sifat-sifat yang berlaku dalam logaritma tersebut dapat diterapkan kedalam soal. Perhatikan beberapa consoh soal berikut.
1. Hitunglah nilai – nilai logaritma berikut :
a. 6log 9 + 6log 8 – 6log 2
b. 9log 135 – 9log 5
Jawab :
Berdasarkan sifat logaritma glog (axb) = glog a + glog b dan glog (a:b) = glog a – glog b maka
a. 6log 9 + 6log 8 – 6log 2
= 6log (9.8 /2)
= 6log 36
= 6log 6²
= 2 6log 6 (berdasarkan sifat glog an = n glog a )
=2 . 1
=2
b. 9log 135 – 9log 5
= 9log ( 135 / 5 )
= 9log 27
=3^2log 33
= 3/2 3log 3 ( berdasarkan sifat g^nlog am = m/n glog a )
= 3/2
2. Jika nilai log 3= a dan log 5 = b, tentukan nilai
a. log 75
b. log 1.500
Jawab
Berdasarkan sifat logaritma glog (axb) = glog a + glog b
a. log 75 = log (3 × 5²)
= log 3 + log 5²
= a + 2b
b. log 1500 = log ( 3 × 5 × 100 )
= log 3 + log 5 + log 100
= a + b + log 10²
= a + b + 2
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