If `a^2+ b^2=7ab`, prove that `log(1/3 (a+ b)) =1/2 (log a+ log b)`

1.	If `a^2+ b^2=7ab`, prove that `log(1/3 (a+ b)) =1/2 (log a+ log b)`
Answer» `a^2+b^2=7ab` adding 2ab on both sides `a^2+b^2+2ab=7ab+2ab` `(a+b)^2=9ab` `(a+b)=3sqrt(ab)` `(a+b)/3=(ab)^(1/2)` taking log both sides `log((a+b)/3)=log(ab)^(1/2)` `log;1/3(a+b)]=1/2logab` `log[1/3(a+b)]=1/2[loga+logb]`.

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