Bab 4 Hukum Indeks,Surd dan Logaritma

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tracker-software 4BAB Indeks, Surd dan Logaritma tracker-software
Modul
Indices, Surds and Logarithms PBD

4.1 Hukum Indeks/ Laws of Indices

1 Cari nilai setiap yang berikut. i-Kuiz
Find the value of each of the following. SP 4.1.1

CONTOH (a) 34 = 81 (b) (−5)2 = 25

⎝⎛⎜121 ⎞2 = ⎛ 3 ⎞2 = 9
⎠⎟ ⎝⎜ 2 ⎠⎟ 4

(c) ⎛ − 2 ⎞3 = 8 (d) ⎛ 2 1 ⎞3 = ⎛ 9 ⎞3 = 729 (e) (0.03)2 = 0.0009
⎜⎝ 3 ⎟⎠ 27 ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎠⎟ 64

Tip Pintar

Nombor dalam bentuk indeks dengan asas 10 digunakan untuk mewakili nombor amat besar atau amat kecil.
Numbers in indices form with base 10 are used to represent very large or very small numbers.
Contoh/Example: Jisim Bumi ialah 5.972 × 1024 kg. Jisim elektron ialah 9.109 × 10 31 kg.
The mass of Earth is 5.972 × 1024 kg. The mass of electron is 9.109 ×10 31 kg.

2 Cari nilai setiap yang berikut. (a) (−4)0 = 1
Find the value of each of the following. SP 4.1.1

CONTOH

⎛ 3 2 ⎞−2 = 1 = 1 = 9
⎝⎜ 3 ⎟⎠ ⎛ 11⎞2 ⎛ 121⎞ 121
⎜⎝ 3 ⎠⎟ ⎝⎜ 9 ⎠⎟

(b) (7)−2 = 1 = 1 (c) ⎛ − 3 ⎞−3 = 1 = 1 = 64
49 ⎜⎝ 4 ⎟⎠ 27
(7)2 ⎛ 3 ⎞3 ⎛ − 27 ⎞
⎜⎝ − 4 ⎟⎠ ⎝⎜ 64 ⎟⎠

(d) ⎛⎜⎝153 ⎞−3 = 1 = 1 = 125 (e) (0.2)−2 = 1 = 1 = 25
⎠⎟ ⎛ 8 ⎞3 ⎛ 512 ⎞ 512 0.04
⎜⎝ 5 ⎟⎠ ⎝⎜ 125 ⎠⎟ (0.2)2

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tracker-software 3 Cari nilai setiap yang berikut.

Find the value of each of the following. SP 4.1.1

CONTOH 1

1 (a) (8)3 = 3 8 = 2

⎛ 2 1 ⎞ 2 = 9 = 3
⎜⎝ 4 ⎟⎠ 4 2

( )3 9 3 = 33 = 27 1 1⎞ 1
⎛ 1 ⎞4 81⎠⎟ 3
(b) (9)2 = (c) ⎜⎝ 81⎟⎠ = 4 ⎛ =
⎝⎜

1 (e) ⎛ 64 ⎞− 2 = 1 = 1 = 1
3 ⎛ 4 ⎞2
(d) (−216)3 = 3 216 = – 6 2 ⎝⎜ 5 ⎟⎠
⎝⎜ 125 ⎠⎟ ⎛ 3 64 ⎞2
4 Cari nilai setiap yang berikut. ⎛ 64 ⎞3 ⎜⎝⎜ 125 ⎠⎟⎟
Find the value of each of the following. SP 4.1.2 ⎜⎝ 125 ⎠⎟
CONTOH
35 ×32 = 35+2 = 37 = 2187 = 1 = 25
⎛ 16 ⎞ 16
(b) 32 ÷3−3 = 32−(−3) = 35 = 243 ⎜⎝ 25 ⎠⎟

(a) 27 × 2−5 = 27+(−5) = 22 = 4

( )(c) 24 2 28 = 256

=

642( )(d( 11 64 = 8 (e) ⎛ 4 ⎞3 ÷ ⎛ 4 ⎞5 = ⎛ 4 ⎞3−5 = ⎛ 4 ⎞−2 = 25 = 1196
⎜⎝ 5 ⎠⎟ ⎝⎜ 5 ⎠⎟ ⎜⎝ 5 ⎟⎠ ⎝⎜ 5 ⎟⎠ 16
4 = (64)2 =

Tip Kalkulator

Untuk mencari nilai bagi/To find the value of ⎛ 4 ⎟⎠⎞−2, tekan/press ( 4 ab 5 ) ^ ( – 2 ) =
Paparan/The display 1196 akan ⎜⎝ 5 c

muncul/will appear.

