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