13

   

4 3

4

−

1

−

27.

17

8

= 4m 12 n 3

2

n3

− ( −2 )

1

÷m6

1 3

   

= m 4 n 3 ÷ 2 −2 m 3 n −2 ÷ (m 2 )

1

=q •

2 3

 1 = m n ÷ 2 −2 m n −2 ÷  1  2 m

23 12

3 4

3 4

−2

 2   1 m n ÷  2m 3 n  ÷     m

2 3

1 2 1 − + − 3 3 6

= 9a 6 b 6

1 4

23 12

1

q •

2 3

1 3 2 − − +( − ) 2 4 3

t

(q 2 ) 4

−

1

=

3 4

−

2

1

26.

1 2

1

1

• (33 ab) 3 ÷ (a 3 b) 2

1 3

13

−

2

b2

2

1

1

= a 3 b 2 • 32 a 3 b 3 ÷ a 6 b 2

=s5 ÷ 4 t3 •t

1

b4

a

3 5

5 1 3 − + 2 5

1 2

1 1 − 4

b2

2a 4

= s2

−

3 2

1

5 2

t

5

−

= 2 −1 a 4 b 4

• s 5 = ( s 5 ) 2 • ( s 2 ) −1 • s 5

20.

1

1 1 3 − +( − ) 4 2

19 60

3

1

= 2 −1 a 2

=p 1 = 19 p 60

19.

1

= a 2 b 2 ÷ 2a 4 b 4 • a

3 2 2 − − 4 5 3 −

1

1 1 3 −  3 ab ÷ 4 16ab •  a 4  = (ab) 2 ÷ ( 2 4 ab) 4 • a 2  

( p )2 =

−

1

÷ ( p2 )3

1 5

p4

1

= m 6n 6 1 = 1 1 m6n6

61 60

24. 3 4

b

= a 2b

1

=

18.

1 1 1 1 + +1 1− + 2 2 2

2 1 3 − + − 3 4 5 −

1

= a2

3 5

x =x

1

1 4

x

•

1

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

(x3 ) 5 −

1

1

1 2

x4

−

−

1 3

1 3

b

2 1 2+ − 3 2

5 1

−

1

2

1

1

−

1

2

xy 4 ÷ 3 64 x 2 y ÷ x 3 y 5 = xy 4 ÷ (4 3 x 2 y ) 3 ÷ x 3 y 5 2

1

1

−

1

6.

3(3x) – 3x – 1 = 216 3(3x) – 3x(3– 1) = 216 3x(3 –3–1) = 216 1 3x(3 – ) = 216 3 8  3x   = 216 3 3x = 81 3x = 34 ∴ x=4

7.

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

2

= xy 4 ÷ 4 x 3 y 3 ÷ x 3 y 5 = 4 −1 x

2 1 1− − ( − ) 3 3

2

=

1 1 2 − − 3 5

y4

29

1 3 − 60 x y 4 2

= 28.

Exponential and Logarithmic Functions

x3 29

4 y 60

2

Exercise 5B (p.214) Level 1 x

2

5 2 = 125 3 1.

2

x

5 2 = (5 3 ) 3

8.

42x = 16(2x) (22)2x = 24(2x) 24x = 24 + x ∴ 4x = 4 + x 4 x= 3

9.

23x = 8(4x) 23x = 23(22)x 23x = 23(22x) 23x = 23 + 2x ∴ 3x = 3 + 2x x =3

10.

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

x

∴

5 2 = 52 x =2 2 x=4 x 3

8 = 16

2.

x

(23 ) 3 = 2 4 2 x = 24 ∴ x=4 6

3.

6 ∴

x +1 2 x +1 2

= 36 = 62

x +1= 2 2 x=2 x 3

4 = 32 4.

x 3

Level 2

2 5

(2 2 ) = (25 )

2 5

2x

2 3 = 22 ∴

2x =2 3 x =3

5.

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

11.

4x + 1 + 4x – 4x – 1 = 19 (4)4x + 4x – (4– 1)4x = 19 4x(4 + 1 – 4– 1) = 19 1 4x(5 – ) = 19 4 19 4x( ) = 19 4 4x = 4 ∴ x =1

2

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 1 6x = 6 6x = 6–1 ∴ x = −1

121

Certificate Mathematics in Action Full Solutions 4A 13.