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tracker-software 5 Ringkaskan yang berikut.

Simplify the following. SP 4.1.2

CONTOH (a) 12p2q5 × pq3 = 12p2 1q5 3 (b) (2ef 3)3 = 8e3f 9
= 12p3q8
23n u24n = 23n+4n−5n
25n

22n

(c) 2m4n5 = (d) a5b × a2b = a5b+2b−(−4b) (e) 521(e+7) u 521(3e−1) =
6m5n2 a−4b 521(e+7+3e−1)
= 521(4e+6)
( )1 a11b
= 52e 3
3
m4−5n5−2

( )=1 n3
3 m−1n3 = 3m

6 Buktikan setiap yang berikut. ( )(a) 33m+4 = 81 27m
Prove each of the following. SP 4.1.2

CONTOH

9x = 81⎣⎢⎡ m⎤
27 ( )( ) ( ) ( )33m ⎦⎥
( )32x−3= 32 x 34 33 = 81 27m
27
32x−3 = 32x = = 9x
33 27

(b) 23a−1 = 8a (c) 52−3x = 25
2 125x
( )52 25 25
( )23a 23 a 8a = 53 x = 125x
2 2 53x
21 = =

7 Ringkaskan setiap yang berikut. (a) 5x+2 − 5x =
Simplify each of the following. . SP 4.1.2
CONTOH ( )5x+2 − 5x = 5x 52 − 5x

2x − 2x+3 = = 5x(25 − 1)

( )2x − 2x+3 = 2x − 2x 23 = 2x(1− 8) ( )= 24 5x
( )= −7 2x

(b) 2y + 2y+2 − 2y+3 = (c) 9b − 32b+1 =

2y + 2y+2 − 2y+3 ( )9b − 32b+1 = 9b − 32b 31

( ) ( )= 2y + 2y 22 − 2y 23 9b( ) ( )=−3 ⎡ 32 b⎤ = 9b −3 9b
⎢⎣ ⎥⎦
= 2y(1+ 4 − 8)
= 9b(1− 3)
( )= −3 2y
( )= −2 9b

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tracker-software 4.3 Hukum Logaritma/ Laws of Logarithms tracker-software

1 Tulis setiap yang berikut dalam bentuk logaritma. Tip Pintar
Write each of the following in logarithmic form. SP 4.3.1 · log10 x = log x lg x
· loge x ln x
CONTOH 1 (b) 10−2 = 1
100
53 125 (a) 83 2
log5125 3
log8 2 1 log10 ⎛ 1 ⎞ = −2
3 ⎜⎝ 100 ⎟⎠

(c) e1.5 4.4817 (d) px q (e) ey z Tip Pintar
loge 4.4817 1.5 logpq x logez y
llooggaerditikmeanjaalti is.eNbilaagi abiagi
e ialah 2.718281828...
(nombor bukan nisbah).
llooggaeriisthkmno. Twhneavsanluaetuorfael is
2.718281828...
(an irrational number).

2 Tulis setiap yang berikut dalam bentuk indeks.
Write each of the following in index form. SP 4.3.1

CONTOH (a) log9 27 3 (b) log10 (0.00001) (c) logab c (d) logem n
2 en m
log2128 7 = −5 ac b
27 128 3 10−5 = 0.00001

92 27

3 Cari nilai setiap yang berikut dengan menggunakan kalkulator.
Find the value of each of the following using the calculator. SP 4.3.1

CONTOH (a) log101 0 (b) loge 8 2.0794 (c) log1010 1

log108 0.9031

(d) loge 20 2.9957 (e) log10 23.4 1.3692 (f) log10 (0.00082) = (g) loge 0.009 =
−4.7105
−3.0862

(h) log10 ⎛ 1 ⎞ = Tip Pintar
⎝⎜ 5 ⎟⎠
N = ax adalah sentiasa positif bagi semua nilai x jika a > 0. Maka,
log10 0.2 = −0.6990 N = ax is always positive for all values of x if a > 0. Therefore,

t loga 0 adalah tak tertakrif/is undefined.
t logaritma bagi nombor negatif adalah tak tertakrif.

logarithm of a negative number is undefined.

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tracker-software 4 Cari nilai setiap yang berikut.

