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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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sodelete | Student, College Freshman | (Level 1) Honors

Posted May 5, 2011 at 7:35 AM via web

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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 = 1 . 

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted May 5, 2011 at 11:41 AM (Answer #2)

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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.

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