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Question
secondary 3 | A Maths
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= a x log5(6)
= a x log5(3 x 2)
= a x (log5(3) + log5(2))
= a x (b + log5(2))
= ab + alog5(2)
so log5(2) = ab + alog5(2)
ab = log5(2) - a(log5(2))
ab = (1 - a)(log5(2))
So log5(2) = ab/(1 - a)
Alternatively,
log5(2)
= log3(2)/log3(5)
= 1/log2(3) x log5(3)
= log5(3)/log2(3)
= b/(log2(6/2))
= b/(log2(6) - log2(2))
= b/(1/log6(2) - 1)
= b/(1/a - 1)
ab/(1 - a)
See 1 Answer
We obtain log 2 3 from log 6 2, just we need to swap the order of log 6 2 into 1 divided by log 2 6 so that split the log 2 6 into log 2 2 + log 2 3.
Once we obtain log 2 3 (and we are oven log 5 3) we can get log 5 2.
So the number we have to settle with is either the 2 or the 3.
Somehow we need to obtain log 2 3 or log 3 2 from log 2 6 alone as a result. The only way is to split log 2 6 into log 2 2 + log 2 3. The question connects onwards from there.
Fortunately, these questions are unlikely to be seen in the O Levels.
Log 5 2 = log 3 2 / log 3 5 by the change of base rule
We have log 5 3 which is the reciprocal of log 3 5 and we have log 2 3 which is the reciprocal of log 3 2.