Answer
$${\text{ }}y = 6x - \ln 27 + 3$$
Work Step by Step
$$\eqalign{
& y = {e^{2x}}{\text{ at the point }}x = \frac{1}{2}\ln 3 \cr
& y\left( {\frac{1}{2}\ln 3} \right) = {e^{2\left( {\frac{1}{2}\ln 3} \right)}}{\text{ }} \cr
& y\left( {\frac{1}{2}\ln 3} \right) = 3 \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{e^{2x}}} \right] \cr
& \frac{{dy}}{{dx}} = 2{e^{2x}} \cr
& {\text{Calculate the slope at }}x = \frac{1}{2}\ln 3 \cr
& \frac{{dy}}{{dx}} = 2{e^{2\left( {\frac{1}{2}\ln 3} \right)}} \cr
& \frac{{dy}}{{dx}} = 6 \cr
& {\text{The equation of the tangent line is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 3 = 6\left( {x - \frac{1}{2}\ln 3} \right) \cr
& y - 3 = 6x - 3\ln 3 \cr
& y - 3 = 6x - \ln 27 \cr
& {\text{ }}y = 6x - \ln 27 + 3 \cr} $$
