# Bab 5 Indeks dan Logaritma

5.6 Indeks dan Logaritma, SPM Praktis (Soalan Panjang)
Soalan 3:
Diberi bahawa p= 3r dan q = 3t, ungkapkan yang berikut dalam sebutan rdan/ atau t.

(b)  log9p – log27 q.

Penyelesaian:
(a)
Diberi p = 3r, log3 p = r
q= 3t, log3 q =t

${\mathrm{log}}_{3}\frac{p{q}^{2}}{27}$
= log3 pq2 – log327
= log3 p + log3 q2 – log3 33
= r + 2 log3 q – 3 log3 3
= r + 2 log3 q – 3(1)
= r + 2t – 3

(b)
log9 p– log27 q
$\begin{array}{l}=\frac{{\mathrm{log}}_{3}p}{{\mathrm{log}}_{3}9}-\frac{{\mathrm{log}}_{3}q}{{\mathrm{log}}_{3}27}\\ =\frac{r}{{\mathrm{log}}_{3}{3}^{2}}-\frac{t}{{\mathrm{log}}_{3}{3}^{3}}\\ =\frac{r}{2{\mathrm{log}}_{3}3}-\frac{t}{3{\mathrm{log}}_{3}3}\\ =\frac{r}{2}-\frac{t}{3}\end{array}$

Soalan 4:
(a)  Permudahkan:
log2(2x + 1) – 5 log4 x2 + 4 log2 x
(b)  Seterusnya, selesaikan persamaan:
log2(2x + 1) – 5 log4 x2 + 4 log2 x = 4

Penyelesaian:
(a)
log2 (2x + 1) – 5 log4 x2 + 4 log2 x
$\begin{array}{l}={\mathrm{log}}_{2}\left(2x+1\right)-\frac{5{\mathrm{log}}_{2}{x}^{2}}{{\mathrm{log}}_{2}4}+4{\mathrm{log}}_{2}x\\ ={\mathrm{log}}_{2}\left(2x+1\right)-\frac{5}{2}{\mathrm{log}}_{2}{x}^{2}+{\mathrm{log}}_{2}{x}^{4}\\ ={\mathrm{log}}_{2}\left(2x+1\right)-{\mathrm{log}}_{2}{\left({x}^{2}\right)}^{\left(\frac{5}{2}\right)}+{\mathrm{log}}_{2}{x}^{4}\end{array}$
$\begin{array}{l}={\mathrm{log}}_{2}\left(2x+1\right)-{\mathrm{log}}_{2}{x}^{5}+{\mathrm{log}}_{2}{x}^{4}\\ ={\mathrm{log}}_{2}\frac{\left(2x+1\right)\left({x}^{4}\right)}{{x}^{5}}\\ ={\mathrm{log}}_{2}\frac{2x+1}{x}\end{array}$

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

# Bab 5 Indeks dan Logaritma

5.6 Indeks dan Logaritma, SPM Praktis (Soalan Panjang)
Soalan 1
(a)  Cari nilai bagi
i.       2 log2 12 + 3 log25 – log2 15 – log2 150.
ii.       log832
(b) Tunjukkan bahawa 5n  + 5n + 1 + 5n + 2 boleh dibahagi dengan 31 bagi semua nilai nyang merupakan integer positif.

Penyelesaian:
(a)(i)
2 log2 12 + 3 log2 5 – log215 – log2 150
= log2 122 + log2 53– log2 15 – log2 150
$={\mathrm{log}}_{2}\frac{{12}^{2}×{5}^{3}}{15×150}$
= log2 8
= log2 23
= 3

(a)(ii)

(b)
5n   + 5n + 1 + 5n + 2
= 5n   + (5 × 5n ) + (52 × 5n )
= 5n  (1 + 5 + 52)
= 31 × 5n
Oleh itu, 5n   + 5n + 1 + 5n + 2 boleh dibahagi dengan 31 bagi semua nilai nyang merupakan integer positif.

Soalan 2:
(a)  Diberi log10 x = 3 dan log10y = –2. Tunjukkan bahawa 2xy – 10000y2 = 19.
(b)  Selesaikan persamaan log3 x = log9(x + 6).

Penyelesaian:
(a)
log10x = 3      → (x = 103)
log10y = –2    → (y = 10-2)
2xy – 10000y2 = 19
Sebelah kiri:
2xy – 10000y2
= 2 × 103 × 10-2 – 10000 (10-2)2
= 20 – 10000 (10-4)
= 20 – 1
= 19
= sebelah kanan

(b)
$\begin{array}{l}{\mathrm{log}}_{3}x={\mathrm{log}}_{9}\left(x+6\right)\\ {\mathrm{log}}_{3}x=\frac{{\mathrm{log}}_{3}\left(x+6\right)}{{\mathrm{log}}_{3}9}\\ {\mathrm{log}}_{3}x=\frac{{\mathrm{log}}_{3}\left(x+6\right)}{{\mathrm{log}}_{3}{3}^{2}}\\ {\mathrm{log}}_{3}x=\frac{{\mathrm{log}}_{3}\left(x+6\right)}{2}\end{array}$
2log3 x= log3 (x + 6)
log3 x2= log3 (x + 6)
x2 = x + 6
x2 x – 6 = 0
(x + 2) (x – 3) = 0
x = – 2 atau 3.
log3 (– 2) tidak wujud.

