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Exercise 2.12 · Q5

Q.If a2+b2=7aba^2+b^2=7ab, show that log⁡(a+b3)=12(log⁡a+log⁡b)\log\left(\dfrac{a+b}3\right)=\dfrac12(\log a+\log b).

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Step 1. a2+b2=7ab⇒a2+2ab+b2=9ab⇒(a+b)2=9aba^2+b^2=7ab\Rightarrow a^2+2ab+b^2=9ab\Rightarrow(a+b)^2=9ab.

Step 2. Divide by 9: (a+b3)2=ab\left(\dfrac{a+b}3\right)^2=ab.

Step 3. Take log⁡\log of both sides: 2log⁡(a+b3)=log⁡(ab)=log⁡a+log⁡b2\log\left(\dfrac{a+b}3\right)=\log(ab)=\log a+\log b. …

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