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EXERCISE 6.1 · Q82

Q.If log⁡(x+y3)=12log⁡x+12log⁡y\log\left(\dfrac{x+y}{3}\right)=\dfrac12\log x+\dfrac12\log y, show that xy+yx=7\dfrac{x}{y}+\dfrac{y}{x}=7.

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log⁡(x+y3)=12log⁡x+12log⁡y=12log⁡(xy)=log⁡xy\log\left(\dfrac{x+y}{3}\right)=\dfrac12\log x+\dfrac12\log y=\dfrac12\log(xy)=\log\sqrt{xy}.

Since log⁡\log is one-one, x+y3=xy\dfrac{x+y}{3}=\sqrt{xy}.

Square both sides: (x+y)29=xy  ⟹  (x+y)2=9xy\dfrac{(x+y)^2}{9}=xy \implies (x+y)^2=9xy. …

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