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secondary 3 | A Maths
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Hi Mr Eric, could you help me please? Thanks!
Date Posted: 4 years ago
Views: 220
Good evening Candice! Doing this for you now.
yes ! Thanks Mr Eric :)
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done
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Good evening Candice! Here is an idea to do this question. Let me know if you need more explanation.
Date Posted:
4 years ago
awesome ! Thanks a lot Mr Eric. I need to bother u again for another question. Very sorry but pls help me. I will send out shortly.
Alternative working :
log5(2)
= log5(6 ÷ 3)
= log5(6) - log5(3)
= log5(2)/log6(2) - b
= log5(2)/a - b
So log5(2) - log5(2)/a = -b
(1 - 1/a)log5(2) = -b
log5(2) = -b/(1 - 1/a)
= -ab/(a - 1)
= ab/(1 - a)
Good identities to remember :
① loga(b) = logc(a) / logc(b)
② loga(b) = loga(c)/logb(c)
log5(2)
= log5(6 ÷ 3)
= log5(6) - log5(3)
= log5(2)/log6(2) - b
= log5(2)/a - b
So log5(2) - log5(2)/a = -b
(1 - 1/a)log5(2) = -b
log5(2) = -b/(1 - 1/a)
= -ab/(a - 1)
= ab/(1 - a)
Good identities to remember :
① loga(b) = logc(a) / logc(b)
② loga(b) = loga(c)/logb(c)
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