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secondary 3 | A Maths
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Shan Kuek
Shan Kuek

secondary 3 chevron_right A Maths chevron_right Singapore

Help me out plz

Date Posted: 6 days ago
Views: 5
J
J
6 days ago
log15(3) = p
log7(5) = q



log7(3)

= log5(3)/log5(7)

= 1/log3(5) ÷ 1/log7(5)

= 1/(log3(15 ÷ 3)) x log7(5)

= 1/(log3(15) - log3(3)) x q

= q/(1/log15(3) - 1)
= q/(1/p - 1)

= pq/(1 - p)

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Eric Nicholas K
Eric Nicholas K's answer
1639 answers (Tutor Details)
1st
The last part is a bit blurry.
Shan Kuek
Shan Kuek
5 days ago
May I ask why log5(3) need to be found first? I don’t quite understand?
Eric Nicholas K
Eric Nicholas K
5 days ago
We are required to find log 7 3. We are given log 7 5, but we cannot form a link solely based on these two alone. We need the connector log 5 3 or log 3 5 to link these two. This is why I work on the log 15 3 to transform it into some log 3 5 or log 5 3.
Eric Nicholas K
Eric Nicholas K
5 days ago
We make use of the fact that 3 x 5 = 15 to do this question.
Shan Kuek
Shan Kuek
5 days ago
Okayyy thank you