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Exercise 2.12 · Q6

Q.Prove 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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Step 1. By the product rule, the sum equals log⁡(a2bc⋅b2ca⋅c2ab)\log\left(\dfrac{a^2}{bc}\cdot\dfrac{b^2}{ca}\cdot\dfrac{c^2}{ab}\right).

Step 2. Multiply out the argument: a2b2c2bc⋅ca⋅ab=a2b2c2a2b2c2=1\dfrac{a^2b^2c^2}{bc\cdot ca\cdot ab}=\dfrac{a^2b^2c^2}{a^2b^2c^2}=1. …

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