Indices, Surds And Logarithms Part 3 - Buncee

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Additional Mathematics Chapter 3: Indices, Surds And Logarithms Matematik Tambahan Bab 4: Indeks, Surds dan Logaritma

Part 3: Logarithms Bahagian 3: Logarithma

1 Laws of Logarithms / Hukum-hukum Logaritma

Law 1

If ab = c, then loga c = b

Petua ingat: a kuasa b ialah c log a c is b

Law 2

loga bc = c loga b

Law 3

loga a = 1 for any positive values of a

Self Practice 4.9 Latih Diri 4.9

1 Convert the following to logarithmic form.

Tukar setiap yang berikut kepada bentuk logaritma.

(a) 34 = 81

log3 81 = 4

(b) 27 = 128

log2 128 = 7

(c) 53 = 125

log5 125 = 3

(d) 63 = 216

log6 216 = 3

2 Convert the following to index form.

Tukar setiap yang berikut kepada bentuk indeks.

(a) log10 10 000 = 4

104 = 10 000

(b) log10 0.0001 = – 4

10-4

= 0.0001

(c) log2 128 = 7

27 = 128

(d) Log4 64 = 3

43 = 64

3 Find the value of each of the following:

Cari nilai bagi setiap yang berikut:

(a) log10 9 = 0.9542 (b) log10 99 = 1.9956 (c) log10 (

)

3 = – 0.2375

(d) log2 64 = log2 26

= 6 log2 2

= 6

(e) log3 81 = log3 34

= 4 log3 3

= 4

(f) log4 256 = log4 44

= 4 log4 4

= 4

(g) log10 100 000 = log10 105

= 5 log10 10

= 5

4 Solve the following equations.

Selesaikan setiap persamaan berikut.

(a) log2 x = 5

25 = x

x = 32

(b) log8 x = 3

83 = x

x = 512

(c) log2 x = 8

28 = x

x = 256

5 Determine the value of each of the following.

Tentukan nilai bagi setiap yang berikut.

(a) antilog 2.1423 = 138.78 (b) antilog 1.3923 = 24.68 (c) antilog 3.7457 = 5568.01

(d) antilog (– 3.3923)

= 0.0004052

(e) antilog (– 2.5676)

= 0.002706

(f) antilog (– 4.5555)

= 0.00002783

2

Remember the three basic laws of logarithms Ingat tiga asas bagi hukum logaritma

If a, x and y are positive and a 1, then / Jika a, x dan y adalah positif dan a 1, maka

(a) loga xy = loga x + loga y [Product law / Hukum darab]

(b) loga

= loga x – loga y [Division law / Hukum bahagi]

(c) loga xn = n loga x [Power law / Hukum kuasa]

(d) loga (x + y – z) = log a (x + y – z) Don’t open the bracket until finish for + and –

(e) loga (2x – y) = loga (2x – y) Don’t open the bracket until finish for + and –

(f) loga 1 = 0 for any positive values of a

(g) loga a = 1 for any positive values of a.


1 Given that log7 4 = 0.712 and log7 5 = 0.827. Evaluate each of the following:

Diberi bahawa log7 4 = 0.712 dan log7 5 = 0.827. Nilaikan setiap yang berikut:

(a) log7 1

= log7

= log7 5 – log7 4

= 0.827 – 0.712

= 0.115

(b) log7 28 = log7 (4 7)

= log7 4 + log7 7

= 0.712 + 1

= 1.712

(b) log7 100 = log7 (4 5 5)

= log7 4 + log7 5 + log7 5

= 0.712 + 0.827 + 0.827

= 2.366

(d) log7 0.25 = log7

= log7 1 – log7 4

= 0 – 0.712

= – 0.712

2 Evaluate each of the following without using a calculator.

Nilaikan setiap yang berikut tanpa menggunakan kalkulator.

(a) log3 21 + log3 18 – log3 14 = log3 [

]

= log3 27

= log3 33

= 3 log3 3

= 3

(b) 2 log4 2 –

log4 9 + log4 12 = log4 2

2 – log4

+ log4 12

= log4 4 – log4 3 + log4 12

= log4 [

]

= log4 16

= log4 42

= 2 log4 4

= 2

(c) log2 7 + log2 12 – log2 21 = log2 [

]

= log2 4

= log2 22

= 2 log2 2

= 2

3


1 Write the following expressions as single logarithms.

Tulis setiap ungkapan berikut sebagai logaritma tunggal.

