Prove that (b/c)^lga*(c/a)^lgb*(a/b)^lgc=1

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.