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 x

14.

x

∴ 2 =1

or

2 =2

2x = 20 ∴ x =0

or or

2x = 21 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 x

∴

3 =3 x =1

or 1

or or

3x = 9 x

3 =3 x=2

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 1 y= or y = 5 5 ∴ 5x =

16.

1 5

or

5x = 5

5x = 5–1 or ∴ x = −1 or

5x = 51 x =1

24x – 20(22x) + 64 = 0 (22x)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 ∴ 22x = 4

or

22x = 16

2 = 2 or ∴ 2x = 2 or x = 1 or

22x = 24 2x = 4 x=2

2x

2

 x +2 y = 125  (1) 17. 5 2 x+ y = 1  ( 2) 25 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

122

x = –1 By substituting x = –1 into (4), we have y = –2(–1) = 2 ∴ The solution is x = –1, y = 2. x+ y 18. 3 = 27  (1) 3x− y = 4  ( 2) 4 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.  2 x + y = 729  (1) 19. 3 x −2 y = 3  ( 2) 9 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 5 x= 2 5 ∴ The solution is x = , y = 1. 2 3 x −2 y = 4  (1) 20. 2 x+ y 5 = 625  ( 2)  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)

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

∵

1 = 4–2 16  1 ∴ log4   = −2  16  1 = 10–2 100  1  −2 = ∴ log   100 

4.

∵

5.

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

Exponential and Logarithmic Functions

log 9 log 32 = log 27 log 33 2 log 3 10. = 3 log 3 2 = 3 log 64 log 4 3 = log16 log 4 2 3 log 4 11. = 2 log 4 3 = 2  1   1  log 2   log 2  2   25  = 5  1 log 2 5 log 2 5 2 =

12.

log 2 5 −2 1

log 2 5 2 − 2 log 2 5 1 log 2 5 2 = −4 =

13. log 6 = log (2 × 3) = log 2 + log 3 = x+ y 1

6.

7.

8.

9.

x

∵ 10 = 88 ∴ x = log 88 = 1.94 (cor. to 3 sig. fig.) 200 2 = log 100 = log 102 = 2 log 10 =2

14. log 18 = log ( 2 × 32 ) 2 1

= log ( 2 2 × 3) 1

= log 2 2 + log 3 1 log 2+ log 3 2 1 = x+ y 2 =

log 200 – log 2 = log

log 4 + log 25 = log(4 × 25) = log 100 = log 102 = 2 log 10 =2  1 1 log3   + log3 108 = log3 ( × 108) 12  12  = log3 9 = log3 32 = 2 log3 3 =2

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

1

b3 = 8 b = 83 = 512

123

Certificate Mathematics in Action Full Solutions 4A

Let a =

1 , then 2

1

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

x

1   =8 2 x

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

−

∴

1 3

5

−3

20.

=8

b = 8− 3 1 = 512 ∴ A possible solution is a = 2, b = 512 1 1 or a = , b = . (or any other reasonable answers) 512 2

16.

log a =3 log b 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)

log 5 + 2 log 2 − 1 log 5 + log 2 2 − log10 = 2 log 4 − log 0.5 log 4 2 − log 0.5

21.

Level 2  2   2  17. 3 log 5 + log   = log 53 + log    25   25   3 2  = log  5 ×  25   = log 10 =1

3  2 3 19. log4 2 – log4 18 + 2 log4   = log4   + log4   2  18  2  1  3 2  = log4  ×     9  2   1 9 = log4  ×  9 4 1 = log4   4 = log4 4–1 = –log4 4 = −1

 5 × 22 log  10 = 42 log 0.5 log 2 = log 32 log 2 = log 2 5 log 2 = 5 log 2 1 = 5

  

log 8 + 3 log 5 − 2 log 8 + log 5 3 − 2 log10 = 2 log 5 − log 0.25 log 5 2 − log 0.25 log 8 + log125 − log100 = log 25 − log 0.25 =

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

124

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 =

22. = 2

= = =

 8 × 125  log   100  25 log 0.25 log10 log100 1 log10 2 1 2 log10 1 2

5 log x 4 4 log x = 3 log x 2 + 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

1 (3 log 4 + log 5) 2 1 = (3 x + y ) 2 =

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

2x = 10 log 2x = log 10 x log 2 = 1 1 x= log 2 = 3.32 (cor. to 3 sig. fig.)

2.

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

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

24.

