Answer
$${\bf{N}}\left( 2 \right) = \frac{1}{{\sqrt 5 }}{\bf{i}} - \frac{2}{{\sqrt 5 }}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = t{\bf{i}} + \ln t{\bf{j}},{\text{ }}t = 2 \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {t{\bf{i}} + \ln t{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = {\bf{i}} + \frac{1}{t}{\bf{j}} \cr
& {\text{Find the unit Tangent Vector }}{\bf{T}}\left( t \right) \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{\left\| {{\bf{r}}'\left( t \right)} \right\|}},{\text{ }}{\bf{r}}'\left( t \right) \ne 0 \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{i}} + \frac{1}{t}{\bf{j}}}}{{\left\| {{\bf{i}} + \frac{1}{t}{\bf{j}}} \right\|}} = \frac{{{\bf{i}} + \frac{1}{t}{\bf{j}}}}{{\sqrt {1 + \frac{1}{{{t^2}}}} }} = \frac{{t\left( {{\bf{i}} + \frac{1}{t}{\bf{j}}} \right)}}{{\sqrt {{t^2} + 1} }} = \frac{{t{\bf{i}} + {\bf{j}}}}{{\sqrt {{t^2} + 1} }} \cr
& {\text{Find }}{\bf{T}}'\left( t \right) \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{t}{{\sqrt {{t^2} + 1} }}{\bf{i}} + \frac{1}{{\sqrt {{t^2} + 1} }}{\bf{j}}} \right] \cr
& {\text{Differentiate by hand and replace}} \cr
& {\bf{T}}'\left( t \right) = \frac{1}{{{{\left( {{t^2} + 1} \right)}^{3/2}}}}{\bf{i}} - \frac{t}{{{{\left( {{t^2} + 1} \right)}^{3/2}}}}{\bf{j}} \cr
& {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = 2 \cr
& {\bf{T}}'\left( 2 \right) = \frac{1}{{{{\left( {{2^2} + 1} \right)}^{3/2}}}}{\bf{i}} - \frac{2}{{{{\left( {{2^2} + 1} \right)}^{3/2}}}}{\bf{j}} \cr
& {\bf{T}}'\left( 2 \right) = \frac{1}{{{5^{3/2}}}}{\bf{i}} - \frac{2}{{{5^{3/2}}}}{\bf{j}} \cr
& {\text{Finding the Principal Unit Normal Vector}} \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left\| {{\bf{T}}'\left( t \right)} \right\|}} \cr
& {\text{Evaluate }}{\bf{N}}'\left( t \right){\text{ at }}t = 2 \cr
& {\bf{N}}\left( 2 \right) = \frac{{{\bf{T}}'\left( 2 \right)}}{{\left\| {{\bf{T}}'\left( 2 \right)} \right\|}} = \frac{{\frac{1}{{{5^{3/2}}}}{\bf{i}} - \frac{2}{{{5^{3/2}}}}{\bf{j}}}}{{\left\| {\frac{1}{{{5^{3/2}}}}{\bf{i}} - \frac{2}{{{5^{3/2}}}}{\bf{j}}} \right\|}} \cr
& {\bf{N}}\left( 2 \right) = \frac{{\frac{1}{{{5^{3/2}}}}{\bf{i}} - \frac{2}{{{5^{3/2}}}}{\bf{j}}}}{{1/5}} \cr
& {\bf{N}}\left( 2 \right) = \frac{1}{{\sqrt 5 }}{\bf{i}} - \frac{2}{{\sqrt 5 }}{\bf{j}} \cr} $$
