Question 1129637
 If {{{x^2+y^2=7xy}}}, prove that {{{log(x+y/3)=(1/2)(log(x)+log(y))}}} 


{{{x^2+ y^2 = 7xy}}}  .......... add {{{2xy}}} both sides

{{{ x^2 + y^2 + 2xy = 9xy }}}

{{{(x + y)^2 / 3^2 = xy }}}

 {{{((x + y) / 3)^2 = xy }}}


take log both sides 

{{{ log(((x + y) / 3)^2) =log( xy )}}}

{{{2 log ((x + y) / 3) = log (xy) }}}

{{{2 log ((x + y) / 3)= log (x) + log( y )}}}

{{{log ((x + y) / 3) = (log x + log y) / 2}}}

{{{log ((x + y) / 3) = (1/2)(log x + log y) }}}.....(Proved!)



or this way



{{{log((x + y)/3) = (1/2)*log(xy) }}}

{{{log((x + y)/3) = log(((xy)^(1/2))) }}}

{{{(x + y)/3 = (xy)^(1/2) }}}

{{{(x + y)/3 = sqrt(xy) }}}

{{{((1/3)(x + y))^2 = (sqrt(xy) )^2}}}

{{{(1/9) * (x + y)^2 = xy }}}

{{{x^2 + 2xy + y^2 = 9xy }}}

{{{x^2 + y^2 = 7xy }}}.....(Proved!)