Find the value of each of the following. SP 4.3.1

CONTOH (a) log4 32 + log4 2 = (b) 2log26 − log29 =

log654 + log6 4 log4 (32× 2) log262 −log29

= log6 (54 × 4) =log 4 64 = log2 36
= log4 43 9
=log6 216 = log663
= 3log4 4 = 3(1) = 3 = log2 4 = log2 22 = 2
= 3log66 = 3(1) = 3

(c) log6 30 − log6 1 − log6 45 = (d) log318 + log36 − 2log3 2 = (e) 3log36 + log312 − 5log3 2 =
54
log3 18 × 6 log3 63 ×12
4 25
log6 30

⎛ 1 ⎞ × 45 = log3 27 = log333 = 3 ( )= log3 33 × 3 = log334 = 4
⎜⎝ 54 ⎟⎠

= log6 30 × 54
45

= log6 (6 × 6) = 2log6 6 = 2

5 Diberi log25 2.322 dan log27 2.807. Cari nilai setiap yang berikut.

Given log2 5 2.322 and log2 7 2.807. Find the value of each of the following. SP 4.3.2

CONTOH (a) log2 20 = (b) log2 8 =
7
log270 = log2 (2× 5× 7) log2 (4 × 5)
log28 − log27
= log22 log25 log27 log2 4 + log25
= 1 + 2.322 + 2.807 = log223 − 2.807
= 6.129 = log2 22 + 2.322
= 2 + 2.322 = 3 − 2.807 = 0.193
= 4.322

(c) log21.4 = 1 (e) log2700 = log2 (4 × 25× 7)

⎛ 7 ⎞ (d) log2 35 = log2 (35)2 = log2 22 log252 log27
⎝⎜ 5 ⎠⎟
log 2 = 1 log2 35 = 1 log2 (7 × 5) = 2 + 2log25 + 2.807
2 2 = 4.807 + 2(2.322)
log27 − log25 = 9.451
= 2.807 − 2.322 = 1 ( 2.807 + 2.322)
2
= 0.485 = 2.5645

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tracker-software 6 Dberi log23 m dan log25 n. Ungkapkan setiap yang berikut dalam sebutan m dan n.
Given log2 3 m and log2 5 n. Express each of the following in terms of m and n. SP 4.3.3

CONTOH (a) log215 = log2 (3× 5) (b) log20.6 = log2 3
5
log245 = log2 (9 × 5) = log23 + log25
=m+n = log23 − log25
= log232 + log25
=m−n

= 2log23 + log25

= 2m + n

(c) log275 = log2 (25× 3) (d) log2150 = (e) log2 6 =
125
= log252 + log23 log2 (25× 3× 2)
= 2log25 + m log26 − log2125
= 2n + m = log252 + log23 + log2 2
= 2log25 + m + 1 = log2 (3× 2) − log253
= 2n + m + 1
= log23 + log2 2 − 3log25

= m − 3n + 1

7 Dengan menggunakan kaedah penukaran asas, cari nilai setiap yang berikut.
By using the method of changing the base, find the value of each of the following. SP 4.3.4

CONTOH (a) log813 log3 3 (b) log16 0.125 =
log381
log8 2 log 2 2 log16 ⎛ 1 ⎞
log28 1 ⎝⎜ 8 ⎠⎟
log3 34
1 1 = log161− log16 8
log 2 23
1 4 = 0 − log28 = − log2 23
log216 log2 24
3
= − 3
4

(c) log 6 216 = log6 216 (d) log717 log1017 (e) log 5 8 log10 8
log6 6 log10 7 log10 5
1.230 0.9031
= log6 63
0.845 0.3495
1 1.456 2.584

log6 6 2

= ⎛ 3 AWAS!
1⎞
⎝⎜ 2 ⎠⎟
b logab
=6 loga c ϶ logac

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tracker-software 8 Diberi log35 a. Ungkapkan setiap yang berikut dalam sebutan a.
Given log3 5 a. Express each of the following in terms of a.

CONTOH (a) log9625 log3625 (b) log9 25 =
log39 27
log33 log9 25 −log9 27
log53 log35 log354 log3 25 log3 27
log332 log39 log3 9
1 4a = −

log35 2 = log3 52 − log3 33
1 2a log3 32 log3 32
2a − 3
a = 2

9 Tanpa kalkulator, cari nilai bagi setiap yang berikut.
Without using a calculator, find the value of each of the following.

