of a^2+b^2=2ab, prove log[(a+b)/2]=1/2(loga+logb)

Answer 1

systems of equations by graphing

Answer 2

I have made some editorial changes to reflect my interpretation of the question:
"If a^2+b^2=2ab,
prove log[(a+b)/2]=(1/2)(loga+logb)"

If a²+b²=2ab, then
a²+b²-2ab=0
(a-b)²=0 after factoring
So we conclude that a=b

Substituting a=b into the left-hand side of the equation:

log[(a+b)/2]
=log((a+a)/2)
=log(a)

The right-hand side:
(1/2)(loga+logb)
=(1/2)(log(a)+log(a))
=log(a)

Therefore:
log[(a+b)/2]=(1/2)(loga+logb)