So the problem started off as the following

$99n = \log(n)^2 - log(n)$

And I want to solve for n. My thought is to raise both sides to the 10 but I don't think that would work... Expanded the equation looks like this

$99n = \log(n)\log(n) - log(n)$

So intuition tells me it wouldn't work. How would I go about solving for n? More generally, how do I get rid of the logs?

  • 1
    $\begingroup$ Mmm... rarely do equations like these have nice solutions. Wolfram itself only offers an approximation solution $n≈0.0357533$. (link) $\endgroup$ – Eevee Trainer Sep 9 at 4:45
  • $\begingroup$ Okay good to know, thank you! $\endgroup$ – financial_physician Sep 9 at 5:01
  • $\begingroup$ I assumed $\log_e(n)$ in my answer. $\endgroup$ – Claude Leibovici Sep 9 at 5:16

As said in comments, no hope for a closed form result.

You are looking for the zero's of function $$f(n)=99n + \log(n)- \log^2(n)$$ for which $$f'(n)=\frac{1}{n}-\frac{2 \log (n)}{n}+99$$ $$f''(n)=\frac{2 \log (n)}{n^2}-\frac{3}{n^2}$$

The first derivative cancels at $$n_*=-\frac{2}{99} W\left(-\frac{99 \sqrt{e}}{2}\right)$$ where appears Lambert function but, since its argument is $ < - \frac 1e$, this is not a real. This makes the derivative to be positive for any value of $n$.

For inspection, let $n=\frac{1}{e^{k}}$ and you will notice that the function change sign between $k=2$ and $k=3$. So, start Newton method with $n_0=\frac{1}{e^{5/2}}$. You should get the following iterates $$\left( \begin{array}{cc} m & n_m \\ 0 & 0.08208499862 \\ 1 & 0.08570849302 \\ 2 & 0.08575340627 \\ 3 & 0.08575341274 \end{array} \right)$$


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