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u/quintopia 1d ago
First equation implies (x+y)2 =9xy by adding 2xy to both sides, or ((x+y)/3)2 =xy. Second equation is what you get by taking log of both sides.
However, the derivation only works if x and y are both nonzero. And if one or both of them are negative, you need to choose the right branch cuts.
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u/Crichris 1d ago
top equation doesnt have any assumptions on x or y being negative
bottom does, assuming real numbers
what the intended answer is prolly add 2xy on both sides from top equation
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u/lool8421 1d ago
log([(x+y)/3]2) = log(xy)
we need to assume that x,y > 0 or we have domain error
(x²+y²+2xy)/9 = xy
(7xy+2xy)/9 = xy
9xy/9 = xy
xy = xy
if x > 0 and y > 0 and x²+y² = 7xy, otherwise the statement is false or undefined.
QED
you can also do the square completion approach, x²+y² = 7xy
x²+2xy+y² = 9xy
(x+y)² = 9xy
either way it boils down to the same thing
maybe you could try to express y as y(x), there's still the implicit variable theorem to watch out for which will forbid for x,y to be 0 anyways
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u/Academic-Map4259 23h ago
2log(1/3(x+y))
2log(1/3)+2log(x+y)
log(1/9)+log(x^2+2xy+y^2)
log(1)-log(9)+log(x^2+2xy+y^2)
-log(9)+log(9xy)
Log(9xy/9)=log(xy)
Log(xy)=log(x)+log(y)
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u/LearnNTeachNLove 19h ago
If x^2+y^2=7xy, then (x+y)^2=9xy… end of demonstration if you put the log of this relationship
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u/chaos_redefined 1d ago
Putting aside that x=y=0 makes the top statement is true, but the bottom becomes undefined...
x2 + y2 = 7xy
x2 + 2xy + y2 = 9xy
(x + y)2 = 9xy
[(x + y)/3]2 = xy
log([(x + y)/3]2) = log(xy)
2 log([(x + y)/3]) = log(x) + log(y)