<s>[INST] 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?
 [/INST] $(2^{x/2}-2^{(2-x/2)})^2 \ge 0;$
$2^x +2^{4-x} -2(2^{x/2}2^{(2-x/2)}) \ge 0;$
$2^x +2^{4-x} \ge 8;$
Equality for $2^{x/2}= 2^{(2-x/2)}$ (Why?);
$2^x=2^2$, or $x=2$.
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