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

Question

I encountered an equation of type $$ ax + b\log(x) + c=0$$ Here a, b, and c are constants. Does anyone know how to solve these type of equations? I guess this way: $$\log(x)= \frac{c-ax}{b}$$ $$x= 10^{(c-ax)/b}$$ But I do not even know how to solve this too. Please help!

Answer 1

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 $$ x=(b/a)\mathrm{W}\left((a/b)e^{-c/b}\right) $$

Answer 2

This doesn't have a solution in the 'normal' functions.
http://en.wikipedia.org/wiki/Lambert_W_function tells you all about it.
If you substitute $x=by/a$, you can rearrange to get $y+\log(y) = d$, so $ye^y=e^d$