Apabila 7log2=a dan ²log3=b maka 6log 98

Kelas 10 Matematika
Bab Logaritma

⁶log 98 
= (²log 98) / (²log 6)
= (²log (49 . 2)) / (²log (3 . 2))
= (²log 49 + ²log 2) / (²log 3 + ²log 2)
= (²log 7² + 1) / (b + 1)
= (2 . 1/⁷log 2 + 1) / (b + 1)
= (2/a + 1) / (b + 1)
= (2 + a)/2 . 1/(b + 1)
= (a + 2) / (2b + 2)