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MISCELLANEOUS EXERCISE 6 (II) · Q167

Q.Show that, log⁡a2bc+log⁡b2ca+log⁡c2ab=0\log\dfrac{a^2}{bc}+\log\dfrac{b^2}{ca}+\log\dfrac{c^2}{ab}=0.

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log⁡a2bc+log⁡b2ca+log⁡c2ab=log⁡[a2bc⋅b2ca⋅c2ab]\log\dfrac{a^2}{bc}+\log\dfrac{b^2}{ca}+\log\dfrac{c^2}{ab}=\log\left[\dfrac{a^2}{bc}\cdot\dfrac{b^2}{ca}\cdot\dfrac{c^2}{ab}\right] (product rule, combining all three into one log). …

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