Working through a problem, I found an inequality that I can't seem to find a closed form solution for. The inequality is as follows:
$$\alpha (1+r+cx)^{-4x} \leq \beta,$$
for which I'm trying to find a closed form solution to $x$. We can simplify this a little by taking logs on both sides,
$$\ln\alpha - 4x\ln(1+r+cx) \leq \ln\beta$$ $$\iff x\ln(1+r+cx) \geq \frac{\ln \alpha - \ln \beta}{4}$$
But I don't think I can simplify this much further.
Now, since I know the value of all parameters except $x$, I can try write some code to figure out close-enough numerical solutions. However, I suspect there must be some prior art on figuring out a closed-form approximation to expressions like these—can anyone point me towards the right ideas?
You may assume that $c>0, r>0, \alpha>0, \beta>0$.
Let us consider the equation $$ x\log(1+r+cx) = \frac 14 \log \left(\frac{\alpha }{\beta }\right)$$ Now, let $$y=cx\qquad \qquad s=r+1 \qquad\qquad k=\frac c4 \log \left(\frac{\alpha }{\beta }\right)$$ and so, we look for the zero of function$$f(y)=y \log(s+y)-k$$ which as a formal solution in terms of the generalized Lambert function (have a look at equation $(4)$ in the linked paper).
This being said, from pratical point of view, it is not very useful. For sure the solution is upper bounded by $$y_{\text{max}}=\frac{k}{W(k)}=a$$
Using a series expansion of $f(y)$ around $y=a$ and series revesion would give $$y=a+t-\frac{ (a+2 s)}{2 ((a+s) ((a+s) \log (a+s)+a))}t^2+O\left(t^3\right)$$ where $$t=\frac{k-a \log (a+s)}{\frac{a}{a+s}+\log (a+s)}$$
Trying with $s=12.345$ and $k=2.3456$, this would give, as an approximation $y=0.914937$ while the solution, given using Newton method, is $y=0.907670$ (relative error $= 0.80$%).
Adding one more to term to the expansion would lead to the estimate $y=0.908780$ (relative error $= 0.12$%).
$$\color{red} {\text{This needs to be seriously tested on a lot of cases}}$$