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.
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.