document.write( "Question 1119107: If a^2 + b^2 = 11ab and a>b >0, prove that {(a-b)/3}=1/2 (log a + log b) \n" ); document.write( "

Algebra.Com's Answer #734567 by ikleyn(52781) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "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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\n" ); document.write( "\n" ); document.write( " The statement in the post IS NOT CORRECT. IT IS INCORRECT (!) (! !) (! ! !).\r
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\n" ); document.write( "\n" ); document.write( " What is the correct statement, you will see from my solution.\r
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document.write( "If a^2 + b^2 = 11ab,  then\r\n" );
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document.write( " = 11ab - 2ab = 9ab,   which implies\r\n" );
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document.write( " = 9ab,   or\r\n" );
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document.write( " = ab,   or\r\n" );
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document.write( " = ab\r\n" );
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document.write( "Take the logarithm of both sides. Account that a > b > 0, according to the condition. You wull get\r\n" );
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document.write( " = log(a) + log(b),\r\n" );
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document.write( "which implies\r\n" );
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document.write( " = .    <<<---=== It is the correct statement.\r\n" );
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