bagaimana menyelesaikan ini? tolong dibantu yaaa.. (:

2'log 3 = p dan 7'log 2 = q

maka:
18'log 84
= log84 / log18
= (log 2*2*3*7) / (log 2*3*3)
= (7'log 2 + 7'log 2 + 2'log 3 + 2'log 7) / (7'log 2 + 2'log 3 + 2'log 3)
= (q + q + p + 1/q) / (q + p + p)
= (p + 2q + 1/q) / (2p + q)
= [(p + 2q + 1/q) / (2p + q)] * q/q
= (pq + 2q² + 1) / (2pq + q²)

aku dapetnya kayak gitu, padahal jawabannya seharusnya A -_-

[tex]^{2}log_{3} = p \\ ^{7}log_{2} = q \\ ^{18}log_{84} = ...[/tex]

Penyelesaian :

[tex] = \frac{^{2}log_{84}}{^{2}log_{18}} \\ = \frac{^{2}log_{(2*2*3*7)}}{^{2}log_{9*2}} \\ = \frac{^{2}log_{(2*2*3*7)}}{^{2}log_{9+^{2}log{2}}} \\ = \frac{^{2}log_{(2*2*3*7)}}{^{2}log_{3^{2}}+^{2}log{2}}} \\ = \frac{^{2}log_{2}+^{2}log_{2}+^{2}log_{3}+^{2}log_{7}}{2.^{2}log_{3}+^{2}log{2}}} \\ = \frac{1+1+p+(1/q)}{2p+1}} \\ = \frac{2+p+(1/q)}{2p+1}}[/tex]

Setelah itu Kalikan q/q :

[tex]= \frac{q}{q} *\frac{2+p+(1/q)}{2p+1}}[/tex]

[tex]= \frac{2q+pq+1}{2pq+q}}[/tex]

Jadi Jawabannya A