CONTOH (a) log3125× log5 243 =

log56 × log45× log616 = log10125 × log10 243
log10 6 log10 5 log1016 log10 3 log10 5
log10 5 × log10 4 × log10 6
log10 53 log10 35
log10 42 = log10 3 × log10 5
log10 4
=

2log10 4 = 3×5
log10 4
= = 15

=2

(b) log7 64 × log4 9 × log3 49 = (c) log2 27 × log312 =
log212
log10 64 × log10 9 × log10 49
log10 7 log10 4 log10 3 log10 27 log1012
log10 2 × log10 3
log10 43 log10 32 log10 72
= log10 7 × log10 4 × log10 3 ⎛ log1012 ⎞
⎜ log10 2 ⎟
= 3× 2× 2 ⎝ ⎠

= 12 = log10 27 × log1012 × log10 2
log10 2 log10 3 log1012

= log10 33
log10 3

=3

4.4 Aplikasi Indeks, Surd dan Logaritma
Applications of Indices, Surds and Logarithms

1 Selesaikan persamaan berikut. ( )(a) 3x 32x = 9
Solve the following equations. SP 4.4.1

CONTOH

( )52n+1 5n = 1 3x+2x = 32
25
3x = 2
52n+1+n = 5−2
x = 2
3n + 1= −2 3

n = −3
3
= −1

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tracker-software ( )(b) 92x 43x = 72 +1 tracker-software
(c) 4x+4 − 8 = 64

( )3 2x 2x+4 − 2x+3 = 64

92x 42 = 72 ( ) ( )24 2x − 23 2x = 64

( )92x 82x = 72 (16 − 8) 2x = 64

(9 × 8)2x = 72 2x = 8
x =3
2x = 1

x = 1
2

2 Cari nilai a dan b bagi setiap persamaan yang berikut.
Find the values of a and b of the following equations. SP 4.4.1

CONTOH 1 3−3 2 = a+b
( )(a) 3 2 3
( )2 2 − 1 2 = a + b 2
( )( )2 2 − 1 2 2 − 1 = a + b 2 ( )1 = a+b 3

8 − 4 2 +1= a + b 2 3
9−4 2 =a+b 2 12 − 12 3+9

∴ a = 9, b = −4 ( )1 3 = a+b 3

3
21− 12

7−4 3 =a+b 3

∴ a = 7, b = −4

(b) 4+ 2 =a+b 2 (c) 7 − 1 = a + b 7
2− 2 7 + 3

(4+ 2)(2+ 2) ( 7 −1)( )7 − 3
) =a+b 2 ) =a+b 7
(2− 2)(2+ ( 7 + 3)(
2 7 −3

10 + 6 2 = a+b 2 10 − 4 7 = a+b 7
4−2 7−9

5+3 2 = a+b 2 −5 + 2 7 = a + b 7

∴ a = 5, b = 3 ∴ a = −5, b = 2

3 Selesaikan persamaan berikut. (a) 3x − 9x2 − 32 = 4
Solve the following equations. SP 4.4.1
3x − 4 = 9x2 − 32
CONTOH Kuasaduakan kedua-dua belah/Squaring both
sides:
1+ x + 3 = 2x − 1
Kuasaduakan kedua-dua belah/Squaring 9x2 − 24x + 16 = 9x2 − 32
both sides: 24x = 48
x=2
1+ 2 x + 3 + ( x + 3) = 2x − 1

2 x+3 = x−5
Kuasaduakan kedua-dua belah/Squaring
both sides:

4( x + 3) = x2 − 10x + 25
x2 − 14x + 13 = 0

( x − 13)( x − 1) = 0

x 13 atau/or x 1
Semak/Check: x 1 tidak memuaskan
persamaan./does not satisfy the equation.
? Jawapannya ialah/The solution is x 13.

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(b) 4x − 15 = 2 x − 1 (c) 2x + 3 − x + 1 = x − 2 tracker-software
Kuasaduakan kedua-dua belah/Squaring both Kuasaduakan kedua-dua belah/Squaring both
sides: sides:

4x − 15 = 4x − 4 x + 1 (2x + 3) −2 (2x + 3)( x + 1) + ( x + 1) = x − 2

4 x = 16 x + 3 = (2x + 3)( x + 1)

x =4 Kuasaduakan kedua-dua belah/Squaring both
x = 16 sides:

x2 + 6x + 9 = 2x2 + 5x + 3

x2 − x − 6 = 0

(x − 3)(x + 2) = 0

x 3 atau/or x = −2
Semak: Apabila x = −2 digantikan dalam
persamaan itu, suatu nombor khayalan akan
diperoleh.
Chec k: When x = – 2 is substituted into the
equation, an imaginary number is formed.

