tight estimate for a log-linear inequality

Question

Given $q>0$ and $p$, how do we get a tight estimate for the smallest $x$ such that $x\log(x)+px \geq q$? (such an $x$ always exists).

Answer 1

Consider the function $$f(x)=x\log(x)+px - q$$ The first derivative $$f'(x)=\log (x)+1+p$$ cancels at a single place $x=e^{-(p+1)}$ and at this point the value of the function is $-(q+e^{-(p+1)})$. The second derivative being always positive, then this point corresponds to a minimum.

So, the function (which is defined for $x \geq 0$) starts from $-q$, goes to a minimum and increases again. This means that the inequality $$x\log(x)+px \geq q$$ will hold for any $x$ larger than the root of $f(x)=0$.

The solution of this equation is given by $$x=\frac{q}{W\left(q \,e^p \right)}$$ where appears Lambert function.

Since $q>0$, the solution is larger than $x_0=e^{-p}$ and the first iterate of Newton method would be $x_1=q+e^{-p}$. But $f(x_0)f''(x_0)<0$ makes $x_1$ an overestimate of the solution (from Darboux theorem).

Edit

Since, in a comment, you tell that your problem is to find $n$ such as $$\frac{2^n}{n!} < \epsilon$$ If ound in old notes of mine a similar problem that I adapted to your. Considering $\epsilon=10^{-k}$, I was able to find this simple correlation $$n\approx 3.45433 + 2.52846\, k^{0.746311}$$ which I established for $5 \leq k \leq 50$.

So, if you want to refine more, use this as an estimate from which you start Newton method using Stirling approximation for $n!$.