=

(2 + 3) log 3 x 8 log 3 x − 3 log 3 x

=

5 log 3 x (8 − 3) log 3 x

5 5 =1 =

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

  

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

25.

 log 6  + 1 ÷ 2 x =   log 3  = 1.32 (cor. to 3 sig. fig.) 3.

2 log x 3 − 3 log x 4 2(3 log x) − 3( 4 log x ) = log x 2 y − log xy  x2 y   log  xy  6 log x − 12 log x log x (6 − 12) log x = log x = −6

26.

=

1

1 log 320 2 1 = log (43 × 5) 2 1 = (log 43 + log 5) 2 =

41 – 3x = 5 log 41 – 3x = log 5 (1 – 3x) log 4 = log 5 log 5 1 – 3x = log 4 log 5 3x = 1 – log 4  log 5   ÷ 3 x = 1 −  log 4  = −0.0537 (cor. to 3 sig. fig.)

4.

27. log 200 = log (4 × 5 × 10) = log 4 + log 5 + log 10 = x + y +1 28. log 320 = log 320 2

Exponential and Logarithmic Functions

5.

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 x+2= log 3 log 5 − log 2 x= –2 log 3 = −1.17 (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 log 9 x= log 8 − log 9 = −18.7 (cor. to 3 sig. fig.)

125

Certificate Mathematics in Action Full Solutions 4A 6.

7.

8.

9.

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 log 6 x= 3 log 6 − 2 log 4 = 0.688 (cor. to 3 sig. fig.) log (6x – 4) = 2 log (6x – 4) = log 100 6x – 4 = 100 6x = 104 1 x = 17 3 log3 (4x – 1) = 3 log3 (4x – 1) = 3 log3 3 log3 (4x – 1) = log3 33 4x – 1 = 27 4x = 28 x =7 log (3x – 2) = 0 log (3x – 2) = log 1 3x – 2 = 1 3x = 3 x =1

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

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 12. log (2x – 3) + log 2 = –1 log [(2x – 3) × 2] = –log 10 log (4x – 6) = log 10–1 4x – 6 = 10–1 1 4x = +6 10 61 4x = 10 61 x= 40 13. log4 (x + 3) + log4 (x – 3) = 2 log4 [(x + 3)(x – 3)] = 2log4 4 log4 (x2 – 9) = log4 42 x2 – 9 = 16

126

x2 = 25 x=5

or

x = –5 (rejected)

∴ x =5 14. log3 (4x + 9) – log3 (3x – 2) = 1  4x + 9   = log3 3 log3   3x − 2  4x + 9 =3 3x − 2 4x + 9 = 9x – 6 5x = 15 x =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 x=3

or

x=−

3 (rejected) 2

∴ x =3 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 3 x= 2 For y = 2, log2 (2x – 1) = 2 2x – 1 = 4 5 x= 2 3 5 ∴ x = or 2 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 1 x+1= 1000 999 x=− 1000 For y = 4, log (x + 1) = 4 x + 1 = 10 000 x = 9999 999 ∴ x=− or 9999 1000 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

5 For y = 1, log (x – 1) = 1 x – 1 = 10 x = 11 For y = 2, log (x – 1) = 2 x – 1 = 100 x = 101 ∴ x = 11 or 101

 I2 3 = log  2  I1

∴ The sound intensity in restaurant B is 31.6 times to that in restaurant A. 2.

−

   

   

  

∴ The sound intensity after the party is 0.000 631 times to that during the party. 3.

Let I1 and I2 be the sound intensities before adjustment and after adjustment respectively. Then the sound intensity levels before and after  I1   I2  adjustment are 10 log   dB and 10 log   dB  I0   I0  respectively. ∵ The sound intensity level increases by 5 dB after the adjustment.  I2   I1  ∴ 5 = 10 log   – 10 log   I  0  I0    I2 5 = 10 log   I 0  I2 1 = log  2  I1

   

 I  − log 1  I   0

   

  

I2 1 = 2 I1 10 = 3.16 (cor. to 3 sig. fig.)

   

 I  − log 1  I   0

 I1 16 = log  5  I2

 I  − log 2  I   0

   

I1 16 = −5 I 2 10 = 0.000 631 (cor. to 3 sig. fig.)

