Posted by Anonymous on Wednesday, January 6, 2010 at 9:49pm.
of a^2+b^2=2ab, prove log[(a+b)/2]=1/2(loga+logb)

math  kayla, Wednesday, January 6, 2010 at 11:01pm
systems of equations by graphing

math  MathMate, Thursday, January 7, 2010 at 7:04am
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+b²2ab=0
(ab)²=0 after factoring
So we conclude that a=b
Substituting a=b into the lefthand side of the equation:
log[(a+b)/2]
=log((a+a)/2)
=log(a)
The righthand side:
(1/2)(loga+logb)
=(1/2)(log(a)+log(a))
=log(a)
Therefore:
log[(a+b)/2]=(1/2)(loga+logb)
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