2^x - 6 + 2 ^( 3-x) = 0

2^x - 6 + 2 ^( 3-x) = 0

$$2^x - 6 + 2 ^{3-x}= 0\\ 2^x - 6 + \frac{2 ^3}{2^x}= 0\\\\ (2^{x})^2 - 6*2^x + 8= 0\\\\ let\;\;y=2^x\\\\ y^2-6y+8=0\\\\ (y-2)(y-4)=0\\\\ y=2 or 4\\\\ 2^x=2\;\;or\;\;2^x=4\\\\ x=1\;\;or\;\;x=2$$