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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.

mrtaurho
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The Contextual Path
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    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

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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.

ajotatxe
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