(a) log2 x + log2 y2 = log2 [x y

2]

= log2 xy2

(b) logb x – 3 logb y = logb x – logb y3

= logb (

)

= logb

(c) log2 x + 3 log2 y = log2 x + log2 y3

= log2 [x y3]

= log2 xy3

(d)

log4 x + 2 – 3 log4 y

= log4

+ 2 log4 4 – log4 y3

= log4

+ log4 42 – log4 y

3

= log4 + log4 16 – log4 y3

= log4 (

)

= log4 (

)

(e) log3 m4 + 2 log3 n – log3 m = log3 m

4 + log3 n

2 – log3 m

= log3 (

)

= log3 m3n

2

2 Given log2 3 = p and log2 5 = q, express each of the following in terms of p and q.

Diberi log2 3 = p dan log2 5 = q, ungkapkan setiap yang berikut ke dalam sebutan-sebutan p dan q.

(a) log2 10 = log2 (2 5)

= log2 2 + log2 5

= 1 + q

(b) log2 45 = log2 (3 3 5)

= log2 3 + log2 3 + log2 5

= p + p + q

= 2p + q

(c) log2 = log2

=

log2 15

=

log2 (3 5)

=

[ log2 3 + log2 5]

=

(p + q)

Changing the base of logaritms Menukarkan asas logaritma

Formula: loga b =

Example / Contoh 1:

Change the log3 7 into log10

log3 7 =

Example / Contoh 2:

Change the log4 9 into log10

log4 9 =

Logarithms with base e are known as natural logarithms and are written as loge or ln Logaritma dengan asas e dikenali sebagai logaritma jati dan selalu ditulis sebagai loge atau ln

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1 Evaluate each of the following by converting it to base 10.

Nilaikan setiap yang berikut dengan menukarkannya kepada asas 10.

(a) log3 22 =

=

= 2.8137

(b) log6 1.32 =

=

= 0.1550

(c) log5 18 =

=

= 1.7959

(d) log4 0.815

=

=

= – 0.1475

2 Convert each of the following to natural logarithms and evaluate them.

Tukar setiap yang berikut kepada logaritma jati dan nilaikan mereka.

(a) log7 225 =

=

= 2.7833

(b) log9 8 =

=

= 2.6309

(c) log20 379 =

=

= 1.9820

3 Given log3 2 = t, express each of the following in terms of t.

Diberi log3 2 = t, ungkapkan setiap yang berikut ke dalam sebutan t.

(a) log2 9 =

=

=

=

=

(b) log9 8 =

=

=

=

(c) log2 18 = log2 (2 9) = log2 2 + log2 9

= 1 + log2 9

= 1 +

=

+

=

(d) log2

= log2 9 – log2 4

= log2 9 – log2 22

= log2 9 – 2 log2 2

= log2 9 – 2

=

– 2

=

–

=

4 If log2 m = a and log2 n = b, express each of the following in terms of a and b.

Jika log2 m = a dan log2 n = b, ungkapkan setiap yang berikut ke dalam sebutan-sebutan a dan b.

(a) log4 m2n

3 =

=

=

=

(b) log8

=

=

=

=

(c) logmn 8n =

=

=

=

=

5


1 Solve the following equations by giving answers in three decimal places.

Selesaikan seetiap persamaan berikut dengan memberi jawapan betul kepada tiga tempat perpuluhan.

(a) 42x – 1

= 7x

Taking log10 for both sides Ambil log10 untuk kedua-dua belah

log10 42x – 1

= log10 7x

(2x – 1) log10 4 = x log10 7

2x log10 4 – log10 4 = x log10 7

2x log10 4 – x log10 7 = log10 4

x(2 log10 4 – log10 7) = log10 4

x =

= 1.677

(b) 52x – 1

= 79x – 1


log10 52x – 1

= log10 79x – 1

(2x – 1) log10 5 = (x – 1) log10 79

2x log10 5 – log10 5 = x log10 79 – log10 79

2x log10 5 – x log10 79 = – log10 79 + log10 5

x(2 log10 5 – log10 79) = – log10 79 + log10 5

x =

= 2.399

(c) 73x – 1

= 50x


log10 73x – 1

= log10 50x

(3x – 1) log10 7 = x log10 50

3x log10 7 – log10 7 = x log10 50

3x log10 7 – x log10 50 = log10 7

x(3 log10 7 – log10 50) = log10 7

x =

= 1.011

2 Solve the following equations by using natural logarithms. Give the answer in three decimal places. Selesaikan setiap persamaan berikut dengan menggunakan logaritma jati. Beri jawapan betul kepada tiga tempat

perpuluhan.

