If $P = 2x$ and $Q = 2^x$, then $P^a \cdot Q^{-a}$ in $x$ is—

$P^aQ^{-a}=\dfrac{(2x)^a}{(2^x)^a}=\dfrac{2^a x^a}{2^{ax}}=\dfrac{x^a}{2^{a(x-1)}}$.