<s>[INST] $2^{10 - x} \cdot 2^{10 - x} = 4^{10-x}$ $$2^{10 - x} \cdot 2^{10 - x} = 4^{10-x}$$
Is that correct?
I would've done
$$
2^{10 - x} \cdot 2^{10 - x}\;\; = \;\;
(2)^{10 - x + 10 - x} \; = \; (2)^{2 \cdot (10 - x)} \;=\; 4^{10 - x}\tag{1}
$$
Is that allowed?

If so, can I say that
$$
\frac{4^x}{2^y} =  2^{x - y} \tag{2}
$$
 [/INST] $$\text{Yes:}\;\;\;\large 2^{10 - x} \cdot 2^{10 - x} =\; 4^{10-x}\tag{1.a}$$

$$\underbrace{\large 2^{10 - x} \cdot 2^{10 - x} 
= (2)^{10 - x + 10 - x} = (2)^{2 \cdot (10 - x)} = 4^{10 - x}}\tag{1.b}
$$
  $$\text{Yes, that is that is allowed.}$$



$$\underbrace{
\large \frac{4^x}{2^y} =  2^{x - y}}\;\;\; ?\tag{2 ?}$$
  $$\text{NOT correct. See below:}$$ 

$$(2)\quad \large \frac{4^x}{2^y} \;=\; 4^x \cdot 2^{-y} \;= \;(2^2)^x \cdot 2^{-y} \;=\; 2^{(2x)} \cdot 2^{(-y)} \;=\;2^{(2x-y)}$$
</s>