# Bab 5 Indeks dan Logaritma

5.3 Persamaan yang Melibatkan Indeks (Contoh Soalan)

Contoh 5 [Asas tidak sama – tukar kepada bentuk lagaritma biasa (log10) bagi kedua-dua belah]:
Selesaikan setiap persamaan yang berikut.
(a) 3x + 1 = 7
(b) 2 (3x ) = 5
(c) 2x .3x = 9x 4
(d) 5x 1.3x + 2 = 10

Penyelesaian:

# Bab 5 Indeks dan Logaritma

5.2 Logaritma

 N = ax   ↔   loga N = x

loga N = x ialah bentuk logaritma manakala N = ax ialah bentuk index.

Takrif Logaritma:
1.      Jika aialah nombor positif, maka axjuga nombor positif. Logaritma bagi nombor negatif adalah tidak tertakrif.
2.      Logaritma yang asasnya 10 dikenali sebagai logaritma biasa. Simbolnya ialah log10, singkatannya lg.
3.      Nilai logaritma biasa dapat dicari dengan menggunakan kalkulator saintifik.
4.      Bagi asas logaritma bukan 10, asanya perlu dinyatakan, misalnya log381.
5.      Logaritma boleh berasas sebarang nombor positif, kecuali 1,
(a)    loga1 = 0
(b)   logaa= 1

Contoh 1:
Cari persamaan-persamaan berikut:
(a) log2 64
(b) log3 1
(c) log5 5
(d) log½ 4
(e) log8 0.25

Penyelesaian:

Contoh 2:
Cari persamaan-persamaan berikut:
(a) log3 5x = 2
(b) logx + 1 81 = 2
(c) logx 5x – 6 = 2

Penyelesaian:

# Bab 5 Indeks dan Logaritma

5.4 Persamaan yang Melibatkan Logaritma (Contoh 4 & 5)

Contoh 4:
Selesaikan setiap persamaan yang berikut.

Penyelesaian:

Contoh 5:
Selesaikan setiap persamaan yang berikut.
(a) log4 x = 25 logx 4
(b) log2 5x log4 16x = 6

Penyelesaian:

# Bab 5 Indeks dan Logaritma

5.4 Persamaan yang Melibatkan Logaritma (Contoh 2 & 3)

Contoh 2:
Selesaikan setiap persamaan yang berikut.

Penyelesaian:

Contoh 3:
Selesaikan setiap persamaan yang berikut.
(a) log3 [log2 (2x 1)] = 2
(b) log16 [log2 (5x 4)] = log9 √3

Penyelesaian:

# Bab 5 Indeks dan Logaritma

5.4 Persamaan yang Melibatkan Logaritma

Kaedah:
1. Bagi dua logaritma yang sama asas, jika loga m = loga n, maka m = n .
2. Menukar logaritma kepada bentuk index, jika loga m = n, maka m = an.

Contoh 1:
Selesaikan setiap persamaan yang berikut.
(a) log3 2 + log3 (x + 5) = log3 (3x 1)
(b) log2 8 x 3 = log2 (2x 1)
(c) 3 logx 2 + 2 logx 4 logx 256 = −1

Penyelesaian:

# Bab 5 Indeks dan Logaritma

5.5 Indeks dan Logaritma, SPM Praktis (Soalan Pendek)
Soalan 12
Selesaikan persamaan, ${\mathrm{log}}_{2}5\sqrt{x}+{\mathrm{log}}_{4}16x=6$

Penyelesaian:
$\begin{array}{l}{\mathrm{log}}_{2}5\sqrt{x}+{\mathrm{log}}_{4}16x=6\\ {\mathrm{log}}_{2}5\sqrt{x}+\frac{{\mathrm{log}}_{2}16x}{{\mathrm{log}}_{2}4}=6\\ {\mathrm{log}}_{2}5\sqrt{x}+\frac{{\mathrm{log}}_{2}16x}{2}=6\\ 2{\mathrm{log}}_{2}5\sqrt{x}+{\mathrm{log}}_{2}16x=12\\ {\mathrm{log}}_{2}{\left(5\sqrt{x}\right)}^{2}+{\mathrm{log}}_{2}16x=12\\ {\mathrm{log}}_{2}\left(25x\right)+{\mathrm{log}}_{2}16x=12\\ {\mathrm{log}}_{2}\left(25x\right)\left(16x\right)=12\\ {\mathrm{log}}_{2}400{x}^{2}=12\\ 400{x}^{2}={2}^{12}\\ {x}^{2}=10.24\\ x=3.2\end{array}$

Soalan 13
Diberi bahawa 2 log2 (xy) = 3 + log2 x + log2y. Buktikan x2 + y2– 10xy = 0.

