Question 1119109
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<pre>
If log (x+y)/2 = 1/2 (log x + log y), then


{{{(x+y)/2}}} = {{{sqrt(xy)}}}  ====>


x + y = {{{2*sqrt(xy)}}}    ====>  square both sides  ====>


{{{x2 + 2xy + y^2}}} = 4xy


{{{x^2 - 2xy + y^2}}} = 0  ====>


{{{(x-y)^2}}} = 0  ====>   x = y.
</pre>

Proved ans solved.


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<U>Notice</U>: &nbsp;&nbsp;The given condition is equivalent to  {{{(x+y)/2}}} = {{{sqrt(xy)}}},


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;and it is &nbsp;<U>VERY &nbsp;WELL &nbsp;KNOWN &nbsp;FACT</U>&nbsp; that the last equality is equivalent to &nbsp;x = y.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;My solution gives you the proof of this fact.