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

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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)

(1)+(2)+(3) gives:

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.

 Or (b/c)^lga*(c/a)^lgb*(a/b)^lgc=1.

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