(a) ln (5x + 2) = 15

loge (5x + 2) = 15

(5x + 2) = e15

5x + 2 = 3269017.375

5x = 3269017.375 – 2

5x = 3269015.375

x = 653803.075

(b) 30e2x + 3

= 145

Taking loge for both sides Ambil loge untuk kedua-dua belah

loge 30e2x + 3

= loge 145

loge [30 e2x + 3

] = loge 145

loge 30 + loge e2x + 3

= loge 145

loge 30 + (2x + 3) loge e = loge 145

loge 30 + (2x + 3) = loge 145

(2x + 3) = loge 145 – loge 30

2x = loge 145 – loge 30 – 3

x =

= – 0.712

(c) 5e3x – 4

= 35


loge 5e3x – 4

= loge 35

loge [5 e3x – 4

] = loge 35

loge 5 + loge e3x – 4

= loge 35

loge 5 + (3x – 4) loge e = loge 35

loge 5 + (3x – 4) = loge 35

(3x – 4) = loge 35 – loge 5

3x = loge 35 – loge 5 + 4

x =

= 1.982

(d) ln (3x – 2) = 4

loge (3x – 2) = 4

(3x – 2) = e4

3x – 2 = 54.598

3x = 56.598

x = 18.866

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(e) 41 – e2x

= 5

41 – 5 = e2x

36 = e2x

e2x

= 36


loge e2x

= loge 36

2x loge e = loge 36

2x = loge 36

x =

= 1.792

(f) ln (x + 1)2 = 4

loge (x + 1)2 = 4

(x + 1)2 = e

4

(x + 1) =

(x + 1) =

(x + 1) = e2 and x + 1 = – e

2

x = e2 – 1 x = – e

2 – 1

= 6.389 = – 8.389

3 The price of a house after n years is given by RM260 000 (

)

n. Determine the minimum number of years for

the price of the house to exceed RM300 000 for the first time.

Harga bagi sebuah rumah selepas n tahun diberi oleh RM260 000 (

)

n. Tentukan bilangan tahun minimum untuk harga

rumah itu melebihi RM300 000 untuk kali pertama.

Solution / Penyelesaian

RM260 000 (

)

n > RM300 000

260 000 (

)

n > 300 000

Taking log10 for both sides / Ambil log10 untuk kedua-dua belah

log10 260 000 (

)

n > log10 300 000

log10 [260 000 (

)

n] > log10 300 000

log10 260 000 + log10 (

)

n > log10 300 000

log10 260 000 + n log10 (

) > log10 300 000

n log10 (

) > log10 300 000 – log10 260 000

n >

n > 1.2150

Minimum year = 2 years

4 A company’s saving after n years is RM2 000(1 + 0.07)n. Determine the minimum number of years for their

savings to exceed RM4 000. Satu simpanan syarikat selepas n tahun ialah RM2 000(1 + 0.07)

n. Tentukan bilangan tahun minimum untuk

simpanannya melebihi RM4 000.


RM2 000(1 + 0.07)n > RM4 000

2 000(1.07)n > 4 000

(1.07)n > 2


log10 (1.07)n > log10 2

n log10 (1.07) > log10 2

n >

n > 10.2448

Minimum years = 11 years

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5 After n years, Mr. Chong’s money in a bank is RM4 000(1.1)n. Calculate the number of years for Mr.Chong’s

money to exceed RM5100 for the first time. Selepas n tahun, wang Mr. Chong di dalam bank ialah RM4 000(1.1)

n. Hitung bilangan tahun untuk wang Mr.Chong

melebihi RM5100 untuk kali pertama.


RM4 000(1.1)n > RM5100

4 000(1.1)n > 5100

(1.1)n > 1.275


log10 (1.1)n > log10 1.275

n log10 (1.1) > log10 1.275

n >

n > 2.5490

Minimum number of years = 3 years

6 The air pressure, in Hg, at a height of 10 km above sea level is given by P = 760e – 0.125h

, where h is the height,

in km and e = 2.718. Determine the height above sea level if the pressure at that height is 380 mm Hg. Tekanan udara, dalam Hg, pada ketinggian 10 km di atas paras laut diberi oleh P = 760e

– 0.125h, di mana h ialah

ketinggian, dalam km dan e = 2.718. Tentukan ketinggian di atas paras laut jika tekanan pada ketinggian itu ialah 380

mm Hg.


