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

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(52780)   (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


 = 11ab - 2ab = 9ab,   which implies


 = 9ab,   or


 = ab,   or


 = ab


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


 = log(a) + log(b),


which implies


 = .    <<<---=== It is the correct statement.

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