# 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.