P = 760e – 0.125h

380 = 760e – 0.125h

760e – 0.125h

= 380

e – 0.125h

= 0.5

Taking loge for both sides / Ambil loge untuk kedua-dua belah

loge e – 0.125h

= loge 0.5

(– 0.125h) loge e = loge 0.5

– 0.125h = loge 0.5

h =

h = 5.545

Intensive Practice 4.3 Latihan Pengukuhan 4.3

1 Given log5 3 = 0.683 and log5 7 = 1.209. Without using a calculator or four-figure tables, calculate log5 1 and

log7 75. Diberi log5 3 = 0.683 dan log5 7 = 1.209. Tanpa menggunakan kalkulator atau jadual sifir empat angka, hitung log5 1

dan log7 75.


log5 1 = 0

log7 75 =

=

=

=

= 2.219

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2 Given loga 3 = x and loga 5 = y, express loga (

) in terms of x and y.

Diberi loga 3 = x dan loga 5 = y, ungkapkan loga (

) ke dalam sebutan-sebutan x dan y.


loga (

) = loga 45 – loga a3

= loga [5 3 3] – 3 loga a

= loga 5 + loga 3 + loga 3 – 3

= y + x + x – 3

= 2x + y – 3

3 Determine the value of log4 8 + logr .

Tentukan nilai bagi log4 8 + logr .


log4 8 + logr =

+ logr

=

+ logr

=

+

logr r

=

+

= 2

4 Without using a calculator or four figure tables, simplify

Tanpa menggunaakan kalkulator atau jadual empat angka, ringkaskan


=

=

=

=

=

=

=

=

5 Given log10 x = 2 and log10 y = – 1, prove that xy – 100y2 = 9

Diberi log10 x = 2 dan log10 y = – 1, buktikan bahawa xy – 100y2 = 9


log10 x = 2 log10 y = – 1

102 = x 10

– 1 = y

x = 100 y =

Therefore, xy – 100y2 = (100)(

) – 100(

)

2

= 10 – 1

= 9 [Proven]

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6 Given log5 2 = m and log5 7 = p, express log5 4.9 in terms of m and p.

Diberi log5 2 = m dan log5 7 = p, ungkapkan log5 4.9 ke dalam sebutan-sebutan m dan p.


log5 4.9 = log5 (

)

= log5 49 – log5 10

= log5 72 – log5 (2 5)

= 2 log5 7 – [ log5 2 + log5 5 ]

= 2p – [ m + 1 ]

= 2p – m – 1

7 Simplify log2 (2x + 1) – 5 log4 x2 + 4 log2 x

Ringkaskan log2 (2x + 1) – 5 log4 x2 + 4 log2 x


log2 (2x + 1) – 5 log4 x2 + 4 log2 x = log2 (2x + 1) – 5

+ 4 log2 x

= log2 (2x + 1) – 5

+ 4 log2 x

= log2 (2x + 1) – 5

+ 4 log2 x

= log2 (2x + 1) – 5 log2 x + 4 log2 x

= log2 (2x + 1) – log2 x5 + log2 x

4

= log2 [

]

= log2

= log2 (

)

= log2 (2 +

)

8 Given that log2 xy = 2 + 3 log2 x – log2 y, express y in terms of x. Diberi bahawa log2 xy = 2 + 3 log2 x – log2 y, ungkapkan y ke dalam sebutan x.


log2 xy = 2 + 3 log2 x – log2 y

log2 xy = 2log2 2 + log2 x3 – log2 y

log2 xy = log2 22 + log2 x

3 – log2 y

log2 xy = log2 4 + log2 x3 – log2 y

log2 xy = log2 [

]

log2 xy = log2 [

]

Comparing both sides / Bandingkan kedua-dua belah:

xy =

xy2 = 4x

3

y2 = 4x

2

y2 = (2x)

2

y = 2x

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9 Given log2 b = x and log2 c = y, express log4 (

) in terms of x and y.

