How to solve this logarithmic inequality? I've started a data structure course and I need some help with solving these 
logarithmic inequalities. It would also be helpful because later on these
kind of calculation won't pose a problem later on:
for example:
$$log_2(n)^2\le\sqrt{n}$$
$$or$$
$$log_2(n)^{100}\gt n^2$$
Thank you!
 A: Since these relate to a data structure, I'll assume that $n>1$ so the logarithms are positive. Both sides of your inequalities are continuous functions of $n$, so the solutions are unions of intervals between the solutions of the corresponding equalities. Since the logarithms are positive, we can take roots of both sides of the equations. So, changing your inequalities to equalities and taking the square root of the first and $100$'th root of the second, we want to solve
$$\log_2(n)=n^{1/4}$$
$$\log_2(n)=n^{1/50}$$
Both these equalities are in the form
$$\log_2(n)=n^{1/a}$$
where $a=4$ in the first inequality, $a=50$ in the second. Solving that equality,
$$\frac{\ln n}{\ln 2}=e^{(\ln n)/a}$$
$$\ln 2=\ln n \cdot e^{-(\ln n)/a}$$
$$-\frac{\ln 2}a=-\frac{\ln n}a \cdot e^{-(\ln n)/a}$$
$$-\frac{\ln n}a=W\left(-\frac{\ln 2}a \right)$$
where $W(x)$ is the Lambert W function, which is double-valued for $-\frac 1e<x<0$. We end up with
$$n=e^{-a\cdot W(-(\ln 2)/a)}$$
For your first inequality, that gives us the two values of
$$n=e^{-4\cdot W(-(\ln 2)/4)}$$
The first, lower value of that is approximately $2.361375959$. The upper value is the integer $65536$, which is surprising but easily checked. We can see that your inequality is true for the intervals outside those two points, so the solution to your first inequality, given $n>1$, is

$$1<n\le 2.361375959 \quad \text{or} \quad 65536\le n$$

For the second inequality, the solution is between the two points given by
$$n=e^{-50\cdot W(-(\ln 2)/50)}$$
which my calculator gives as

$$2.019724503<n<1.310793842\times 10^{132}$$

again, assuming $n>1$.
