Solve $x^2 - \frac{1}{\ln(x) - 1} + \frac{1}{x^2(\ln(x)-1)} = 0$

Question

How would I find the roots of this function?

\begin{equation} x^2 - \frac{1}{\ln(x) - 1} + \frac{1}{x^2(\ln(x)-1)} = 0 \end{equation}

I don't really know how i would tackle this problem because we're working with natural logarithms. Does anyone have an idea?

Answer 1

If you plot the fuction, you see solutions around $x=1$ and $x=3$ from which you can start Newton method.

You can have approximations using series expansions.

Around $x=0$, $$x^2 - \frac{1}{\ln(x) - 1} + \frac{1}{x^2(\ln(x)-1)} = 1+4 (x-1)+O\left((x-1)^3\right)$$ and then the solution is close to $$x=\frac{3}4$$

Around $x=3$, the expansion would be $$\left(9-\frac{8}{9 (\log (3)-1)}\right)+\frac{2 (x-3) \left(86+81 \log ^2(3)-163 \log (3)\right)}{27 (\log (3)-1)^2}+O\left((x-3)^2\right)$$ and the the solution is close to $$x=\frac{3 \left(83+81 \log ^2(3)-156 \log (3)\right)}{2 \left(86+81 \log ^2(3)-163 \log (3)\right)}\approx 3.00039$$

Edit

With regard to the first root, you could have better and better rational approximations building the $[1,n]$ Padé approximant. This would give $$\left( \begin{array}{ccc} n & x_{(n)} & x_{(n)} \approx \\ 1 & \frac{3}{4} & 0.750000 \\ 2 & \frac{25}{33} & 0.757576 \\ 3 & \frac{1247}{1643} & 0.758977 \\ 4 & \frac{5179}{6822} & 0.759162 \\ 5 & \frac{11947}{15737} & 0.759166 \\ 6 & \frac{1339361}{1764260} & 0.759163 \end{array} \right)$$

Answer 2

Hint: By a numeric method we find $$x\approx 0.75916167335311382164$$ or $$x\approx 3.00039175573599420937$$

Answer 3

I usually do a quick plot like Plot[x^2-1/(Log[x]-1)+1/(x^2(Log[x]-1)),{x,-10,+10}] which shows the above roots quickly and that there are no others.