Diberi log2 b = x dan log2 c = y, ungkapkan log4 (

) ke dalam sebutan-sebutan x dan y.


log4 (

) = log4 8b – log4 c

=

–

=

–

=

–

=

–

=

log2 8b –

log2 c

=

log2 [8 b] –

y

=

[ log2 8 + log2 b] –

y

=

[ log2 2

3 + log2 b] –

y

=

[ 3 log2 2 + log2 b] –

y

=

[ 3 + log2 b] –

y

=

[ 3 + x] –

y

=

[ 3 + x – y]

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10 The intensity of a sound, in decibel, is calculated by using the formula d = 10 log10 (

) where d is the

intensity of sound, in decibel, P is the intensity of sound, in Watt and Po is the weakest intensity of sound that

can be detected by the human ears, in Watt and it is a constant. In a house, a hot water pump has an intensity

of sound of 50 decibels and a wattage of 10 – 7

Watts while a dishwasher has a sound intensity of 62 decibels.

Kuasa bagi satu bunyi, dalam desibel, dikira dengan menggunakan formula d = 10 log10 (

) di mana d ialah kuasa

bunyi dalam unit desibel, P ialah kuasa bunyi, dalam Watt dan Po ialah kuasa terendah bagi bunyi yang boleh dikesan

oleh telinga manusia, dalam Watt dan ianya adalah malar. Di dalam sebuah rumah, satu pam air panas mempunyai

kadaran bunyi 50 desibel dan kadaran kuasa 10 – 7

manakala sebuah mesin pencuci pinggan mempunyai kadaran bunyi

62 desibel.

(a) Calculate the value of Po.

Hitung nilai Po.

(b) Determine the ratio of wattage, in Watts, for the dishwasher to the hot water pump.

Cari nisbah kadar bunyi, dalam Watts, untuk mesin pencuci pinggan kepada pam air panas.

(c) A wattage for sound that exceeds 100 Watts is said to be painful to the human ears. State the minimum

intensity of sound, in decibel, that is considered to be painful to the human ears. Kuasa bagi satu bunyi yang melebihi 100 Watt dikatakan menyakiti telinga manusia. Nyatakan kuasa minimum

bagi satu bunyi, dalam desibel, yang dianggap menyakiti telinga manusia.


(a) From d = 10 log10 (

) d = 50 decibels P = 10

-7 Watts

50 = 10 log10 (

)

5 = log10 (

)

5 = log10 10 – 7

– log10 Po

5 = – 7 log10 10 – log10 Po

5 = – 7 – log10 Po

log10 Po = – 7 – 5

log10 Po = – 12

Po = 10 – 12

(b) For dishwasher For hot water pump

d = 10 log10 (

) d = 10 log10 (

)

62 = 10 log10 (

) 50 = 10 log10 (

)

Ratio of dishwasher to hot water pump = 62 : 50

= 31 : 25

(c) From d = 10 log10 (

) P = 100 Watts

d = 10 log10 (

)

d = 10 log10 (

)

d = 10 log10 102 + 12

d = 10 log10 10 14

d = 10 (14) log10 10

d = 140 decibels

12

11 The population growth in a certain country is P = 2 500 000 e0.04t

where t is the number of years after year

2020 and e = 2.718. Pertumbuhan populasi di sebuah negara ialah P = 2 500 000 e

0.04t di mana t ialah bilangan tahun selepas tahun 2020

dan e = 2.718.

(a) What is that country’s population in 2020?

Berapakah populasi sebuah negara pada tahun 2020?

(b) What is that country’s population in 2030?

Berapakah populasi sebuah negara pada tahun 2030?

(c) In which year will that country’s population exceed 50 000 000?

Dalam tahun manakah sesebuah populasi negara melebihi 50 000 000?


(a) In 2020, t = 0

From P = 2 500 000 e0.04t

= 2 500 000 e0.04(0)

= 2 500 000 e0

= 2 500 000

(b) In 2030, t = 10

From P = 2 500 000 e0.04t

= 2 500 000 e0.04(10)

= 2 500 000 e0.4

= 3729407.0685

= 3729407 [nearest integer]

(c) From P = 2 500 000 e0.04t

50 000 000 = 2 500 000 e

0.04t

20 = e

0.04t

Taking loge for both sides

loge 20 = loge e0.04t

loge 20 = 0.04t loge e

loge 20 = 0.04t

0.04t = loge 20

t =

= 74.8933

= 75 [Nearest year]

So in year = 2020 + 75

= 2095

13


1 A gardener observes a bug infestation towards plants in his garden. He finds out that the area of insect

infestation towards his plants is A = 1000 20.7n

acres, where n is the amount of weeks after the first week of

initial observation. How long will it take for the insects to infest an area of 5 000 hectares? Seorang pekebun memantau serangan serangga terhadap tanaman di kebunnya. Dia mendapati bahawa serangan

serangga terhadap luas tanaman diberi oleh persamaan A = 1000 20.7n

hektar, dengan n ialah bilangan minggu

selepas minggu pertama pemantauan dibuat. Berapakah tempoh masa yang diambil oleh serangga untuk menyerang

kawasan seluas 5000 hektar?


