Question

(1/2)^x+2+(1/2)^x-2=17

Answer 1

$(\frac{1}{2})^{x+2}+(\frac{1}{2})^{x-2} =17\\\\(\frac{1}{2})^{x-2}*((\frac{1}{2})^{4}+1)=17\\\\(\frac{1}{2})^{x-2} *(\frac{1}{16}+1)=17\\\\(\frac{1}{2})^{x-2}*\frac{17}{16}=17\\\\(\frac{1}{2})^{x-2}=16\\\\(\frac{1}{2})^{x-2}=(\frac{1}{2})^{-4}$

$x-2=-4\\\\x=-2$