Since
[tex]17^a = 16 = 4^2 \implies 17^{a/2} = 4[/tex]
it follows that
[tex]17^b = 4 \implies 17^{a/2} = 17^b \implies \dfrac a2 = b \implies a = 2b[/tex]
Then
[tex]2^{a - b} = 2^{2b - b} = 2^b[/tex]
so that
[tex]\dfrac1{2^{a-b}} = 2^{-b}[/tex]
We also have
[tex]17^b = 4 \implies b = \log_{17}(4)[/tex]
so we can go on to say
[tex]\dfrac1{2^{a-b}} = \boxed{2^{-\log_{17}(4)}}[/tex]
