## Question

secondary 3 | A Maths

Anyone can contribute an answer, even non-tutors.

##### Shan Kuek

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Help me out plz

Date Posted: 2 months ago
Views: 14
J
2 months 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)

clear {{ downvoteCount * -1 }} Downvotes
1st
The last part is a bit blurry.
Date Posted: 2 months ago
Shan Kuek
2 months ago
May I ask why log5(3) need to be found first? I don’t quite understand?
Eric Nicholas K
2 months 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
2 months ago
We make use of the fact that 3 x 5 = 15 to do this question.
Shan Kuek
2 months ago
Okayyy thank you