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


*

*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. 
 A: writing $$\frac{2^x}{8}+\frac{15}{8}2^x=2^8$$ so
$$2^x\left(\frac{1}{8}+\frac{15}{8}\right)=2^8$$
Can you finish?
A: hint
If we put $$t=2^{x-3}, $$
the equation becomes
$$t+15t=256$$
or
$$t=16$$
$$2^{x-3} = 16$$
$$x=7$$
A: Hint:
As $2^{-y}=\dfrac1{2^y}$
$$\dfrac1{2^{3-x}}=2^{-(3-x)}=?$$
A: $2^{x-3} + 15\cdot 2^{x-3} = 256;$
$2^{x-3}(1+15) =256;$
$2^{x-3} =256/(16)= 16;$
$x-3=4$ , by inspection. 
$x=7.$
Note:
$\dfrac{1}{2^{3-x}}= 2^{x-3}$.
A: "$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8"$
"From here, we get"
"$2^x + 15 = 2^8$"
Uh....What?    No, you don't.
You get 
$2^x . 2^{-3}*2^3 + 15. 2^{-3}2^3 . 2^x = 2^8*2^3"$
$2^x + 15*2^x = 2^{8+3}$
So $16 *2^x = 2^{11}$
$2^{x+4} = 2^{11}$
$x+4 = 11$
$x = 7$.
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I'd find it easier to do. Let $y = x-3$ so 
$2^y + \frac 15{2^{-y}} = 256$
$2^y + 15*2^y = 256$
$2^y(1 + 15) = 256$
$2^y(16)= 256$
$2^y = \frac {256}{16} = \frac {2^8}{2^4} = 2^4$
$y = x-3 = 4$
$x = 3$.
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Someone else suggested: Let $t = 2^{x-3}$ so
$t + \frac {15}{\frac 1t} = 256$ so
$t + 15t = 256$ etc....