Let I1 and I2 be the sound intensities in restaurants A and B respectively. By the definition of sound intensity level, we have  I1  60 = 10 log    I0 

  I2 15 = 10 log   I 0

 I  – 10 log  2  I   0

  I1 –32 = 10 log   I 0

Level 1

 I  – 10 log  1  I   0

   

 I1 ∴ 35 – 67 = 10 log   I0

Exercise 5E (p.236)

 I2 ∴ 75 – 60 = 10 log   I0

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  35 = 10 log    I0   I2 and 67 = 10 log   I0

 (1) 20.  x − y = 1  ( 2) log y = log x − 1 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 1 y= 9 1 By substituting y = into (3), we have 9 1 10 x = 10( ) = 9 9 10 1 ∴ The solution is x = ,y= . 9 9

 I2 and 75 = 10 log   I0

  

I2 3 = 2 I 1 10 = 31.6 (cor. to 3 sig. fig.)

 (1) 19.  x − 2 y = 32  ( 2) log x = log y + 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.

1.

Exponential and Logarithmic Functions

∴ The sound intensity after adjustment is 3.16 times to that before adjustment. 4.

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

127

Certificate Mathematics in Action Full Solutions 4A   0.6 I   I   − log  β – 90 = 10 log  I    I 0   0  β – 90 = 10 log 0.6 β = 90 + 10 log 0.6 = 87.8 (cor. to 3 sig. fig.)

E1 10 8 = ∴ E 2 10 2 = 106 ∴ The strength of an earthquake measured 8 is 106 times to that measured 2. 5.

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

∴ The sound intensity level of the Hi-Fi is 87.8 dB after adjustment. 8.

 1.1I ∴ 10 log   I0

Level 2

 2I   I β – 30 = 10 log   – 10 log  I  0  I0   2I   I  β – 30 = 10 log  − log    I 0   I 0  β – 30 = 10 log 2 β = 30 + 10 log 2 = 33.0 (cor. to 3 sig. fig.)

   

∴ 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.  I  ∴ 90 = 10 log    I0  Let β dB be the sound intensity level of the Hi-Fi after adjustment.  0.6 I   ∴ β = 10 log    I0   0.6 I β – 90 = 10 log   I0

128

   

∴ The increase in the sound intensity level is 0.414 dB.

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.  I  ∴ 30 = 10 log    I0  Let β dB be the sound intensity level when the master of ceremony speaks.  2I  ∴ β = 10 log    I0 

7.

  I  – 10 log   I   0

  1.1I   I   − log  = 10 log  I    I 0   0  = 10 log 1.1 = 0.414 (cor. to 3 sig. fig.)

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

6.

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  I   1.1I   dB respectively. 10 log   dB and 10 log   I  0  I0 

  I  – 10 log   I   0

   

9.

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 magnitude of the earthquake on the Richter scale 1 = log ( × 107.2) 8 6 . 30 = (cor. to 3 sig. fig.)

5

Exponential and Logarithmic Functions 1

x2

6

Revision Exercise 5 (p.244)

x3

Level 1

=

1

(x3 ) 2 1

x3

=

1

1.

(x2 ) 6

7 5 5 7 (a) ( x ) = ( x )

3

(b)

x2

7

1 3 − 2

= x5

= x3 −

1 3

(b)

x

4

1

= =

7

=x 6 1 = 7 x6

1

(x4 ) 3 1 4

x3 =x

−

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

4 3

4

 216     125 

2.

−

2 3

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

−1

(c)

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

(a)

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

 27  −   343 

−

1 3

= x −18 • x 3

=x

−1

=

−1

1

3.

(a)

1

=a

(d)

a c4 b

1

−1 1−

−

1 1 2b 2

1 2

× −1

1

÷ [(ab) 4 ]2 1

÷ (ab) 2 1 1

÷ a 2b 2

1 1 1 1 − − − = a 2 2b 2 2 −1

= a −1 1 = a

−1

2 x = 3 16 4.

1

(a)

= (2 4 ) 3

−1

= a b −1c 4 =

50

1

4

= 23 4 ∴ x= 3 4 x = 64

a 2 b −3 c12 = ( a 2 b −3 c12 ) 3

2 3

1

= a2 b

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

2 3

50 3

1

1

3

−

4 3

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

−1

 3 = −   7 7 =− 3

−18+

x3

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

(b)

=x

1

(b) [(2 2 ) x ] 2 = 2 6 1

(2 2 x ) 2 = 2 6 ∴

2 x = 26 x =6

3 5 3 5(5x) – 2(5x) = 5

(c) 5x + 1 – 2(5x) =

129

Certificate Mathematics in Action Full Solutions 4A 3 5 1 5x = 5 5x = 5–1 x = −1

1  1  (d) log9 9 + log9   = log9  9 ×  81 81     1 = log9   9 = log9 9–1 = −1

3(5x) =

(d)

5.

