Semantics: 'determinant bundle', top exterior power of vector bundle From what I can dig up, given a vector bundle $E\rightarrow X$, the determinant bundle associated to this is $\Lambda^{n}E\rightarrow X$, where $n$ is the rank of $E\rightarrow X$.  Is this the same thing as "the determinant of the vector bundle $E\rightarrow X$"?
If this is true, then does the statement "$E\rightarrow X$ has trivial determinant" mean that $\Lambda^{n}E\rightarrow X$ is trivializable as a vector bundle"?
 A: The answer to both of your questions is yes! This can be seen through the construction of the top exterior bundle via transition functions. 
The top exterior power  of a vector bundle
$\Lambda^n E$ of a rank $n$ vector bundle $E$ has transition functions $\wedge^n g_{UV}:U\cap V\to\text{GL}(\wedge^n\mathbb{C}^n) = \mathbb{C}^*$ where $g_{UV}$ are the transition functions coming from $E$. This immediately implies what we are looking at is a line bundle and the expansion is shown cleanly as;
\begin{align*}
(\wedge^ng_{UV})(e_1\wedge\dots\wedge e_n) 
&= \wedge_{i=1}^ng_{UV}(e_i) \\
    &= \wedge_{i=1}^n\left(\sum_{j=1}^n g_{ji}e_j\right)\\
    &= \left[\sum_{\sigma \in S_n}(-1)^{|\sigma|}\prod_{i=1}^n g_{i\sigma(i)}\right] \wedge_{i=1}^ne_i \\
    &= \det(g_{UV})\cdot e_1\wedge\dots\wedge e_n.
\end{align*}
where the $e_i$'s denote the standard basis vectors and we have made use of the relations 
\begin{align*}
\alpha\wedge\alpha &= 0\\
 \alpha\wedge\beta &= -\beta\wedge\alpha
\end{align*} for any $\alpha,\beta\in \Omega^1(E)$.
This line bundle is, for now apparent reasons, known as the determinant bundle of $M$ and is also denoted $\det(E)$.
Finally, saying that $E$ having trivial determinant means that the determinant bundle of $E$ is a trivial bundle.
Note that the notion of top exterior power (for a vector space) is in fact the formal definition of determinant. That is the determinant is the unique multi-linear functional acting on $n$ vectors in an $n$-dimensional space which is alternating and whose evaluation on the standard basis is 1 (i.e. preserves the volume of the unit cube). 
