Consider $xe^{-x}=\epsilon$ for $\epsilon\ll1$. If $\epsilon=0$ then $x=0$ is a solution. Taking logs we find
$$\log{x} -x - \log{\epsilon} = 0$$
where each term is labelled $(a),(b)$ and $(c)$ respectively. Now my tutor has stated that
$$(a)\sim(b) \implies x\sim1$$
but how is this implication found?
So $x \sim 1$ is really not quite the right way to write what's going on here. The point is that to have $\log(x)$ balance with $x$, you need $x$ to be bounded away from $0$ and away from infinity. If $x$ gets too small or too large then $\log(x)-x$ will become a large negative number. The maximum of this function is at $x=1$ where it is equal to $-1$. So if you want $\log(x)-x$ to not contribute to your leading order term, then you will need $x=O(1)$ as $\epsilon$ goes to zero. Note that this isn't really incompatible with your $\epsilon=0$ situation because your equation with the logarithms makes no sense with $\epsilon=0$.
That said I'm not sure I see why you would want $\log(x)$ to balance with $x$ in the first place in this problem.