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secondary 3 | A Maths
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Hi,
Could you kindly advise me the working solution for this question which I find it challenging to resolve. Many thanks :)
Date Posted: 4 years ago
Views: 200
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done
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clear
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Hope the explanation is clear enough.
Date Posted:
4 years ago
Thanks for your reply, Mr Wong but this is not the correct answer :(
What is the answer given? We can express it several ways.
Ans = ab/a+b
To continue from this working,
c = (alog101 + blog103) / (2log10403)
c = (alog101 + blog103) / (2log(101x103))
c = (alog101 + blog103) / (2(log101 + log103))
c = (a + b log103/log101) / (2(1+ log103/log101))
c = (a + b(a/b)) / (2(1 + a/b))
c = 2a / (2(1 + a/b))
= a/(1 + a/b)
= ab/(a + b)
c = (alog101 + blog103) / (2log10403)
c = (alog101 + blog103) / (2log(101x103))
c = (alog101 + blog103) / (2(log101 + log103))
c = (a + b log103/log101) / (2(1+ log103/log101))
c = (a + b(a/b)) / (2(1 + a/b))
c = 2a / (2(1 + a/b))
= a/(1 + a/b)
= ab/(a + b)
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