Question

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

Answer 1

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

implies that

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

implies that

${\left(a+b\right)}^2=9ab$

$\log {\left(a+b\right)}^2=\log 9ab$

implies that

$2\log \left(a+b\right)=\log 9+\log a+\log b$

implies $2\log \left(a+b\right)-\log 9=\mathrm{loga}+\log b$

$2\log \left(a+b\right)-\log 3^2=\mathrm{loga}+\log b$

$2\log \left(a+b\right)-2\log 3=\mathrm{loga}+\log b$

$2[\log \left(a+b\right)-\log 3]=\mathrm{loga}+\log b$

$\log \frac{a+b}{3}=\frac{1}{2}\left(\log a+\log b\right)$