Answer
$ f\circ g(x)=\log_{2}(x-2),\qquad(2,+\infty$)
$ g\circ f(x)=\log_{2}x-2,\qquad(0,+\infty$)
Work Step by Step
Restriction on the domain for logarithmic functions:
argument must be positive.
$f\circ g(x)=f(g(x))=\log_{2}$(g(x))=$\log_{2}(x-2),$
restriction$:\quad x-2>0$
$x>2$
domain$=(2,+\infty$)
$g\circ f(x)=g(f(x)=f(x)-2=\log_{2}x-2$
restriction$:\quad x>0$
domain=$(0,+\infty$)
