SOLUTION: If a^2 + b^2 = 11ab and a>b >0, prove that {(a-b)/3}=1/2 (log a + log b)

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Question 1119107: If a^2 + b^2 = 11ab and a>b >0, prove that {(a-b)/3}=1/2 (log a + log b)
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
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If a^2 + b^2 = 11ab and a>b >0, prove that {(a-b)/3}=1/2 (log a + log b)
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            The statement in the post  IS  NOT  CORRECT.   IT  IS  INCORRECT   (!)   (! !)   (! ! !).

            What is the correct statement,  you will see from my solution. 


If a^2 + b^2 = 11ab,  then


a%5E2+-+2ab+%2B+b%5E2 = 11ab - 2ab = 9ab,   which implies


%28a-b%29%5E2 = 9ab,   or


%28a-b%29%5E2%2F9 = ab,   or


%28%28a-b%29%2F3%29%5E2 = ab


Take the logarithm of both sides. Account that a > b > 0, according to the condition. You wull get


2%2Alog+%28%28%28a-b%29%2F3%29%29 = log(a) + log(b),


which implies


log+%28%28%28a-b%29%2F3%29%29 = %281%2F2%29%2A%28log%28%28a%29%29+%2B+log%28%28b%29%29%29.    <<<---=== It is the correct statement.