Question

If $a^2+b^2=7ab$, then prove that $\log \left(\frac{a+b}{3}\right)=\frac{1}{2}(\log a+\log b)$

Answer 1

$a^2+b^2=7ab$

$\Rightarrow {(a+b)}^2-2ab=7ab$
$\Rightarrow {(a+b)}^2=9ab$

$\Rightarrow {\left(\frac{a+b}{3}\right)}^2=ab$

Taking log on both sides, we get
$\Rightarrow \log {\left(\frac{a+b}{3}\right)}^2=\log ab$
$\Rightarrow 2\log \left(\frac{a+b}{3}\right)=\log a+\log b$ ($\because \log x^m=m\log x;\log mn=\log m+\log n$)
$\Rightarrow \log \left(\frac{a+b}{3}\right)=\frac{1}{2}(\log a+\log b)$