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

(a) ∵

(e)

log 625

1 = 10–4 10 000

log 5

 1   = −4 ∴ log   10 000 

(f)

(b) ∵ 0.000 000 1 = 10–7 ∴ log 0.000 000 1 = −7 (c) ∵ 64 = 82 ∴ log8 64 = 2 (d) ∵ ∴ 6.

7.

1 5

=5

8. −

(b) ∵ ∴

102x = 20 2x = log 20 x = 0.651 (cor. to 3 sig. fig.) 400 + log 2 8 = log 50 + log 2 = log (50 × 2) = log 100 =2

(a) log 400 – log 8 + log 2 = log

(b) log 125 + log 80 = log (125 × 80) = log 10 000 = log 104 =4 (c) log6 108 + log6 2 = log6 (108 × 2) = log6 216 = log6 63 =3

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 1 x= log 5 = 1.43 (cor. to 3 sig. fig.)

(b)

2x – 1 = 6 log 2x – 1 = log 6 (x – 1) log 2 = log 6 log 6 x–1= log 2 log 6 x= +1 log 2 = 3.58 (cor. to 3 sig. fig.)

(c)

32x + 3 = 33 log 32x + 3 = log 33 (2x + 3) log 3 = log 33 log 33 2x + 3 = log 3 log 33 2x = –3 log 3 x = 0.0913 (cor. to 3 sig. fig.)

 1  1 =− log 5   2  5 10x = 45 x = log 45 = 1.65 (cor. to 3 sig. fig.)

=

(a)

1 2

(a) ∵ ∴

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

130

log 256 log 4 4 = log16 log 4 2 4 log 4 = 2 log 4 =2

x  log  −1 = –1 2 

5

Exponential and Logarithmic Functions   I2 10 log   I 0

x   1 log  −1 = log   2    10  1 x –1= 10 2 x = 1.1 2 x = 2.2

 I  − log 1  I   0

  = 8  

 I2  4 log   =  I1  5 I2 4 = 5 I 1 10 4

I2 = 10 5 I1

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

∴ The percentage increase of the sound intensity: I 2 − I1 ×100% = I1 4

9.

5 = 10 I 1 − I 1 ×100% I1

(a) log 144 = log (24 × 32) = log 24 + log 32 = 4 log 2 + 2 log 3 = 4x + 2 y (b) log 288 = log 288

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

1 2

1 log (25 × 32) 2 1 = (log 25 + log 32) 2 1 = [5 log 2 + 2 log 3] 2 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  I1  45 = 10 log    I0   I2 and 53 = 10 log   I0

   

 I2 ∴ 53 – 45 = 10 log   I0   I2 8 = 10 log   I 0  I2 4 = log  5  I1

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

   

   

  

I2 4 = 5 I 1 10 = 6.31 (cor. to 3 sig. fig.) ∴ The sound intensity of group B is 6.31 times to that of group A. 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   I1  ∴ 10 log   – 10 log   = 8 I  0  I0 

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 7.5 E1 10 = ∴ E 2 10 4.6 = 102.9 = 794 (cor. to 3 sig. fig.) ∴ The strength of the earthquake in Indonesia is 794 times to that in Pakistan. 14. (a) log (4x – 2) – log 2x = y  4x − 2  =y log   2x  2x −1 = 10y x 2x – 1 = x10y x (2 – 10y) = 1 1 x= 2 − 10 y 1 1 = =1 2 − 10 0 2 − 1 1 1 When y = 1, x = =− 8 2 − 101

(b) When y = 0, x =

∴ A possible solution is x = 1, y = 0 or x = −

1 , 8

y = 1. (or any other reasonable answers)

131

Certificate Mathematics in Action Full Solutions 4A Level 2

3 4

a ab ÷ ( a b ) •   b

1

(4 a ) 2

•a

a

3

−

1 3

=

(a 4 ) 2 a

=

1 2

a

•a

1 3

a

15. (a)

−

1 3

3 4

= ab ÷ [(a b) ] •

= a •a

3 5

•

6

p3

5

p2

=p

−

=a

1 6

3 5

=p

−

3 5

•

•

( p3 )

