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Prove that (b/c)^lga*(c/a)^lgb*(a/b)^lgc = 1 .Prove that (b/c)^lga*(c/a)^lgb*(a/b)^lgc...
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We'll take logarithms both sides:
lg[(b/c)^lga*(c/a)^lgb*(a/b)^lgc] = lg1
We'll use the product rule of logarithms:
lg[(b/c)^lga] + lg[(c/a)^lgb] + lg[(a/b)^lgc] = 0
We'll use the quotient rule and power rule of logarithms:
lga(lg b - lg c) + lgb(lg c - lg a) + lg c(lg a - lg b) = 0
We'll remove the brackets:
lga*lg b - lg a*lg c + lgb*lg c - lg b*lg a + lg c*lg a - lg c*lg b = 0
We'll eliminate like terms and we'll get:
0 = 0 q.e.d.
It is obvious that the identity (b/c)^lga*(c/a)^lgb*(a/b)^lgc=1 is true.
Posted by giorgiana1976 on May 5, 2011 at 11:41 AM (Answer #2)
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