# Equation with x,y as exponential functions

by cng99
Tags: exponential equation, olympiad, polynomial exponent, two variable, x y
P: 44
 Quote by epenguin Can you show us what the family looks like, so we see what happens as you reduce this something to 1?
I can't.
 HW Helper P: 3,323 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.
 P: 44 Holy Menials! Mentallic, you're a genius!!! I love you! I just thought of something new (credits go to you of course) We know that, Arithmetic mean >= geometric mean. Thus, [16^(x*x + y) + 16^(y*y + x)]/2 >= { [16^(x*x + y)]*[16^(y*y + x)] } ^1/2 Hence g(x,y) >= 2*[16^( x*x + y*y + x + y)]^1/2 Now (x*x + y*y + x + y) >= -1/2 Thus g(x,y) >= 2 *[16^(-1/4)] And g(x,y)>= 1 But at (x,y) = (-1/2,-1/2) g(x,y)=1 SOLVED!!!! I LOVE YOU!!!!
 HW Helper P: 3,323 Oh awesome! You found the answer without using any crude assumptions like I did And I'm hardly a genius, I just gave you the kick start you needed. Using the geometric mean <= arithmetic mean was the real genius here

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