x² + y² = 23xy and x,y > 0 prove that log() = [log(x) + log(y)]

Notice that (x+y) appears in the equation we are
to prove. The square  of that is x²+2xy+y² which 
is 2xy more than the left side, so we add
2xy to both sides:

          x² + 2xy + y² = 23xy + 2xy

Factor the left and combine terms on the right:

               (x + y)² = 25xy 

take logs of both sides

            log(x + y)² = log(25xy)

Use rules of logarithms to rewrite both sides:

           2·log(x + y) = log(25) + log(x) + log(y)

Change 25 to 5²

           2·log(x + y) = log(5²) + log(x) + log(y)

Change log(5²) to 2·log(5)

           2·log(x + y) = 2·log(5) + log(x) + log(y)

2·log(x + y) - 2·log(5) = log(x) + log(y) 

Factor out 2 on the left

2[log(x + y) - log(5)] = log(x) + log(y) 

Use a rule of logarithms on the bracketed expression on the left

           2[log()] = log(x) + log(y)

Multiply both sides by 


             log() = [log(x) + log(y)]

Edwin