Answer
$x = t$, $y = t$, $z = 1 + \ln(2)t$
Work Step by Step
Given: $x = \ln(t+1)$, $y=t\cos2t$, $z=2^{t}$
In order to find the tangent vector $r'(t)$ we have to take the derivative of $r(t)$.
Thus, $r'(t) = \frac{1}{t+1}i + (\cos(2t) - 2t\sin(2t))j + 2^{t}\ln(2)k$
By setting one component of $r(t)$ equal to the corresponding given point $(0,0,1)$, we get $t=0$.
$r'(0) = \frac{1}{0+1}i + (\cos(0t) - 2*0\sin(0t))j + 2^{0}\ln(2)k = i + j + (\ln2)k$
Equation of tangent line passing through the point $(0,0,1)$ is: $$r(t) =<0,0,1>+t<1,1,\ln 2> $$
Therefore, the parametric equations of the line are: $$ x=t,y=t, z=1+\ln(2)t$$
