Bab 5 Indeks dan Logaritma

Posted on May 14, 2020 by Myhometuition

Soalan 15 (4 markah):

\begin{array}{l} (a) Diberi P = \log_{a} Q, nyatakan \\ syarat-syarat bagi a . \\ (b) Diberi \log_{3} y = \frac{2}{\log_{x y} 3}, ungkapkan \\ y dalam sebutan x . \end{array}

Penyelesaian:
(a)
a > 0, a ≠ 1

(b)

\begin{array}{l} \log_{3} y = \frac{2}{\log_{x y} 3} \\ \frac{\log_{x y} y}{\log_{x y} 3} = \frac{2}{\log_{x y} 3} \\ \log_{x y} y = 2 \\ y = {(x y)}^{2} \\ y = x^{2} y^{2} \\ \frac{1}{x^{2}} = \frac{y^{2}}{y} \\ y = \frac{1}{x^{2}} \end{array}

Soalan 16 (2 markah):
Diberi 2^p + 2^p = 2^k. Ungkapkan p dalam sebutan k.

Penyelesaian:

\begin{array}{l} 2^{p} + 2^{p} = 2^{k} \\ 2 (2^{p}) = 2^{k} \\ 2^{p} = \frac{2^{k}}{2^{1}} \\ 2^{p} = 2^{k - 1} \\ p = k - 1 \end{array}

Soalan 17 (3 markah):

\begin{array}{l} Diberi \frac{25^{h + 3}}{125^{p - 1}} =1, ungkapkan p dalam \\ sebutan h . \end{array}

Penyelesaian:

\begin{array}{l} \frac{25^{h + 3}}{125^{p - 1}} = 1 \\ 25^{h + 3} = 125^{p - 1} \\ {(5^{2})}^{h + 3} = {(5^{3})}^{p - 1} \\ 5^{2 h + 6} = 5^{3 p - 3} \\ 2 h + 6 = 3 p - 3 \\ 3 p = 2 h + 9 \\ p = \frac{2 h + 9}{3} \end{array}

Soalan 18 (3 markah):

\begin{array}{l} Selesaikan persamaan: \\ \log_{m} 324 - \log_{\sqrt{m}} 2 m = 2 \end{array}

Penyelesaian:

\begin{array}{l} \log_{m} 324 - \log_{\sqrt{m}} 2 m = 2 \\ \log_{m} 324 - \frac{\log_{m} 2 m}{\log_{m} m^{\frac{1}{2}}} = 2 \\ \log_{m} 324 - 2 (\frac{\log_{m} 2 m}{\log_{m} m}) = 2 \\ \log_{m} 324 - 2 \log_{m} 2 m = 2 \\ \log_{m} 324 - \log_{m} {(2 m)}^{2} = l o g_{m} m^{2} \\ \log_{m} (\frac{324}{4 m^{2}}) = l o g_{m} m^{2} \\ \frac{324}{4 m^{2}} = m^{2} \\ 4 m^{4} = 324 \\ m^{4} = 81 \\ m = \pm 3 (- 3 ditolak) \end{array}

Bab 5 Indeks dan Logaritma

Posted on October 15, 2016 by user

5.6 Indeks dan Logaritma, SPM Praktis (Soalan Panjang)

Soalan 3:

Diberi bahawa p= 3^r dan q = 3^t, ungkapkan yang berikut dalam sebutan rdan/ atau t.

(a) \log_{3} \frac{p q^{2}}{27},

(b) log₉p – log₂₇ q.

