Answer
$$\eqalign{
& \left( {\text{a}} \right){e^{ - x\ln \pi }} \cr
& \left( {\text{b}} \right){e^{2x\ln x}} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right){\pi ^{ - x}} \cr
& {\text{Express as a power of }}e,{\text{ recall that }}{x^y} = {e^{y\ln x}},{\text{ then}} \cr
& {\pi ^{ - x}} = {e^{ - x\ln \pi }} \cr
& \cr
& \left( {\text{b}} \right){x^{2x}},{\text{ }}x > 0 \cr
& {\text{Express as a power of }}e,{\text{ recall that }}{x^y} = {e^{y\ln x}},{\text{ then}} \cr
& {x^{2x}} = {e^{2x\ln x}},{\text{ }}x > 0 \cr} $$