# Solving Problems of the Type $\log(x) = ax^2 + bx + c$

I have recently been struggling to solve the following advanced mathematics problem which is presented as $$|0.1 x^2 + 2 x + 3| = \log(x)$$. I know that before solving the problem, it must be taken into account that the initial equation must have both positive and negative aspects on the left side since there exists a modulus, therefore:

\begin{align} 0.1 x^2 + 2 x + 3 = \log(x)\\ -0.1 x^2 - 2 x - 3 = \log(x) \end{align}

But from then on I do not have any idea as to how to solve such problems which are of the form $$\log (x) = a x^2 + b x + c$$. I think that it has something to do with getting rid of the log, but I do not know how to do it cleanly. If I take everything to the exponential of 10, I would have $$x$$ by itself but then I would have the problem $$10^{-0.1 x^2 + 2 x + 3} = x$$ and I am stumped by that as well. I would be much obliged if anyone could provide advice as to how such a problem could be solved.

• First of all I guess this has nothing to do with the Polylogarithm. It would be more likely that the Lambert W-function will appear somewhere since it is defined as the inverse function of $f(x)=xe^x$ for the purpose to solve similiar equations like yours. But I am not sure if there is a closed-form solution for general $a,b,c$. In your case the solutions are given by $x=-1.19812\mp 1.31258i...$ for the first equation and $x=0.0454515...$ for the second one according to WolframAlpha. – mrtaurho Oct 20 '18 at 18:13
• I doubt there is any closed-form solution to this type of equation. I am not sure that even the Lambert W would help. – Jair Taylor Oct 20 '18 at 18:27

Since $$0.1x^2+x+3=0.1(x+5)^2+0.5>0$$ we have that $$|0.1x^2+2x+3|\ge 0.1x^2+2x+3>x>\log x$$ The equation has no real solution.