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Question Number 10064 by PradipGos. last updated on 22/Jan/17

if a^2 +b^2 =3ab then prove that log((a+b)/(√5))=(1/2)(loga+logb)

$$\mathrm{if}\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{3ab}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{log}\frac{\mathrm{a}+\mathrm{b}}{\sqrt{\mathrm{5}}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}{a}+\mathrm{log}{b}\right) \\ $$$$ \\ $$

Answered by ridwan balatif last updated on 22/Jan/17

a^2 +b^2 =3ab→(a+b)^2 −2ab=3ab→a+b=±(√(5ab)) we take a+b=+(√(5ab)) log((a+b)/(√5))=log(((√5)×(√(ab)))/(√5)) =log(√(ab)) =log(ab)^(1/2) =(1/2)log(ab) =(1/2)(loga + logb) (Terbukti)

$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{3ab}\rightarrow\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\mathrm{2ab}=\mathrm{3ab}\rightarrow\mathrm{a}+\mathrm{b}=\pm\sqrt{\mathrm{5ab}} \\ $$$$\mathrm{we}\:\mathrm{take}\:\mathrm{a}+\mathrm{b}=+\sqrt{\mathrm{5ab}} \\ $$$$\mathrm{log}\frac{\mathrm{a}+\mathrm{b}}{\sqrt{\mathrm{5}}}=\mathrm{log}\frac{\sqrt{\mathrm{5}}×\sqrt{\mathrm{ab}}}{\sqrt{\mathrm{5}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{log}\sqrt{\mathrm{ab}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{log}\left(\mathrm{ab}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{ab}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{loga}\:+\:\mathrm{logb}\right)\:\left(\mathrm{Terbukti}\right) \\ $$

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