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secondary 3 | E Maths
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Date Posted: 6 years ago
Views: 256
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Here you go. Hope you understand
Date Posted:
6 years ago
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there is a formular, log(a/b)=log(a)-log(b).
so,
log5(25/sqr(n))=log(5)(25)-log(5)(sqr(n))
there is another formular, log(a^b)=b*log(a)
so,
log(5)(sqr(n))=log(5)(n^(1/2)）=1/2log(5)(n)
from the question, we know
log(5)(n)=m
therefore,
log5(25/sqr(n))=log(5)(25)-log(5)(sqr(n))
=2-1/2m
notice, 25=5^2, so log(5)(25)=2
so,
log5(25/sqr(n))=log(5)(25)-log(5)(sqr(n))
there is another formular, log(a^b)=b*log(a)
so,
log(5)(sqr(n))=log(5)(n^(1/2)）=1/2log(5)(n)
from the question, we know
log(5)(n)=m
therefore,
log5(25/sqr(n))=log(5)(25)-log(5)(sqr(n))
=2-1/2m
notice, 25=5^2, so log(5)(25)=2
Date Posted:
6 years ago
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