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
