SOLUTION: if x*x+y*y=7xy, prove log(x+y)=1/2(logx+logy)+log3

     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

   √(x+y)² = √9xy

       x+y = √9xy
 
       x+y = 3√xy
 
Take logs of both sides:

     log(x+y) = log(3√xy)

     log(x+y) = log(3) + log(√xy)

     log(x+y) = log(3) + log(√x√y)

     log(x+y) = log(3) + log(√x) + log(√y)

Change square roots to 1/2 powers:

     log(x+y) = log(3) + log(x1/2) + log(y1/2)

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

if x*x+y*y=7xy, prove
log(x+y)=1/2(logx+logy)+log3

If x*x+y*y=7xy, prove log(x+y)=1/2(logx+logy)+log3
   Prove: 
    ----- Squaring x + y
   
    --- Substituting 7xy for  (GIVEN)
   
 ------- Taking the square root of both sides
      <=== ONLY the POSITIVE square root on the right-side is needed here 

 ----- Taking the log of both sides of the above equation



 <=== QED!