Question 815452
if x*x+y*y=7xy, prove
log(x+y)=1/2(logx+logy)+log3 
<pre>
     x²+y² = 7xy     <--given

The left side would be a perfect square if it had
2xy between the x² and y².  So we add 2xy to both
sides to make it into a perfect square:

 x²+2xy+y² = 7xy+2xy
(x+y)(x+y) = 9xy
    (x+y)² = 9xy

Take positive square roots of both sides

   &#8730;<span style="text-decoration: overline">(x+y)²</span> = &#8730;<span style="text-decoration: overline">9xy</span>

       x+y = &#8730;<span style="text-decoration: overline">9xy</span>
 
       x+y = 3&#8730;<span style="text-decoration: overline">xy</span>
 
Take logs of both sides:

     log(x+y) = log(3&#8730;<span style="text-decoration: overline">xy</span>)

     log(x+y) = log(3) + log(&#8730;<span style="text-decoration: overline">xy</span>)

     log(x+y) = log(3) + log(&#8730;<span style="text-decoration: overline">x</span>&#8730;<span style="text-decoration: overline">y</span>)

     log(x+y) = log(3) + log(&#8730;<span style="text-decoration: overline">x</span>) + log(&#8730;<span style="text-decoration: overline">y</span>)

Change square roots to 1/2 powers:

     log(x+y) = log(3) + log(x<sup>1/2</sup>) + log(y<sup>1/2</sup>)

Move eponents in front as multipliers:

     log(x+y) = log(3) + (1/2)log(x) + (1/2)log(y)

Foctor out (1/2) from the last two terms:

     log(x+y) = log(3) + (1/2)[log(x) + log(y)]

Rearrange the terms to what we were to prove:

     log(x+y) = (1/2)[log(x) + log(y)] + log(3) 


Edwin</pre>