Question

If $a^2+b^2=23ab$, show that :

$log\frac{a+b}{5}=\frac{1}{2}(loga+logb)$

Answer 1

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

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

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

Taking log on both sides;

$log\frac{{(a+b)}^2}{25}$ = $logab$

$\log (\frac{a+b}{5}{)}^2$ = $logab$
2 $\log \left(\frac{a+b}{5}\right)$ = $loga$ + $logb$
$\log \left(\frac{a+b}{5}\right)$ = $\frac{1}{2}$($loga$ + $logb$)