Answer
$$L = 9$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {{e^{2t}},2{e^{2t}} + 5,2{e^{2t}} - 20} \right\rangle {\text{, for }}0 \leqslant t \leqslant \ln 2 \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {{e^{2t}},2{e^{2t}} + 5,2{e^{2t}} - 20} \right\rangle \cr
& {\text{differentiating}} \cr
& {\bf{r}}'\left( t \right) = \left\langle {2{e^{2t}},4{e^{2t}},4{e^{2t}}} \right\rangle \cr
& {\text{Use the Definition of Arc Length for Vector Functions }} \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& L = \int_0^{\ln 2} {\left| {\left\langle {2{e^{2t}},4{e^{2t}},4{e^{2t}}} \right\rangle } \right|} dt \cr
& L = \int_0^{\ln 2} {\sqrt {{{\left( {2{e^{2t}}} \right)}^2} + {{\left( {4{e^{2t}}} \right)}^2} + {{\left( {4{e^{2t}}} \right)}^2}} } dt \cr
& {\text{simplifying}} \cr
& L = \int_0^{\ln 2} {\sqrt {4{e^{4t}} + 16{e^{4t}} + 16{e^{4t}}} } dt \cr
& L = \int_0^{\ln 2} {\sqrt {36{e^{4t}}} } dt \cr
& L = \int_0^{\ln 2} {6{e^{2t}}} dt \cr
& {\text{integrating}} \cr
& L = \left( {3{e^{2t}}} \right)_0^{\ln 2} \cr
& L = 3{e^{2\ln 2}} - 3{e^0} \cr
& L = 12 - 3 \cr
& L = 9 \cr} $$
