Question

If a^2+b^2=23ab show that log a+b/5=1/2(log a+log b )

Answer 1

We have,

${a}^{2} + {b}^{2} = 23ab$


Adding 2ab both sides,

${ a}^{2} + {b}^{2} + 2ab = 25ab \\ \\ {(a + b)}^{2} = 25ab \\ \\ a + b = 5 \sqrt{ab}$


We have,

$log(a + b)$

Substituing the values,
$log(5 \sqrt{ab} ) \\ \\ = log(5) + log( \sqrt{ab} ) \\ \\ = log(5) + log( {ab}^{ \frac{1}{2} } ) \\ \\ = log(5) + \frac{1}{2} log(ab) \\ \\ = log(5) + \frac{1}{2} ( loga + logb )$


Hence your answer.