x^2+y^2=7xy then log [1/3 (x+y) ]=
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If x^2 + y^2 = 7xy, show that log [(x + y)/3] = (log x + log y)/2?
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Answer:
If x^2 + y^2 = 7xy, show that log [(x + y)/3] = (log x + log y)/2?
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