? x = 3.

4 Selesaikan persamaan berikut. (a) loge x 4logxe
Solve the following equations. SP 4.4.1

CONTOH

2x+3 = 3x+2 loge x = logxe4

log10 2x+3 = log10 3x+2 loge x = logee4
loge x
( x + 3)log10 2 = ( x + 2)log103
(loge x)2 = 4
xlog10 2 − xlog10 3 = 2log10 3 − 3log10 2
loge x = ±2
x (log10 2 − log10 3) = 2log10 3 − 3log10 2

x = ( 2log10 3 − 3log10 2) x e2 atau/or x = e−2
(log10 2 − log10 3) x = 7.389 atau/or x = 0.135

= 0.9542 − 0.9031
0.3010 − 0.4771

= −0.29

(b) loge ⎛ x ⎞ = 10 − 3loge x (c) (log 2 3x ) = log236
⎝⎜ e2 ⎠⎟ (log9 3)

loge x − logee2 = 10 − 3loge x log23x = log9 3× log236
4loge x = 10 + 2logee
4loge x = 12 = (log33) × log262
loge x = 3 (log39)

x e3 = 20.086 = 1 × 2log26
2

= log26

3x 6

x2

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3x − 2 52x (e) log35x = 2 + log3 ( x − 4)
(d) = 33x

( )( )33x 3x−2 =52x log35x − log3 ( x − 4) = 2

34x−2 = 52x log3 5x = 2log3 3
x−4

(4x − 2)log103 = 2xlog105 5x = 9
x−4
4xlog10 3 − 2xlog10 5 = 2log10 3
5x = 9x − 36
x (4log103 − 2log105) = 2log103
4x = 36

x = 2log10 3 x=9
4log10 3 − 2log10
5

x = 1.87

Praktis Pentaksiran 4

Kertas 1

1 Selesaikan/Solve 2x − 4x2 − 9 = 1. x 2 atau/or x = − 1
3

2x − 4x2 − 9 = 1 Semak: Apabila x = − 1 digantikan dalam
3
2x − 1= 4x2 − 9 persamaan itu, terdapat nombor khayalan.

Kuasaduakan kedua-dua belah/Squaring both Check: When x = −an31imisasguibnsatrityunteudmibnetor.the
sides: equation, there is

4x2 − 4x + 1= 4x2 − 9

4x = 10 ? x 2.

x = 5
2
3 Selesaikan/solve
2 Selesaikan/Solve y−8 = y −2
4x + 1 − 3x − 2 = x − 1
y−8 = y −2
Kuasaduakan kedua-dua belah/Squaring both
sides: y − y−8 = 2
Kuasaduakan kedua-dua belah/Squaring both
(4x + 1) − 2 (4x + 1)(3x − 2) + (3x − 2) sides:

= x −1 y − 2 y (y − 8) + (y − 8) = 4

6x = 2 (4x + 1)(3x − 2) 2y − 12 = 2 y (y − 8)

3x = (4x + 1)(3x − 2) y − 6 = y (y − 8)

Kuasaduakan kedua-dua belah/Squaring both Kuasaduakan kedua-dua belah/Squaring both
sides: sides:
y2 − 12y + 36 = y2 − 8y
9x2 = 12x2 − 5x − 2
3x2 − 5x − 2 = 0 4y = 36
y =9
( x − 2)(3x + 1) = 0

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tracker-software 4 Diberi 2x m, 23y n dan 2x+3y = 8 + 8y. x ⎛ y lg5⎠⎞⎟ = z (lg2 + lg5)
Ungkapkan n dalam sebutan m. ⎝⎜ x
Klon It is given that 2x m, 23y n and
SPM 2x+3y = 8 + 8y. Express n in terms of m. ⎛ y ⎞
‘15 ⎜⎝ x ⎠⎟

ylg5 = z lg5 + lg5

2x+3y = 8 + 8y y = z ⎛ y + 1⎟⎠⎞
⎝⎜ x
( )2x23y = 8 + 23 y
z ⎛ y + x ⎞ = y
⎜⎝ x ⎠⎟
mn = 8 + n

mn − n = 8 z xy
x+y
n = 8 =
m −1

5 Diberi loga 3 x dan loga7 y, ungkapkan
Klon log7 81a2 dalam sebutan x dan y.
SPM Given loga 3 x and loga 7 y, express log781a2
‘15

in terms of x and y. 7 Diberi logb6 x, ungkapkan dalam sebutan x

log7 81a2 = loga 81a2 Klon dan y,
log a 7 SPM
‘16 Given logb 6 x, express in terms of x and y,
( )=1
y loga 81+ logaa2 (a) logb 216 (b) log6 36b3

