Question 407766
Hmm.  Since you abbreviated please with the letter "z", I'll solve the problem for you, for your "enlightenment", hahaha.

{{{x^2 + y^2 = 6xy }}} <==> {{{x^2 + 2xy + y^2 = 8xy}}}
==> {{{(x+y)^2 = 8xy}}}

Take logs of both sides:

{{{2log((x+y)) = log8  +log x  + logy}}},

after applying additive property of logs.

Hence it follows that 

{{{2log((x+y)) = 3log2 + logx +logy}}}.