I need to solve:

$\displaystyle \frac{1}{x} e^{-\frac{3}{x}} = 0.1086$

My method:

$\displaystyle \frac{1}{x} e^{-\frac{3}{x}} = 0.1086$

$\displaystyle e^{-\frac{3}{x}} = 0.1086x$

$\displaystyle \ln \left(e^{-\frac{3}{x}} \right)= \ln (0.1086x)$

$\displaystyle -\frac{3}{x} = \ln (0.1086) + \ln (x)$

But I'm stuck from here. Don't understand what to do.