Answer
$$\eqalign{
& \left( {\text{a}} \right)\frac{1}{9} \cr
& \left( {\text{b}} \right)\frac{7}{2} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right){\text{ Let }}f\left( x \right) = {e^{ - 2x}},{\text{ evaluate }}f\left( {\ln 3} \right) \cr
& f\left( {\ln 3} \right) = {e^{ - 2\ln 3}} \cr
& {\text{Use the property }}a\ln b = \ln {b^a} \cr
& f\left( {\ln 3} \right) = {e^{\ln {3^{ - 2}}}} \cr
& {\text{Use the property }}{e^{\ln x}} = x \cr
& f\left( {\ln 3} \right) = {3^{ - 2}} \cr
& f\left( {\ln 3} \right) = \frac{1}{9} \cr
& \cr
& \left( {\text{b}} \right){\text{ Let }}f\left( x \right) = {e^x} + 3{e^{ - x}},{\text{ evaluate }}f\left( {\ln 2} \right) \cr
& f\left( {\ln 2} \right) = {e^{\ln 2}} + 3{e^{ - \ln 2}} \cr
& {\text{Use the property }}a\ln b = \ln {b^a} \cr
& f\left( {\ln 2} \right) = {e^{\ln 2}} + 3{e^{\ln {2^{ - 1}}}} \cr
& f\left( {\ln 2} \right) = 2 + 3\left( {{2^{ - 1}}} \right) \cr
& f\left( {\ln 2} \right) = 2 + 3\left( {\frac{1}{2}} \right) \cr
& f\left( {\ln 2} \right) = 2 + \frac{3}{2} \cr
& f\left( {\ln 2} \right) = \frac{7}{2} \cr} $$
