Answer
$$x = 1 + 2t;{\text{ }}y = 2 - t$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = {t^2}{\bf{i}} + \left( {2 - \ln t} \right){\bf{j}};{\text{ }}{t_0} = 1 \cr
& {\text{Calculate }}{\bf{r}}\left( {{t_0}} \right) \cr
& {\bf{r}}\left( 1 \right) = {\left( 1 \right)^2}{\bf{i}} + \left( {2 - \ln 1} \right){\bf{j}} \cr
& {\bf{r}}\left( 1 \right) = {\bf{i}} + 2{\bf{j}} \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( {{t_0}} \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{t^2}{\bf{i}} + \left( {2 - \ln t} \right){\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = 2t{\bf{i}} - \frac{1}{t}{\bf{j}} \cr
& {\bf{r}}{\text{'}}\left( 1 \right) = 2\left( 1 \right){\bf{i}} - \frac{1}{1}{\bf{j}} \cr
& {\bf{r}}{\text{'}}\left( 1 \right) = 2{\bf{i}} - {\bf{j}} \cr
& {\text{Let }}{{\bf{r}}_0} = {\bf{r}}\left( {{t_0}} \right){\text{ and }}{{\bf{v}}_0}{\text{ = }}{\bf{r}}{\text{'}}\left( {{t_0}} \right){\text{ }} \cr
& {\text{The tangent line to the graph of }}{\bf{r}}\left( t \right){\text{ at }}{{\bf{r}}_0}{\text{ is given by the vector }} \cr
& {\text{equation}} \cr
& {\bf{r}} = {{\bf{r}}_0} + t{{\bf{v}}_0} \cr
& {\bf{r}} = {\bf{r}}\left( {{t_0}} \right) + t{\bf{r}}{\text{'}}\left( {{t_0}} \right){\text{ }} \cr
& {\bf{r}} = {\bf{i}} + 2{\bf{j}} + t\left( {2{\bf{i}} - {\bf{j}}} \right) \cr
& {\bf{r}} = {\bf{i}} + 2{\bf{j}} + 2t{\bf{i}} - t{\bf{j}} \cr
& {\bf{r}} = \left( {1 + 2t} \right){\bf{i}} + \left( {2 - t} \right){\bf{j}} \cr
& {\text{Thus}},{\text{the parametric equations of the tangent line at }}t = {t_0}{\text{ are:}} \cr
& x = 1 + 2t;{\text{ }}y = 2 - t \cr} $$
