If log (a+b2)=12(log a+log b) , show that 12(a+b)=ab\log \space \Big(\dfrac{a + b}{2}\Big) = \dfrac{1}{2} (\log \space a + \log \space b)\text{ , show that }\dfrac{1}{2} (a + b) = \sqrt{ab}log (2a+b)=21(log a+log b) , show that 21(a+b)=ab.
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Given,
⇒log (a+b2)=12(log a+log b)⇒log (a+b2)=12log ab⇒log (a+b2)=log ab12⇒(a+b2)=ab12⇒12(a+b)=ab.\Rightarrow \log \space \Big(\dfrac{a + b}{2}\Big) = \dfrac{1}{2} (\log \space a + \log \space b) \\[1em] \Rightarrow \log \space \Big(\dfrac{a + b}{2}\Big) = \dfrac{1}{2} \log \space ab \\[1em] \Rightarrow \log \space \Big(\dfrac{a + b}{2}\Big) = \log \space {ab} ^{\dfrac{1}{2}} \\[1em] \Rightarrow \Big(\dfrac{a + b}{2}\Big) = {ab} ^{\dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{1}{2} (a + b) = \sqrt{ab}.⇒log (2a+b)=21(log a+log b)⇒log (2a+b)=21log ab⇒log (2a+b)=log ab21⇒(2a+b)=ab21⇒21(a+b)=ab.
Hence, proved that 12(a+b)=ab\dfrac{1}{2} (a + b) = \sqrt{ab}21(a+b)=ab.
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