From A = 1000 20.7n

5000 = 1000 20.7n

5 = 20.7n

Taking log10 for both sides / Ambil log10 untuk kedua-dua belah:

log10 5 = log10 20.7n

log10 5 = 0.7n log10 2

0.7n =

0.7n = 2.3219

n = 3.3170

= 3 weeks + 0.3170 7 days

= 3 weeks 2 days

2 The electric current that flows in an electrical circuit for t seconds after its switch is turned off is I = 32 4 – t

amp.

Arus elektrik yang mengalir dalam satu litar elektrik, t saat selepas suisnya ditutup diberi oleh I = 32 4 – t

amp.

(a) Calculate the current flow when the switch is off.

Hitung arus yang mengalir ketika suisnya ditutup.

(b) Calculate the current flow after Berapakah arus yang mengalir selepas

(i) 1 seconds / 1 saat (ii) 2 seconds / 2 saat

(c) How long will it take for the current to reach 0.5 amps?

Berapakah masa yang diambil untuk arus mencapai 0.5 amp?


(a) From I = 32 4 – t

When the switch is just off / Apabila suis baru ditutup: t = 0

I = 32 4 – 0

= 32 amps

(b) (i) From I = 32 4 – t

= 32 4 – 1

= 32

= 8 amps

(ii) From I = 32 4 – t

= 32 4 – 2

= 32

= 32

= 2 amps

(c) From I = 32 4 – t

0.5 = 32 4 – t

0.5 = 32 4 – t

0.015625 = 4 – t


log10 0.015625 = log10 4 – t

log10 0.015625 = – t log10 4

– t log10 4 = log10 0.015625

t log10 4 = – log10 0.015625

t = –

= 3 saat

14

Intensive Practice 4.4 Latihan Pengukuhan 4.4

1 Mr. Ramasamy keeps RM1000 in a bank. The amount of money rises by W = 1000(1.09)t after t years.

Calculate Mr. Ramasamy menyimpan RM1000 di dalam sebuah bank. Jumlah wang menaik dengan W = 1000(1.09)

t selepas t

tahun. Hitung

(a) the amount of money after 5 years,

jumlah wang selepas 5 tahun,

(b) the time taken, t, in years, for the money to rise from RM1000 to RM1200. masa yang diambil, t, dalam tahun, untuk wang itu naik dari RM1000 kepada RM1200.


(a) From W = 1000(1.09)t

= 1000(1.09)5

= 1538.62

(b) From W = 1000(1.09)t

1200 = 1000(1.09)t

1.2 = (1.09)t


log10 1.2 = log10 (1.09)t

log10 1.2 = t log10 1.09

t log10 1.09 = log10 1.2

t =

t = 2.1156 years

2 The remaining radioactive substance of uranium after t years is W(t) = 50 2 – 0.0002t

gram, where t 0.

Bahan radioaktif yang tinggal untuk uranium selepas t tahun ialah W(t) = 50 2 – 0.0002t

gram, di mana t 0.

(a) Determine the initial mass of the uranium.

Tentukan jisim awal bagi uranium itu.

(b) Determine the time that is needed for the uranium to weigh 8 grams.

Tentukan masa yang diperlukan untuk uranium itu berjisim 8 grams.


(a) From W(t) = 50 2 – 0.0002t

gram

At initial mass, t = 0 / Pada jisim awal, t = 0

W(t) = 50 2 – 0.0002(0)

W(t) = 50 2 – 0

W(t) = 50 grams

(b) From W(t) = 50 2 – 0.0002t

gram

8 = 50 2 – 0.0002t

0.16 = 2 – 0.0002t


log10 0.16 = log10 2 – 0.0002t

log10 0.16 = – 0.0002t log10 2

– 0.0002t log10 2 = log10 0.16

t (– 0.0002log10 2) = log10 0.16

t =

= 13219.2810 years

15

3 The mass, J of a bacteria after time t, in hours is J = 25 e0.1t

gram.

