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+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)

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