solving an equation involving exponential and logarithmic function How can we solve this equation for $n$?
It's supposed to be easy, but I couldn't figure it out!
$$2^n \leq \left(\frac{en}{d}\right)^{dk}$$
The answer is $n=O(dk\log(dk)).$
Assume that $n,d,k,$ all are positive integers greater than one.
Thanks for your help,
 A: Since an exponential function grows faster than a power function, for sufficiently large $n$, $LHS \geq RHS$. However for intermediate values of $n$ the RHS grows faster than LHS. The crossover point depends on $d,k$. With minimum $k=2$, the cross over point for $d=2$ is about $n=18.6$ with $d=3$ it's about $28$, and continues increasing with $d$.
As $d$ increases from $2$ to $n$ then the RHS increases, until $d>n$ then RHS decreases with increasing $d$.
With some experimentation I found that $n=2.3d$ solves the inequality. Since $2.3d < 2d\cdot ln(2d)$ it is a more "accurate" answer. However the difference between the LHS and the RHS rapidly increases with increasing $n$. When $d=10,n=23$, the difference is almost $10^16. Hence, "accurate" in quotes.
With all that said here is a solution that makes $LHS=RHS$: $$
n=\frac{-dk}{ln 2}W(\frac{-ln 2}{ek}) \tag{1}$$
where $W$ is Lambert W function. $W(x) \approx \ln x-\ln \left(\ln x\right)+\frac{\ln \left(\ln x\right)}{1.0759\ln x}$. But using $ln( x)$ as a simpler, but coarser approximation, then (1) becomes $n \approx \frac{-dk}{ln 2}ln(\frac{-ln 2}{ek})$, which is vaguely reminiscent of the answer in the OP.
Bottom line, I don't know how to get the answer in the OP, other than it works by being large enough. However, the exact solution, and an approximation to it can be derived. 
Let me know if you need the steps to derive (1).