Penyelesaian:

(a)

Diberi p = 3^r, log₃ p = r

q= 3^t, log₃ q =t

\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{\log_{3} p}{\log_{3} 9} - \frac{\log_{3} q}{\log_{3} 27} \\ = \frac{r}{\log_{3} 3^{2}} - \frac{t}{\log_{3} 3^{3}} \\ = \frac{r}{2 \log_{3} 3} - \frac{t}{3 \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} = \log_{2} (2 x + 1) - \frac{5 \log_{2} x^{2}}{\log_{2} 4} + 4 \log_{2} x \\ = \log_{2} (2 x + 1) - \frac{5}{2} \log_{2} x^{2} + \log_{2} x^{4} \\ = \log_{2} (2 x + 1) - \log_{2} {(x^{2})}^{(\frac{5}{2})} + \log_{2} x^{4} \end{array}

\begin{array}{l} = \log_{2} (2 x + 1) - \log_{2} x^{5} + \log_{2} x^{4} \\ = \log_{2} \frac{(2 x + 1) (x^{4})}{x^{5}} \\ = \log_{2} \frac{2 x + 1}{x} \end{array}

(b)

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

\begin{array}{l} \log_{2} \frac{2 x + 1}{x} = 4 \\ \frac{2 x + 1}{x} = 2^{4} \\ \frac{2 x + 1}{x} = 16 \\ 2 x + 1 = 16 x \\ 14 x = 1 \\ x = \frac{1}{14} \end{array}

Bab 5 Indeks dan Logaritma

Posted on September 28, 2016 by user

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

= \log_{2} \frac{12^{2} \times 5^{3}}{15 \times 150}

= log₂ 8

= log₂ 2³

= 3

(a)(ii)

\begin{array}{l} \log_{8} 32 = \frac{\log_{2} 32}{\log_{2} 8} \\ = \frac{\log_{2} 2^{5}}{\log_{2} 2^{3}} = \frac{5}{3} \end{array}

(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} \log_{3} x = \log_{9} (x + 6) \\ \log_{3} x = \frac{\log_{3} (x + 6)}{\log_{3} 9} \\ \log_{3} x = \frac{\log_{3} (x + 6)}{\log_{3} 3^{2}} \\ \log_{3} x = \frac{\log_{3} (x + 6)}{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.

Jadi, x = 3.

Bab 5 Indeks dan Logaritma

Posted on July 6, 2016 by user

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) 3^x ⁺ ¹ = 7
(b) 2 (3^x ) = 5
(c) 2^x .3^x = 9^x ⁻ ⁴
(d) 5^x ⁻ ¹.3^x ⁺ ² = 10

Penyelesaian:

Bab 5 Indeks dan Logaritma

Posted on July 6, 2016 by user

5.2 Logaritma

N = a^x ↔ log_aN = x

log_a N = x ialah bentuk logaritma manakala N = a^x ialah bentuk index.

Takrif Logaritma:

1. Jika aialah nombor positif, maka a^xjuga 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) log_a1 = 0

(b) log_aa= 1

Contoh 1:

Cari persamaan-persamaan berikut:

(a) log₂ 64

(b) log₃ 1

(d) log_½ 4

(e) log₈ 0.25

Penyelesaian:

Contoh 2:

Cari persamaan-persamaan berikut:

(a) log₃ 5x = 2

(b) log_x_{+ 1} 81 = 2

Penyelesaian:

Bab 5 Indeks dan Logaritma

Posted on July 4, 2016 by user

5.4 Persamaan yang Melibatkan Logaritma (Contoh 4 & 5)

Contoh 4:

Selesaikan setiap persamaan yang berikut.

\begin{array}{l} (a) \log_{9} (x - 2) = \log_{3} 2 \\ (b) \log_{4} x = \frac{3}{2} \log_{2} 3 \end{array}

Penyelesaian:

Contoh 5:

Selesaikan setiap persamaan yang berikut.

(a) log₄ x = 25 log_x 4
(b) log₂5√x + log₄ 16x = 6

Penyelesaian:

Bab 5 Indeks dan Logaritma

Posted on July 4, 2016 by user

5.4 Persamaan yang Melibatkan Logaritma (Contoh 2 & 3)

Contoh 2:

Selesaikan setiap persamaan yang berikut.

\begin{array}{l} (a) \log_{\sqrt{y}} 81 - 3 = \log_{\sqrt{y}} 3 \\ (b) 2 \log_{2} x - \log_{2} (\frac{x}{2} - 1) - 4 = 0 \end{array}

