If the Wronskian for a set of functions is also a function, how does it relate to linear independence?

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I have learnt that if the Wronskian is zero, then the set of functions are linearly dependent.

For example however, for the set of functinos $\{x,e^x,e^{2x}\}$:

This is the Wronskian for the set of functions (By definition) $$ \begin{vmatrix} x & e^{x} & e^{2x} \\ 1 & e^x & 2e^{2x} \\ 0 & e^x & 4e^{2x} \\ \end{vmatrix} $$

By cofactor expansion of the bottom row, it is equal to $-e^x(2xe^{2x}-e^{2x})+4e^{2x}(xe^x-e^x)$ which equals $(2x-3)e^{3x}$.

and this 'new' function has a value of $x$ which makes it equal to zero.

Does this mean that the set of functions are linearly dependent?

Thanks for the help