1

17. (a)

p2 2

1

• p 10

1

2

y

xy 3 • x

÷ x3 y

3 − 4

1 4

1

−

1 2

3

2

= ( xy ) • y 2 • x 4 ÷ x 3 y 3

1

3

1

3

2

= x4 y4 • y2 • x4 ÷ x3 y 1 3 +

3 1 + 2

= x4 4 y4

2

5

2

÷ x3 y

= xy 4 ÷ x 3 y =x

1− 1

2 3

5

y4

−

1 2

1

1

−

1 3

1 −( − ) 3 −

1

−

1

1

• a 2b2 −

1

1

• a 2b2 1

13 1 + 2

−

−

− 1 2

19

1 2

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

(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)

1 2

1 2

1 −( − ) 2

7

16. (a) = x 3 y 4 (d)

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 ∴ 2x = 1

or

2x = 4

2x = 20 ∴ x =0

or or

2x = 22 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 ∴ 4x = 2 2x

or 1

2 =2 2x = 1 1 ∴ x= 2

132

−

(b) = a 6 b 12

1 6

=p 2 1 = 1 p2

4

3

b4

13

5 1 −

p5 −

1 2

= a 3 2 b 12

(b) =p

2

−

2 1− ( − ) 3 5

1 3 5

−

−

= a 3 b 12 • a 2 b 2

( p2 )5 −

1

= ab 4 ÷ a 3 b

1 3

7

−

2

3

1 2

= ab 4 ÷ ( a 3 b 3 ) −1 • a 2 b 2

=a 1 = 1 a6

p

a b

1 − 3

−

1 3 −1

2

3

1 6

−

•a

1 3

−

−1

2

3

4x = 8

or 22x = 23 or 2x = 3 3 or x = 2

5 x +2 y = 64  (1) 18. (a) 4 2 x− y = 1  ( 2) 16

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)   ( 4)  y = 2x By substituting (4) into (3), we have x + 2(2x) = 3 5x = 3 3 x= 5 3 By substituting x = into (4), we have 5 3 6 y = 2( ) = 5 5 3 6 ∴ The solution is x = , y = . 5 5 x+ y 2 x− y  (1) (b) 3 = 9 3x+ y  ( 2) 9 = 9  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) 3 x + y = 1 

By substituting (3) into (4), we have 3x + x = 1 4x = 1 1 x= 4 1 1 By substituting x = into (3), we have y = 4 4 1 1 ∴ The solution is x = , y = . 4 4 19. (a) −

1 1 log 625 – log 4 = log 625 − 2 – log 4 2 1 = log – log 4 625 1 = log 25 4 1 = log 100 = log 10–2 = −2

(b)

Exponential and Logarithmic Functions

32 log log 32 − log 64 64 = log128 log128 1 log 2 = log128 log 2 −1 log 2 7 − log 2 = 7 log 2 1 =− 7 =

(c)

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

1.5 log16 − 2 log 2 1.5 log 2 4 − 2 log 2 = log 36 − 2 log 3 log 36 − log 32 1.5(4 log 2) − 2 log 2 = log 36 − log 9 (6 − 2) log 2 = 36 log 9 (d) 4 log 2 = log 4 4 log 2 = log 2 2 4 log 2 = 2 log 2 =2 log a 4 + log b 8 log a 4 b 8 = log ab 2 log ab 2 20. (a)

=

log(ab 2 ) 4 log ab 2

4 log ab 2 log ab 2 =4 =

133

Certificate Mathematics in Action Full Solutions 4A 2x × 26 = 96 128x = 96 3 x= 4

4 log x 2 + 2 log x 4 8 log x + 8 log x = log x 3 3 log x (8 + 8) log x = 3 log x (b) 16 log x = 3 log x 16 = 3 log xy − log x y log 3 xy − log( x • 3 y )

=

1

1

1

1

log( xy ) 2 − log xy 2 log( xy ) 3 − log xy 3 1

=

1

1

log x 2 y 2 − log xy 2 1

1

1

log x 3 y 3 − log xy 3  12 12 x y log 1  xy 2 

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

(c)

=

log x

− −

1 2 2

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

(d)