Jisim, J untuk satu bakteria selepas masa t, dalam jam ialah J = 25 e0.1t

gram.

(a) Show that the time taken for the bacterial mass to reach 50 grams is 10 ln 2 hours.

Tunjukkan bahawa masa yang diambil untuk jisim bakteria mencapai 50 gram ialah 10 ln 2 jam.

(b) Determine the time taken in two decimal places.

Tentukan masa yang diambil kepada dua tempat perpuluhan.


(a) From J = 25 e0.1t

gram

50 = 25 e0.1t

2 = e0.1t

Taking loge for both sides / Ambil loge untuk kedua-dua belah:

loge 2 = loge e0.1t

loge 2 = 0.1t loge e

loge 2 = 0.1t

0.1t = loge 2

0.1t = ln 2 ingat bahawa loge = ln

t =

ln 2

t = 10 ln 2

(b) From t = 10 ln 2

= 6.93 hours

Mastery Practice Praktis Masteri

1 Solve the equation 42x – 1

+ 42x

= 4 Selesaikan persamaan 4

2x – 1 + 4

2x = 4


42x – 1

+ 42x

= 4

42x 4

– 1 + 4

2x = 4

42x

[4– 1

+ 1] = 4

42x

[

+ 1] = 4

42x

[

] = 4

42x

= 4

42x

=

42x

= 3.2


log10 42x

= log10 3.2

2x log10 4 = log10 3.2

x (2 log10 4) = log10 3.2

x =

= 0.4195

2 Solve the equation 5n + 1

– 5n + 5

n – 1 = 105

Selesaikan persamaan 5n + 1

– 5n + 5

n – 1 = 105


5n + 1

– 5n + 5

n – 1 = 105

5n 5

1 – 5

n + 5

n 5

– 1 = 105

5n [5

1 – 1 + 5

– 1] = 105

5n [5 – 1 +

] = 105

5n [

] = 105

5n = 105

5n = 25

5n = 5

2

Comparing the indices / Bandingkan indeks: n = 2

16

3 If x = x + , find the value of x in the form of

.

Jika x = x + , cari nilai x dalam bentuk

.


x = x +

x – x =

x( – ) =

x =

x =

x =

x =

x =

x =

4 If logx a + logx

= t, what is the possible value(s) of t?

Jika logx a + logx

= t, apakah nilai yang mungkin bagi t?


logx a + logx

= t

logx [a

] = t

logx 1 = t

t = 0 since logx 1 = 0 for any positive values of x

5 The diagram below shows three circles. Cirlce A has a radius of 2 cm and circle B has a radius of 1 cm.

Rajah bawah menunjukkan tiga bulatan. Bulatan A mempunyai jejari 2 cm dan bulatan B mempunyai jejari 1 cm.

PQ is a common tangent and all circles touch one another. Find the radius of the smallest circle.

PQ ialah tangent sepunya dan semua bulatan bersentuh antara satu sama lain. Cari jejari bulatan yang paling kecil itu.


Let the radius of the smallest circle / Andaikan jejari bulatan yang paling kecil = r

Distance TB : AT

2 + TB

2 = AB

2

12 + TB

2 = (2 + 1)

2

1 + TB2 = 9

TB2 = 8

TB =

17

TC + CD =

+ =

+ =

+ =

+ =

Taking square for both sides: ( + )2 = ( )

2

( + )( + ) = 8

8r + + + 4r = 8

12r + 2 = 8

12r + 2 = 8

12r + 2 = 8

12r + 2r = 8

12r + 2r = 8

12r + 2r (4) = 8

12r + 8r = 8

r (12 + 8 ) = 8

r =

r =

18

6 The temperature of a type of metal decreases from 100C to TC according to T = 100(0.9)x after x second.

Calculate Suhu bagi sejenis logam berkurang daripada 100C kepada TC mengikut T = 100(0.9)

x selepas x saat. Hitung

(a) the temperature of the metal after 5 seconds,

suhu bagi logam itu selepas 5 saat,

(b) The time taken, x, in seconds for the temperature of the metal to decrease from 100C to 80C.

Masa diambil, x, dalam saat untuk suhu logam itu berkurang daripada 100C kepada 80C.

