.
The given condition is not completed. To be complete, it must be re-edited in this way:
If a^2+b^2 = 7ab, where "a" and "b" are positive real numbers, prove that log{1/3(a+b)}=1/2(log a + log b).
With this re-editing, the solution/(the proof) is as follows.
If a^2 + b^2 = 7ab then
a^2 + 2ab + b^2 = 9ab
(a + b)^2 = 9ab
log((a+b)^2) = log(9ab)
2*log(a+b) = log(9) + log(a) + log(b)
2*log(a+b) = 2*log(3) + log(a) + log(b)
2*(log(a+b) - log(3)) = log(a) + log(b)
log(a+b) - log(3) =
= .
QED. The proof is completed.