How to isolate $\alpha$ in $\frac{\alpha}{\log_2(4\alpha)}$ For my master thesis in computer science, I've got stuck at a point where I have to express $\alpha$ as a function of $\varepsilon$, when I know that 
$$\frac{\alpha}{\log_2(4\alpha)} > 96\varepsilon^{-1}$$
for $0 < \varepsilon < 1$. 
I have tried to move this around a lot, but keep having either both $\alpha$ and $\log_2\alpha$ being part of the equation or both $2^\alpha$ and $\alpha$, e.g. I can get it to be 
$$2^\alpha > (4\alpha)^{96\varepsilon^{-1}}$$ 
but do not know how to proceed from here. Any help is appreciated.
For the ones into computer science, this has to do with finding the sparcification constant when $k = 3$, but for everybody else I guess this is just a pure math question.  
Also, I'm not quite sure which tags to give this. Please comment if you know.
 A: The solution of 
$$\frac{\alpha}{\log_2(4\alpha)} = 96\,\varepsilon^{-1}$$ is given in terms of Lambert function
$$\alpha=-\frac{96 }{\varepsilon  \log (2)}W\left(-\frac{\log (2)}{384} \varepsilon  \right)$$
To evaluate it, since the argument is small $(\lt 0.002)$, use the series expansion
$$W(x)=x-x^2+\frac{3 x^3}{2}-\frac{8 x^4}{3}+O\left(x^5\right)$$ or, even better, the Padé Approximant
$$W(x)=\frac{x+\frac{4 }{3}x^2}{1+\frac{7 }{3}x+\frac{5 }{6}x^2}$$
Since $0 \leq \varepsilon\leq 1$, you can have a very good approximation using Taylor again to get
$$\alpha=\frac{1}{4}+\frac{  \log (2)}{1536}\varepsilon+\frac{\log
   ^2(2)}{393216}\varepsilon^2+\frac{ \log ^3(2)}{84934656}\varepsilon^3+O\left(\varepsilon ^4\right)$$
Update
Thinking more about the problem, I think that we could have avoided Lambert function. Let $\beta=4\alpha$ and rewrite the equation as
$$\varepsilon=\frac{384}{\log (2)}\frac{ \log (\beta )}{\beta  }$$ Expanding the rhs as a Taylor series built at $\beta=1$, this would give
$$\varepsilon=\frac{384 (\beta -1)}{\log (2)}-\frac{576 (\beta -1)^2}{\log (2)}+\frac{704 (\beta
   -1)^3}{\log (2)}+O\left((\beta -1)^4\right)$$ and then, using series reversion
$$\beta=1+\frac{\log (2)}{384} \varepsilon  +\frac{ \log
   ^2(2)}{98304}\varepsilon ^2+\frac{\log ^3(2)}{21233664}\varepsilon ^3 +O\left(\varepsilon ^4\right)$$ and, then, the same
$$\alpha=\frac{1}{4}+\frac{  \log (2)}{1536}\varepsilon+\frac{\log
   ^2(2)}{393216}\varepsilon^2+\frac{ \log ^3(2)}{84934656}\varepsilon^3+O\left(\varepsilon ^4\right)$$
