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I am trying to find a closed form solution to an equation of the form $$x^2+ax+b=c\log x$$ I found some resources that express solutions in terms of the Lambert W function if the LHS is linear but I can't find something similar for a quadratic LHS.

I was thinking I might try something along the lines of plugging in $x=1$ since this would give $$1+a+b=0$$ but that really doesn't tell me much other than perhaps a condition on $a$ and $b$?

I know that solutions exist, since this equation came from a problem I was trying to solve where each coefficients has specific values for which a plot shows the graph of $x^2+ax+b+c\log x$ intersecting the $x$-axis.

Edit: a closed form approximation in terms of large $x$ is also fine. For instance, I found that it is possible to approximate the logarithm by $$\log x\approx r x^{1/r}-r$$ where $r$ is a large positive constant. See this post. With this, the equation becomes $$x^2+ax+b=cr(x^{1/r}-1)$$ But if we raise the entire equation to the $r$-th power, the resulting equation will be of order $2r$ and still very difficult (if not impossible) to solve.

IV_
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RKLS
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2 Answers2

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There isn't a closed form solution.

You can complete the square on the left hand side to get something like $$(x-\mu)^{2} = d + c \ln x$$ then take $\exp $ of both sides to get $$n(x;\mu,\sqrt{|c|})= \lambda x^{\pm 1},$$ where $\pm$ depends on the sign of $c$ and $n$ is the Gaussian density. There are no closed form solutions for that sort of thing.

jrbr
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Unless the parameter $c$ is very large your solution won't be large because $x^2$ grows much faster than $\log(x)$. Therefore, large $x$ approximations are not likely to be useful.

The question is what type of solution you are looking for. There might be some generalisation of the Lambert W function that would describe solution to your problem, but even if so, such a solution won't necessarily have to be very useful, depending on what are your intentions with it. If you only need an approximate solution, then you can do the following:

  1. Guess approximate solution $\bar x$.

  2. Approximate the quadratic function $x^2 + ax+b$ by the linear function $$ \bar x^2+2 \bar x(x-\bar x) + ax + b = (2\bar x+a)x + [b-\bar x^2]. $$

  3. You have two options at this step:

    a) Describe the solution of the approximate equation using the Lambert W function.

    b) Even better, you can approximate the logarithm as well as $\log(x) \sim \log(\bar x) + \frac{x-\bar x}{\bar x}$, so you obtain a linear equation $$ (2\bar x+a)x + [b-\bar x^2] = \log(\bar x) + \frac{x-\bar x}{\bar x}. $$ Denote the solution $\phi(\bar x)$.

  4. If you want a better approximation, replace $\bar x$ by the approximate solution that you just obtained and repeat the process.

The above described process gives you extremely easy algorithm to find approximate value of the solution $x$ (if the iterative approximation converge depend on the actual parameters $a$, $b$ and $c$, and it initial guess $\bar x$). What is more, if $\bar x < x$, then $\phi(\bar x)$ gives you the upper bound on $x$, i.e. $x < \phi(x)$. Likewise, if $\bar x > x$, then $x > \phi(\bar x)$.

Bare in mind that the equation has 0,1, or 2 solutions (it has no more than two solutions as the LHS is convex whilst the RHS is concave). Thus, you need to try different stating points to obtain both solutions.

If instead of iterative algorithm you wanted a more precise approximation of $x$ in just one step, then you could approximate $\log(x)$ by its second-order Fourier approximation $\log(\bar x) + \frac{x-\bar x}{\bar x} - \frac{(x-\bar x)^2}{2\bar x^2}$. Equating it to $x^2 + ax+b$ gives you a quadratic equation for $x$. It should have two roots, you are interested in the root that is close to $\bar x$, otherwise the approximation error would be large.

Pavel Kocourek
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