( )=1 (a) logb 216 logb 63
y loga 34 + 2logaa 3logb 6
3x
= 1 ( 4log a 3 + 2)
y
(b) log6 36b3 = log6 36 + log6b3
= 2 ( 2x + 1) = log6 62 + 3log6b
y

= 2 + 3⎛⎜ logbb ⎞
⎝ logb 6 ⎟
⎠
3
= 2 + x

6 Diberi bahawa 2x 5y 10z, ungkapkan z 8x+5
4y − 2
Klon dalam sebutan x dan y. 8 Diberi = 1, ungkapkan y dalam x.
SPM
‘16 It is given that 2x 5y 10z, express z in terms of Klon
SPM
‘17 8x+5
x and y. 4y−2
Given = 1, express y in terms of x.

2x 5y 8x+5 =1
4y − 2
Ambil logaritma kepada asas 10 untuk kedua- ( ) ( )23 (x+5) = 22 (y−2)
dua belah./Take logarithm to base 10 for both
sides.

xlg2 ylg5 23x+15 = 22y−4

lg 2 y lg 5 3x + 15 = 2y − 4
x
2x 10z 2y = 3x + 19

Ambil logaritma kepada asas 10 untuk kedua- y = 3x + 19
dua belah./Take logarithm to base 10 for both 2

sides.

xlg2 zlg10

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tracker-software 9 Selesaikan persamaan: 11 (a) Tukarkan a xb kepada bentuk logaritma. tracker-software
Seterusnya, nyatakan syarat-syarat bagi x.
Klon Solve the equation: Klon Convert a xb to the logarithmic form.
SPM SPM
‘17 loga 625 − log a 4a = 2 ‘18

loga625 − log a 4a = 2 Hence, state the conditions of x.
2
log 625 − loga 4a = 2 (b) Diberi log6 p log pq 6, ungkapkan p
loga a
a dalam sebutan q.

loga625 − loga 4a = 2 Given log6 p 2 6 , express p in terms of q.
⎛ 1⎞ log pq
⎜⎝ 2 ⎟⎠
(a) a xb
loga625 − loga (4a)2 = 2 logxa b dengan keadaan x > 0 and x ≠ 1

loga 625 = 2 (b) log6 p = 2
16a2 logpq 6
625
16a2 = a2 logp p 2
log p 6
16a4 = 625 = ⎛ logp 6 ⎞
⎝⎜⎜ logp pq ⎠⎟⎟
a2 = 25
4 logp p = 2logp pq
5
a = ± 2 a>0 1= 2(logpp + logpq)

∴a 5 1 = 1+ log pq
2 2
1
log pq = − 2

10 Diberi 3w + 3w = 3x, ungkapkan w dalam p− 1
2
Klon sebutan x. = q
SPM Given 3w + 3w = 3x, express w in terms of x.
‘18 p = q−2

3w + 3w = 3x p = 1
q2
( )2 3w = 3x

Ambil logaritma kepada asas 3 untuk kedua-dua
belah./Take logarithm to base 3 for both sides.

( )log3 (2) 3w = log33x

log3 2 + log33w = x
w = x − log32

Genius

Tunjukkan bahawa 3n+1 + 3n + 3n−1 boleh dibahagikan dengan 13 bagi semua nilai integer positif n.
Show that 3n+1 + 3n + 3n−1 is divisible by 13 for all positive integral values of n.

3n+1 + 3n + 3n−1 = 313n + (1)3n + 3−13n

= ⎛ 3 + 1+ 1 ⎞ 3n
⎜⎝ 3 ⎟⎠

= ⎛ 13 ⎞ 3n Terdapat faktor 13 dalam ungkapan 3n+1 + 3n + 3n−1.
⎝⎜ 3 ⎠⎟ There is a factor of 13 in the expression 3n+1 + 3n + 3n−1.
? 3n+1 + 3n + 3n−1 boleh dibahagikan dengan/is divisible by 13.
= (13) ⎛ 1 ⎞ 3n
⎜⎝ 3 ⎠⎟

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