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

(c) log3 (4x – 2) – log3 (x + 1) = 1  4x − 2   = log3 3 log3   x +1  4x − 2 =3 x +1 4x – 2 = 3(x + 1) 4x – 2 = 3x + 3 x =5 (d) log5 (3x + 3) – log5 (x – 1) = 2 log5 5 1  3x + 3   = log5 (5 2 ) 2 log5   x −1  3x + 3 =5 x −1 3x + 3 = 5(x – 1) 3x + 3 = 5x – 5 2x = 8 x=4 (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) 22. (a) log x + 2 log y = 2  ( 2) 3 x − 2 y = 2  From (1), log x + log y2 = log 100 log xy2 = log 100 xy2 = 100 ……(3) From (2), 3x = 2 + 2y 2 + 2y x= ……(4) 3 By substituting (4) into (3), we have  2 + 2y  2   y = 100  3  2y2 + 2y3 = 300 y3 + y2 – 150 = 0 (y – 5)(y2 + 6y + 30) = 0 y = 5 or y =

(b) log 2x + 6 log 2 = log 96 log 2x + log 26 = log 96 log (2x × 26) = log 96

134

− 6 ± 6 2 − 4(1)(30) 2(1) =

− 6 ± − 84 (rejected) 2

5 By substituting y = 5 into (4), we have 2 + 2(5) x= =4 3 ∴ The solution is x = 4, y = 5.

(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.)

 (1) (b) log(6 x − 5 y ) = 1  ( 2) log xy − log 2 = 1 From (1), 6x – 5y = 10 6x = 10 + 5y 10 + 5 y x= ……(3) 6 From (2), xy log = log 10 2 xy = 10 2 xy = 20 ……(4) By substituting (3) into (4), we have  10 + 5 y    y = 20  6  10y + 5y2 = 120 y2 + 2y – 24 = 0 (y – 4)(y + 6) = 0 y = 4 or y = –6 By substituting y = 4 into (3), we have 10 + 5(4) x= =5 6 By substituting y = –6 into (3), we have 10 + 5( −6) 10 x= =− 6 3 ∴ The solution is x = 5, y = 4 or x = −

Exponential and Logarithmic Functions

Multiple Choice Questions (p.246) 1.

Answer: B 1

( x )3 4

x3

=

( x 2 )3 1

(x3 ) 4 3

=

x2 3

x4 3 3 − 4

= x2 3

= x4 2.

 18   log 0.18 = log   100  = 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

10 , y = –6. 3

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  dB = 10 log  I   I0   0  

3.

  1.8 I   I   − log  dB = 10 log  I  I  0    0  = (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 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.)

Answer: C

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 1 y = or y = 1 9 ∴ 3x =

4.

1 9

or

3x = 1

3x = 3–2 or ∴ x = −2 or

3x = 30 x =0

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

135

Certificate Mathematics in Action Full Solutions 4A ∴ 2x – y = 0

∴ The answer is A.

Consider the simultaneous equations:  x + y = 3  (3)  2 x − y = 0  ( 4)

9.

(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.

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.

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 x log 8 = y log 9 log 8 x : y = log 9 6.

7.

8.

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

Answer: C log3 (3x + 12) – log3 (2x + 1) = 1  3 x + 12   = log3 3 log3   2x +1  3 x + 12 =3 2x + 1 3x + 12 = 3(2x + 1) 3x + 12 = 6x + 3 3x = 9 x =3

= 10

HKMO (p. 248)

2

does not cut the y-axis.

∴ C and D cannot be the answer. 1 1 When x = , y = log 1 = 1. 2 2 2 ∴ The graph passes through the point (

136

1 (rejected) 2

1

Answer: A log 1 x Since is undefined for x ≤ 0, the graph of 2

log x = −

x = 10 2

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

y = log 1 x

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

1 , 1). 2

1.

4a = 10 and 25b = 10 log 4a = 1 and log 25b = 1 a log 4 = 1 and 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 60 (xyz) = x60y60z60 = (x6)10(y10) 6(z15)4 = t10t 6t4 = t20 ∴ logxyz t20 = 60 20 logxyz t = 60 logxyz t = 3

5 d =3

∴ 3.

Exponential and Logarithmic Functions

S = log144 3 2 + log144 6 3 1

1

= log144 2 3 + log144 3 6 1

1

= log144 2 2× 6 + log144 3 6 1 1 log144 4 + log144 3 6 6 1 = (log144 4 + log144 3) 6 1 = log144 12 6 1 1 = log144 144 2 6 1 1 = × 6 2 1 = 12 =

137