If $a²+b²=7ab$ where a and b are positive then show that $log(1/3(a+b))=1/2(log a +log b)$ Welcome sir, to the content of my question, please  help me:
If $a²+b²=7ab$ where a and b are positive then show that $log(1/3(a+b))=1/2(log a +log b)$ 

$a^2+b^2=7ab$ follows $(a+b)^2=9ab$   or $a+b=3 (ab)^{\frac{1}{2}}$  or $\frac{1}{3}(a+b)= (ab)^{\frac{1}{2}}.$ Thus
$$
\log\frac{1}{3}(a+b)= \log (ab)^{\frac{1}{2}}=\frac{1}{2}(\log a+\log b).
$$
