Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$



I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from solve this question...
Let $a = 2^x, b = 2^y$. Use some algebraic manipulation to arrive at $(a - 1)^2 + (b - 1)^2 = 2$

This is a circle of radius $\sqrt{2}$ centered at $(1, 1)$, and we want to find the minimum and maximum values of $a + b$.

So $a + b = M \rightarrow b = -a + M$, which is the equation of a line with slope $-1$. The maximum and minimum will be when we find the maximum and minimum y-intercept of this line such that it intersects at least one point on the circle with $a, b &gt; 0$

It's tangent to the circle at the top right when $a = b = 2$ which gives a value of $M = 4$. The smallest y-intercept occurs at $m = 2$, however, we do not include $2$ because it makes one of the coordinates zero. This establishes $(2, 4]$ as the range.
Let $\delta x = y-x$ and $x\ge y$. Then

$$4^x(1+4^{\delta x}) = 2^{x+1}(1+2^{\delta x})$$

then from factorization we can claim that:

$$2x = x+1 ~\text{and}~~ 4^{\delta x} = 2^{\delta x}$$

So, we get $x = y = 1$.