$2^{x-3} + \frac {15}{2^{3-x}} = 256$ $$2^{x-3} + \frac {15}{2^{3-x}} = 256$$


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*Find the unknown $x$. 


My attempt: 
We know that $x^y . x^b = x^{y+b}$.
$$2^x . 2^{-3} + 15. 2^{-3+x} = 2^8$$
and 
$$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8$$
From here, we get
$$2^x + 15 = 2^8$$
However, I'm stuck at here and waiting for your kindest helps. 
Thank you. 

$2^{x-3} + 15\cdot 2^{x-3} = 256;$
$2^{x-3}(1+15) =256;$
$2^{x-3} =256/(16)= 16;$
$x-3=4$ , by inspection. 
$x=7.$
Note:
$\dfrac{1}{2^{3-x}}= 2^{x-3}$.
