Consider the perturbed equation for $\epsilon\ll1$,
$$xe^{-x} = \epsilon, \qquad \tag{1}$$ $$\log{(x)}-x-\log{(\epsilon)}=0\tag{2}$$
By the method of dominant balance, one can find that $x\sim \log{(1/\epsilon )}$. However now my lecturer writes that by substituting $x=B\log{(1/\epsilon)}$ into either equation $(1)$ or $(2)$ we find that
$B=1$ and so $x=\log{(1/\epsilon)}$. However, when I substitute this in I can't seem to come to this conclusion.
See, by $(2)$ I have
$$\log(B\log(1/\epsilon))-B\log(1/\epsilon)-\log(\epsilon)=0$$
$$\log(B) + \log(\log(1/\epsilon))-(1-B)\log(1/\epsilon)=0$$
Now, I don't really know the value of $\log(B)$ so I can't discount that, $\log(\log(1/\epsilon))$ is large so I can't discount that and $\log(1/\epsilon)$ is also large but not as large as the previous term so we may discount this term leaving
$$\log(B) + \log(\log(1/\epsilon))=0$$
or
$$B= 1/\log(1/\epsilon)$$
which is clearly not $B=1$. Where have I gone wrong?