${\ln\left(x\right)}=(t-x)^2$
$\pm\sqrt{\ln\left(x\right)}+x=t$
$\mathrm{e}^{\sqrt{\ln\left(x\right)}+x}=e^t$
And that is as close as I can get it to the form $x\mathrm{e}^x$. What do I do next? Is it possible to solve it this way ? Are there any generalizations?
