# Equation with x,y as exponential functions

 HW Helper P: 3,515 I've just been playing with things for a bit, and just thought I'd dumb things down for simplicity, which could also mean I dumbed things down a little too far. If we look at the exponents only, and take the addition of them we have a function $$f(x,y)=x^2+y+y^2+x$$ $$=\left( x+\frac{1}{2} \right)^2+\left( y+\frac{1}{2} \right)^2-\frac{1}{2}$$ So we clearly have a symmetry about y=x (as we already knew) and the minimum value f can take is -1/2. At this point $(x,y)=(-1/2,-1/2)$ which is where the minimum of f occurs, we found that $$g(x,y)=16^{x^2+y}+16^{x+y^2}=1$$ And so if we take some other value, say, $x=\frac{-1}{2}+k$, $y=\frac{-1}{2}-k$, $k\neq0$ *** then with these values we have $$x^2+y=\left(k-\frac{1}{2}\right)^2-\frac{1}{2}-k$$ $$=\left(k-1\right)^2-\frac{5}{4}$$ $$y^2+x=\left(k+\frac{1}{2}\right)^2-\frac{1}{2}+k$$ $$=\left(k+1\right)^2-\frac{5}{4}$$ Therefore, g becomes $$g(k)=16^{(k-1)^2-5/4}+16^{(k+1)^2-5/4}$$ $$=\frac{1}{32}\left(16^{(k-1)^2}+16^{(k+1)^2}\right)$$ And for all k $$16^{(k-1)^2}+16^{(k+1)^2}>16+16=32$$ So we only have the minimum at (-1/2,-1/2) which means this is the only combination (x,y) that satisfies $g(x,y)=1$ *** I'm uneasy about making this assumption that if we take some small linear increase of k in one variable, we should decrease the other variable with the same magnitude k.