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- anonymous

Solve for (x,y) Such that

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- anonymous

Solve for (x,y) Such that

- chestercat

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- anonymous

\[16^{x^2 + y} + 16^{y^2 + x} = 1\]

- anonymous

are those real numbers?

- ganeshie8

2^4(x^2+y) + 2^4(y^2+x) = 2^-1 + 2^-1

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- ganeshie8

i tempted to start like that... and maybe equate exponents, and getting (x,y) = (-1/2, -1/2) . plz see if any flaws in my logic... guys

- anonymous

That is ONE SOLUTION.
There may be more.
Yes, they're real.

- ganeshie8

@mukushla im clueless plz give hint or something :)

- anonymous

clearly x<0 and y<0 and the equation is symmetric so suppose\[y\ge x \]conclude that just y=x gives us an answer

- anonymous

Howd you make that conclusion?

- anonymous

I know I;m being pretty stupid somewhere. ;/
Still, I dont see it.

- experimentX

|dw:1349455630674:dw|

- anonymous

Both x and y have to be negative.

- experimentX

lol ... i meant that.sorry i forgot to put - on the left.
|dw:1349455681739:dw|

- anonymous

Why does a symmetric situation imply that?
Can you prove that THAT will be the only solution?

- experimentX

if you change x and y in that equation, there isn't any change ... so x and y must be interchangeable.

- experimentX

|dw:1349455922847:dw|

- anonymous

I agree that that case seems highly plausable, but how do we prove it?

- experimentX

that gives one solution ... since x and y are interchangeable at x=y we just have to show that no other solution exist or if it exists then find them.

- experimentX

Let us define a function on R^2
|dw:1349456340587:dw|

- experimentX

Woops!! I made an error, ... earlier ... the f(x, y) goes to heaven for x, y -> -inf ...
there is only one solution ... the minimum value of this surface is at x=-0.5 and y=-0.5
http://www.wolframalpha.com/input/?i=Minimize+16^%28x^2%2By%29+%2B+16^%28x%2By^2%29+-1

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