Penyelesaian:
2 log2 (xy) = 3 + log2x + log2 y
log2 (xy)2 = log2 8 + log2 x + log2y
log2 (xy)2 = log2 8xy
(xy)2 = 8xy
x2 – 2xy + y2 = 8xy
x2 + y2 – 10xy = 0 (terbukti)

Soalan 14
Diberi bahawa 2 log2 (x + y) = 3 + log2 x + log2y. Buktikan x2 + y2= 6xy.

Penyelesaian:
2 log2 (x + y) = 3 + log2x + log2 y
log2 (x+ y)2 = log2 8 + log2 x + log2y
log2 (x+ y)2 = log2 8xy
(x + y)2 = 8xy
x2 + 2xy + y2 = 8xy
x2 + y2 = 6xy  (terbukti)

# Bab 5 Indeks dan Logaritma

5.5 Indeks dan Logaritma, SPM Praktis (Soalan Pendek)
Soalan 9
Selesaikan persamaan, ${\mathrm{log}}_{2}4x=1-{\mathrm{log}}_{4}x$

Penyelesaian:
$\begin{array}{l}{\mathrm{log}}_{2}4x=1-{\mathrm{log}}_{4}x\\ {\mathrm{log}}_{2}4x=1-\frac{{\mathrm{log}}_{2}x}{{\mathrm{log}}_{2}4}\\ {\mathrm{log}}_{2}4x=1-\frac{{\mathrm{log}}_{2}x}{2}\\ 2{\mathrm{log}}_{2}4x=2-{\mathrm{log}}_{2}x\\ {\mathrm{log}}_{2}16{x}^{2}={\mathrm{log}}_{2}4-{\mathrm{log}}_{2}x\\ {\mathrm{log}}_{2}16{x}^{2}={\mathrm{log}}_{2}\frac{4}{x}\\ 16{x}^{2}=\frac{4}{x}\\ {x}^{3}=\frac{4}{16}=\frac{1}{4}\\ x={\left(\frac{1}{4}\right)}^{\frac{1}{3}}=0.62996\end{array}$

Soalan 10
Selesaikan persamaan, ${\mathrm{log}}_{4}x=25{\mathrm{log}}_{x}4$

Penyelesaian:

Soalan 11
Selesaikan persamaan, $2{\mathrm{log}}_{x}5+{\mathrm{log}}_{5}x=\mathrm{lg}1000$

Penyelesaian:

# Bab 5 Indeks dan Logaritma

5.5 Indeks dan Logaritma, SPM Praktis (Soalan Pendek)
Soalan 5:
Selesaikan persamaan, ${\mathrm{log}}_{9}\left(x-2\right)={\mathrm{log}}_{3}2$

Penyelesaian:
$\begin{array}{l}{\mathrm{log}}_{9}\left(x-2\right)={\mathrm{log}}_{3}2\\ \overline{){\mathrm{log}}_{a}b=\frac{{\mathrm{log}}_{c}b}{{\mathrm{log}}_{c}a}}⇒\frac{{\mathrm{log}}_{3}\left(x-2\right)}{{\mathrm{log}}_{3}9}={\mathrm{log}}_{3}2\\ \frac{{\mathrm{log}}_{3}\left(x-2\right)}{2}={\mathrm{log}}_{3}2\\ {\mathrm{log}}_{3}\left(x-2\right)=2{\mathrm{log}}_{3}2\\ {\mathrm{log}}_{3}\left(x-2\right)={\mathrm{log}}_{3}{2}^{2}\\ x-2=4\\ x=6\end{array}$

Soalan 6:
Selesaikan persamaan, ${\mathrm{log}}_{9}\left(2x+12\right)={\mathrm{log}}_{3}\left(x+2\right)$

Penyelesaian:

Soalan 7:
Selesaikan persamaan, ${\mathrm{log}}_{4}x=\frac{3}{2}{\mathrm{log}}_{2}3$

Penyelesaian:
$\begin{array}{l}{\mathrm{log}}_{4}x=\frac{3}{2}{\mathrm{log}}_{2}3\\ \frac{{\mathrm{log}}_{2}x}{{\mathrm{log}}_{2}4}=\frac{3}{2}{\mathrm{log}}_{2}3\\ \frac{{\mathrm{log}}_{2}x}{2}=\frac{3}{2}{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x=2×\frac{3}{2}{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x=3{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x={\mathrm{log}}_{2}{3}^{3}\\ x=27\end{array}$

Soalan 8:
Selesaikan persamaan, $\frac{2}{{\mathrm{log}}_{5}2}={\mathrm{log}}_{2}\left(2-x\right)$

Penyelesaian:
$\begin{array}{l}\frac{2}{{\mathrm{log}}_{5}2}={\mathrm{log}}_{2}\left(2-x\right)\\ 2={\mathrm{log}}_{5}2.{\mathrm{log}}_{2}\left(2-x\right)\\ 2=\frac{1}{{\mathrm{log}}_{2}5}.{\mathrm{log}}_{2}\left(2-x\right)\\ 2{\mathrm{log}}_{2}5={\mathrm{log}}_{2}\left(2-x\right)\\ {\mathrm{log}}_{2}{5}^{2}={\mathrm{log}}_{2}\left(2-x\right)\\ 25=2-x\\ x=-23\end{array}$