Jawaban:

Pembahasan

Sifat logaritma

a log a = 1

a log (b x c) = a log b + a log c

a log (b : c) = a log b - a log c

a log n. ⁿlog b = a log b

a log aⁿ = n

Maka;

Jika ⁷log 2 = p dan ²log 3 = q

⁷log 2 = p → ²log 7 = 1/p

²log 3 = q → ³log 2 = 1/q

⁷log 2 . ²log 3 = ⁷log 3 = pq

Sehingga:

= ⁶log 98

= ²log 98 / ²log 6

= ²log (2 x 7²)/ ²log (2 x 3)

= [ ²log 2 + ²log 7² ] / [ ²log 2 + ²log 3 ]

= [ 1 + 2 (²log 7) ] / [ 1 + ²log 3]

= [ 1 + 2 (1/p) ] / [ 1 + q ]

= [ 1 + 2/p] / [ 1 + q ]

= [ (p + 2)/p ] / [ 1 + q]

= (p + 2)/p : (1 + q)/1

= (p + 2)/p x 1/(1 + q)

= (p + 2)/p(1 + q)

atau

= (p + 2)/(p + pq)

Jadi, nilai untuk ⁶log 98 jika dinyatakan dalam bentuk p dan q adalah (p + 2)/p(1 + q) atau (p + 2)/(p + pq).