Equality of $2^x + 2^{4-x} \geq 8$ 
Want to find the value(s) of $x$ for which equality holds in $$2^x + 2^{4-x} \geq 8$$

I've found it by solving $2^x + 2^{4-x} = 8$: 
$$2^x + 2^{4-x} = 8 \Rightarrow 2^{2x} - 8 \cdot 2^x + 16 = 0 \Rightarrow (2^x - 4)^2 = 0$$
so clearly $x = 2$. 
However, the notes I've been reading goes straight to simply say that equality occurs when $2^x = 2^{4-x}$, i.e. when $x = 4 - x$ and again $x=2$. 

I'm probably missing something really simple here, but why does equality hold simply when $$2^x = 2^{4-x}?$$
Edit: to clarify, I want to know why is this is an 'obvious' condition for equality without writing the equals sign and solving the quadratic?

You could argue it by the arithmetic-geometric mean inequality:
$$\frac{2^x + 2^{4 - x}}{2} \ge \sqrt{2^x \cdot 2^{4 - x}} = \sqrt{2^4} = 4,$$
with equality if and only if $2^x = 2^{4 - x}$.