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

Posted on July 3, 2016 by user

5.4 Persamaan yang Melibatkan Logaritma

Kaedah:

1. Bagi dua logaritma yang sama asas, jika log_a m = log_a n, maka m = n .

2. Menukar logaritma kepada bentuk index, jika log_a 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 log_x 2 + 2 log_x 4 – log_x 256 = −1

Penyelesaian:

Bab 5 Indeks dan Logaritma

Posted on May 14, 2016 by user

5.5 Indeks dan Logaritma, SPM Praktis (Soalan Pendek)

Soalan 12

Selesaikan persamaan,

\log_{2} 5 \sqrt{x} + \log_{4} 16 x = 6

Penyelesaian:

\begin{array}{l} \log_{2} 5 \sqrt{x} + \log_{4} 16 x = 6 \\ \log_{2} 5 \sqrt{x} + \frac{\log_{2} 16 x}{\log_{2} 4} = 6 \\ \log_{2} 5 \sqrt{x} + \frac{\log_{2} 16 x}{2} = 6 \\ 2 \log_{2} 5 \sqrt{x} + \log_{2} 16 x = 12 \\ \log_{2} {(5 \sqrt{x})}^{2} + \log_{2} 16 x = 12 \\ \log_{2} (25 x) + \log_{2} 16 x = 12 \\ \log_{2} (25 x) (16 x) = 12 \\ \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 (x – y) = 3 + log₂ x + log₂y. Buktikan x² + y²– 10xy = 0.

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 – 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

Posted on May 14, 2016 by user

5.5 Indeks dan Logaritma, SPM Praktis (Soalan Pendek)

Soalan 9

Selesaikan persamaan,

\log_{2} 4 x = 1 - \log_{4} x

Penyelesaian:

\begin{array}{l} \log_{2} 4 x = 1 - \log_{4} x \\ \log_{2} 4 x = 1 - \frac{\log_{2} x}{\log_{2} 4} \\ \log_{2} 4 x = 1 - \frac{\log_{2} x}{2} \\ 2 \log_{2} 4 x = 2 - \log_{2} x \\ \log_{2} 16 x^{2} = \log_{2} 4 - \log_{2} x \\ \log_{2} 16 x^{2} = \log_{2} \frac{4}{x} \\ 16 x^{2} = \frac{4}{x} \\ x^{3} = \frac{4}{16} = \frac{1}{4} \\ x = {(\frac{1}{4})}^{\frac{1}{3}} = 0.62996 \end{array}

Soalan 10

Selesaikan persamaan,

\log_{4} x = 25 \log_{x} 4

Penyelesaian:

\begin{array}{l} \log_{4} x = 25 \log_{x} 4 \\ \frac{1}{\log_{x} 4} = 25 \log_{x} 4 \\ \frac{1}{25} = {(\log_{x} 4)}^{2} \\ \log_{x} 4 = \pm \frac{1}{5} \\ \log_{x} 4 = \frac{1}{5} or \log_{x} 4 = - \frac{1}{5} \\ 4 = x^{\frac{1}{5}} 4 = x^{- \frac{1}{5}} \\ x = 4^{5} 4 = \frac{1}{x^{\frac{1}{5}}} \\ x = 1024 x^{\frac{1}{5}} = \frac{1}{4} \\ x = \frac{1}{1024} \end{array}

Soalan 11

Selesaikan persamaan,

2 \log_{x} 5 + \log_{5} x = \lg 1000

Penyelesaian:

\begin{array}{l} 2 \log_{x} 5 + \log_{5} x = \lg 1000 \\ 2. \frac{1}{\log_{5} x} + \log_{5} x = 3 \\ \times (\log_{5} x) \to 2 + {(\log_{5} x)}^{2} = 3 \log_{5} x \\ {(\log_{5} x)}^{2} - 3 \log_{5} x + 2 = 0 \\ (\log_{5} x - 2) (\log_{5} x - 1) = 0 \\ \log_{5} x = 2 or \log_{5} x = 1 \\ x = 5^{2} x = 5 \\ x = 25 \end{array}