I have a parametric curve defined by :
$\gamma\colon\left[ 0,+\infty \right] \to \mathbb{R}^2$ defined: $\gamma(t) = (\ln(t), 3\cdot\ln(6t))$
Now I have to say if the plot of this curve is a line of $\mathbb{R}^2$ and explain why. I don't know what $\mathbb{R}^2$ means, so I can't answer the question. Can anyone help?
By definition: $$\Bbb R^2 := \Bbb R \times \Bbb R = \{(x,y) \mid x,y \in \Bbb R\},$$ that is, the euclidean plane that you're used to, in doing analytic geometry, graphing functions, etc.
For your exercise, call $x(t) = \ln t$ and $y(t) = 3 \ln (6t)$. Use properties of $\ln$ and eliminate $t$ from these relations, get $y$ as a function of $x$. You'll see something familiar.