Solution / Penyelesaian:

(a) From T = 100(0.9)x

= 100(0.9)5

= 59.049 C

(b) From T = 100(0.9)x

80 = 100(0.9)x

0.8 = (0.9)x


log10 0.8 = log10 (0.9)x

log10 0.8 = x log10 0.9

x log10 0.9 = log10 0.8

x =

x = 2.1179

= 2.12 seconds

7 After n years, the price of a car that was brought by Raju is RM60 000(

)

n. Determine the number of years for

the price of the car to be below RM20000 for the first time.

Selepas n tahun, harga sebuah kereta yang dibeli oleh Raju ialah RM60 000(

)

n. Tentukan bilangan tahun

untuk harga kereta itu di bawah harga RM20000 untuk kali pertama.


From RM60 000(

)

n

RM60 000(

)

n < RM20000

60 000(

)

n < 20000

(

)

n < 0.3333


log10 (

)

n < log10 0.3333

n log10 (

) < log10 0.3333

n (– 0.05799) < – 0.4772

n (0.05799) > 0.4772

n > 8.23

After 9 years, the car will be below RM20 000

19

8 Given logx 3 = s and 9 = t, express log9 x3y in terms of s and / or t.

Diberi logx 3 = s dan 9 = t, ungkapkan log9 x3y ke dalam sebutan s dan / atau t.


logx 3 = s 9 = t,

xs = 3 ( )

t = 9

x =

yt = 81

yt = 3

4

y =

log9 x3y =

=

=

=

=

=

=

=

=

=

+

=

+

20

9 Two experiments were carried out to find the relationship between the variables x and y. Both experiments

showed that the relationship between x and y is in accordance to 3(9x) = 27

y and log2 y = 2 + log2(x – 2). Find

the value of x and y that satisfy both experiments. Dua eksperimen dijaalankan untuk mengkaji hubungan antara pembolehubah x dan y. Kedua-dua eksperimen

menunjukkan bahawa hubungan antara x dan y adalah mengikut 3(9x) = 27

y dan log2 y = 2 + log2(x – 2). Cari nilai x dan

y yang memuaskan kedua-dua eksperimen ini.


From 3(9x) = 27

y

3 32x

= 33y

31 + 2x

= 33y

Comparing the indices for both sides / Bandingkan indeks untuk kedua-dua belah:

1 + 2x = 3y ----------(1)

From log2 y = 2 + log2(x – 2)

log2 y = 2log2 2 + log2(x – 2)

log2 y = log2 22 + log2(x – 2)

log2 y = log2 4 + log2(x – 2)

log2 y = log2 [4 (x – 2)]

log2 y = log2 4(x – 2)

Comparing both sides / Bandingkan kedua-dua belah:

y = 4(x – 2) ---------(2)

Substitute (2) into (1): 1 + 2x = 3[4(x – 2)]

1 + 2x = 12(x – 2)

1 + 2x = 12x – 24

1 + 24 = 12x – 2x

25 = 10x

10x = 25

x =

x =

Substitute x =

into (2): y = 4[(

) – 2]

= 2

10 The price of a car drops and can be determined with the equation x log10(

) = log10 p – log10 q. In this

equation, the car with y years of usage and price RMq will drop to RMp after being used for x years. A car is

bought at RM100 000 has 20 years of usage. If the price of the car drops to RM10 000, find the years of usage

for the that car.

Harga sebuah kereta menyusut dan boleh ditentukan dengan persamaan x log10(

) = log10 p – log10 q. Dalam

persamaan ini, kereta dengan y tahun guna dan harga RMq akan jatuh kepada RMp selepas digunakan untuk x tahun.

Sebuah kereta dibeli dengan harga RM100 000 mempunyai 20 tahun guna. Jika harga kereta itu jatuh kepada RM10

000, cari bilangan tahun telah guna bagi kereta itu.


Price RM100 000 drops to RM10 000 q = 100 000 and p = 10 000

RM100 000 has 20 years of usage y = 20

From x log10(

) = log10 p – log10 q

x log10(

) = log10 10 000 – log10 100 000

x log10(0.9) = log10 104 – log10 10

5

x log10 0.9 = 4 log10 10 – 5 log10 10

x log10 0.9 = 4 – 5

x log10 0.9 = – 1

x = –

= 21.85

HAPPY REVISION. SCORE A+ IN ADDITIONAL MATHEMATICS SPM 2021

Indices, Surds And Logarithms Part 3 - Buncee

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