Christmas MT's answer to abc's Secondary 3 A Maths question.
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lg both sides.
(lg 5) (lg 4k) = (lg 7) (lg 5k)
(lg 5) (lg 4 + lg k) = (lg 7) (lg 5 + lg k)
lg 4 lg 5 + lg 5 lg k = lg 7 lg 5 + lg 7 lg k
lg 7 lg k - lg 5 lg k = lg 4 lg 5 - lg 7 lg 5
k = (lg 4 lg 5 - lg 7 lg 5) / (lg 7 - lg 5)
(lg 5) (lg 4k) = (lg 7) (lg 5k)
(lg 5) (lg 4 + lg k) = (lg 7) (lg 5 + lg k)
lg 4 lg 5 + lg 5 lg k = lg 7 lg 5 + lg 7 lg k
lg 7 lg k - lg 5 lg k = lg 4 lg 5 - lg 7 lg 5
k = (lg 4 lg 5 - lg 7 lg 5) / (lg 7 - lg 5)
Date Posted:
6 years ago
Sorry but what happens to the lg attached to the variable k? Do I convert it into exponential form? Like 10 to the power of (lg 4 lg 5 - lg 7 lg 5) / (lg 7 - lg 5) to get k?
Oops, yea.
-1 mark for me, lolz.
-1 mark for me, lolz.
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