How to solve equations with logarithms, like this: $ ax + b\log(x) + c=0$

Rewriting the equation gives $$ ax+b\log(x)+c=0\\ ax+b\log\left(\frac abx\right)+c-b\log\left(\frac ab\right)=0\\ \frac abx+\log\left(\frac abx\right)+\frac cb-\log\left(\frac ab\right)=0\\ \color{#C00000}{(a/b)x}\,e^{\color{#C00000}{(a/b)x}}=(a/b)\,e^{-c/b} $$ The Lambert-W function is the inverse of $xe^x$. Therefore, $$ \color{#C00000}{(a/b)x}=\mathrm{W}\left((a/b)e^{-c/b}\right) $$ and so $$ x=(b/a)\mathrm{W}\left((a/b)e^{-c/b}\right) $$