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Solve for (x,y) Such that
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\[16^{x^2 + y} + 16^{y^2 + x} = 1\]
are those real numbers?
2^4(x^2+y) + 2^4(y^2+x) = 2^-1 + 2^-1
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
That is ONE SOLUTION. There may be more. Yes, they're real.
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@mukushla im clueless plz give hint or something :)
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
Howd you make that conclusion?
I know I;m being pretty stupid somewhere. ;/ Still, I dont see it.
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