Question 478102
{{{log((a+b)/2) = (loga +logb)/2}}}

==> {{{log((a+b)/2) = log(sqrt(ab))}}}

==> {{{(a+b)/2 = sqrt(ab)}}}, since log is 1-to-1

==> {{{(a+b) = 2sqrt(ab)}}}

==> {{{(a+b)^2 = 4ab}}}, after squaring both sides.

==> {{{a^2 + 2ab + b^2 = 4ab}}}

==> {{{a^2 - 2ab + b^2 = 0}}}

==> {{{(a-b)^2 = 0}}} ==> a - b = 0, or a = b.