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MISCELLANEOUS EXERCISE 6 (II) · Q173

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

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Given a2+b2=7aba^2+b^2=7ab.

(a+b)2=a2+2ab+b2=7ab+2ab=9ab(a+b)^2=a^2+2ab+b^2=7ab+2ab=9ab (substituting the given relation).

Taking square roots (both sides positive since a,b>0a,b>0): a+b=3ab  ⟹  a+b3=aba+b=3\sqrt{ab} \implies \dfrac{a+b}{3}=\sqrt{ab}. …

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