Question 977752
<pre>
If log{{x+y}/3}=1/2{logx+log Y}, then find the value of x/y+y/x?
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What the other person who responded did and said, doesn't make any sense, at all, to this author. Plus, nowhere in his response,
is there an answer to the question: What's the value of {{{x/y + y/x}}}?

             {{{log (((x + y)/3)) = (1/2) log ((x)) + log ((y))}}}
        {{{log ((x + y)) - log ((3)) = (1/2) log ((xy))}}}
    {{{2 log (x + y) - 2 log (3) = log ((xy))}}} ---- Multiplying by 2
     {{{log ((x + y))^2 - log ((3))^2 = log ((xy))}}}
      {{{log ((x + y))^2 - log ((9)) = log ((xy))}}}
     {{{log ((x + y))^2 - log ((xy)) = log ((9))}}}
             {{{log ((x + y)^2/xy) = log ((9))}}} 
                {{{(x + y)^2/xy = 9}}} ---- Applying c = d, if {{{log (b, (c)) = log (b, (d))}}}
         {{{(x^2  + 2xy + y^2)/xy = 9}}} ---- FOILing {{{(x + y)^2}}}
           {{{x^2/xy + 2xy/xy + y^2/xy = 9}}} ---- Dividing each expression on left-side by xy
{{{x*cross(x^2)/(cross(x)y) + (2cross(xy))/cross(xy) + y*cross(y^2)/(x*cross(y)) = 9}}}
               {{{x/y + 2 + y/x = 9}}} 
                 {{{x/y + y/x = 9 - 2}}}
                {{{highlight(x/y + y/x) = highlight(7)}}}</pre>