Answer
Magnitude = $2\sqrt{2}$
direction = $\displaystyle \langle\frac{\cos\ln 2-\sin\ln 2}{\sqrt{2}}, \quad \frac{\cos\ln 2+\sin\ln 2}{\sqrt{2}}\rangle$ )
${\bf v}= 2\displaystyle \sqrt{2}\langle\frac{\cos\ln 2-\sin\ln 2}{\sqrt{2}}, \quad \frac{\cos\ln 2+\sin\ln 2}{\sqrt{2}}\rangle$
Work Step by Step
When $t=\ln 2$
${\bf v}=\langle e^{\ln 2}\cos\ln 2-e^{\ln 2}\sin\ln 2, \quad e^{\ln 2}\cos\ln 2+e^{\ln 2}\sin\ln 2\rangle$
$=\langle 2\cos\ln 2-2\sin\ln 2, \quad 2\cos\ln 2+2\sin\ln 2\rangle$
$=\langle 2(\cos\ln 2-\sin\ln 2), \quad 2(\cos\ln 2+2\sin\ln 2)\rangle$
Magnitude:
$|{\bf v}|=\sqrt{[2(\cos\ln 2-\sin\ln 2)]^{2}+[2(\cos\ln 2+\sin\ln 2)]^{2}}$
write $\ln 2$ as $t$, for brevity
$=\sqrt{4(\cos^{2}t-2\cos t\sin t+\sin^{2}t+\cos^{2}t+2\cos t\sin t+\sin^{2}t)}$
$=\sqrt{4(2\cos^{2}t+2\sin^{2}t)}$
$=\sqrt{4\cdot 2(\cos^{2}t+\sin^{2}t)}$
$=\sqrt{8}$
$=2\sqrt{2}$
So
${\bf v}= 2\displaystyle \sqrt{2}\langle\frac{\cos\ln 2-\sin\ln 2}{\sqrt{2}}, \quad \frac{\cos\ln 2+\sin\ln 2}{\sqrt{2}}\rangle$
