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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.
To prove that (b/c)^lga*(c/a)^lgb*(a/b)^lgc=1.
Let (b/c)^ loga = x .
We take logarithms of both sides:
log(b/a) loga = logx.
logb*loga - logc*loga = logx......(1)
Similarly let (c/a)^logb = y.
Then logc*loga - loga*logb = logy......(2)
Similarly let (a/b)^logc = y.
Then loga*logc - logb*logc = logy.....(3)
logb*loga - logc*loga + logc*logb-loga*logb+loga*logb-logb*logc = logx+logy+logz
0 = logx+logy+logz
0 = log(xyz)
Ttaking antilog, we get:
1 = xyz.
Or xyz = 1.
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