Help in solving an inequality involving floor function

Question

I am attempting to solve the following inequality to determine the value of $k$. Note that $n,k\in\mathbb{N}$ with $n>100$. I have been able to reduce the inequality to end up with the floor function of $\log_{2}k$ which is where I am stuck. I need to find the minimum value of $k$ satisfying the inequality. \begin{align*} \lfloor\log_{2}(k)\rfloor(2n^{3}-n^{2})+2n^{2}-n &>4n^{2}+(\lfloor\log_{2}(k)\rfloor-1)n\\ \lfloor\log_{2}(k)\rfloor&>\frac{2n^{2}}{2n^{3}-n^{2}-n} \end{align*}

Answer 1

Since $\lfloor \log_2 (k) \rfloor$ must be an integer, you can start by finding the smallest integer $K$ satisfying

$$K > \frac{2 n^2}{2 n^3 - n^2 - n}$$

If the inequality were $\geq$, then "the least integer greater than or equal to this expression" is exactly what the ceiling function represents, so you could just write $K = \lceil \frac{2 n^2}{2 n^3 - n^2 - n} \rceil$. But because the inequality is a strict one, we have to be careful since if that fraction is an integer then $\lceil \frac{2 n^2}{2 n^3 - n^2 - n} \rceil = \frac{2 n^2}{2 n^3 - n^2 - n} \ngtr \frac{2 n^2}{2 n^3 - n^2 - n}$. In fact, in this case the smallest integer satisfying the inequality is $K = \frac{2 n^2}{2 n^3 - n^2 - n} + 1$.

How to achieve that? Well, you can do it by taking the floor instead of the ceiling, and adding that extra 1, i.e. $K = \lfloor \frac{2 n^2}{2 n^3 - n^2 - n} + 1 \rfloor $ (you can do the +1 inside or outside the floor, the result is the same).

Now, since $K$ is the smallest possible integer satisfying the inequality, we know that $\lfloor \log_2 (k) \rfloor \geq K$. Since $\log$ is a monotonically increasing function, the smallest possible $k$ that provides a solution must be when $\log_2 (k) = K$ (I'll leave the proof of that up to you, but you can first show that the floor of the log must equal $K$, then you can show that the floor is unnecessary when you're looking for the smallest solution).

So $\log_2 (k) = \lfloor \frac{2 n^2}{2 n^3 - n^2 - n} + 1 \rfloor$, and hence $k = 2^{\lfloor \frac{2 n^2}{2 n^3 - n^2 - n} + 1 \rfloor}$.