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in Sets, Relations and Functions by (15 points)
edited by

Prove log a2/bc+logb2/ca+logc2/ab=0

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1 Answer

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by (18.1k points)

L.H.S. = log (a2/bc) + log (b2/ca) + log (c2/ab)

= log (a2/bc x b2/ca x c2/ab)

\(=log\left(\frac{a^2b^2c^2}{a^2b^2c^2}\right)\)

= log 1

= 0 = R.H.S.

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