4.6.1 Indeks, Surd dan Logaritma, SPM Praktis (Soalan Pendek)


Soalan 1:
Selesaikan persamaan, log3 [log2(2x – 1)] = 2

Penyelesaian:
log3 [log2 (2x – 1)] = 2 ← (jika log a N = x, N = ax)
log2 (2x – 1) = 32
log2 (2x – 1) = 9
2x – 1 = 29
x = 256.5


Soalan 2
Selesaikan persamaan, l o g 16 [ l o g 2 ( 5 x 4 ) ] = l o g 9 3  

Penyelesaian:
l o g 16 [ l o g 2 ( 5 x 4 ) ] = l o g 9 3 l o g 16 [ l o g 2 ( 5 x 4 ) ] = 1 4 log 9 3 = log 9 3 1 2 = 1 2 log 9 3 = 1 2 ( 1 log 3 9 ) = 1 2 ( 1 2 ) = 1 4 l o g 2 ( 5 x 4 ) = 16 1 4 l o g 2 ( 5 x 4 ) = 2 5 x 4 = 2 2 5 x = 8 x = 8 5


Soalan 3
Selesaikan persamaan, 5 log 4 x = 125

Penyelesaian:
5 log 4 x = 125 log 5 5 log 4 x = log 5 125 ambil log asas 5 di kedua-dua belah ( log 4 x ) ( log 5 5 ) = 3 ( log 4 x ) ( 1 ) = 3 x = 4 3 = 64



Soalan 4
Selesaikan persamaan, 5 log 5 ( x + 1 ) = 9

Penyelesaian:
5 log 5 ( x + 1 ) = 9 log 5 5 log 5 ( x + 1 ) = log 5 9 log 5 ( x + 1 ) . log 5 5 = log 5 9 log 5 ( x + 1 ) = log 5 9 x + 1 = 9 x = 8

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