Solving Exponential Inequality I would like to know the range of $n$ where this condition is true. Can somebody help me with it. The equation is as follows
$$n^{100}\le2^{n^2}$$
 A: Both functions in your inequality are continuous, so we find the correct intervals by solving for the equality $n^{100}=2^{n^2}$ first. 
Let $x=n^2$. Then
$$x^{50}=2^x$$
$$x=2^{x/50}=e^{x(\ln 2)/50}$$
$$xe^{x(-\ln 2)/50}=1$$
$$x\frac{-\ln 2}{50}\cdot e^{x(-\ln 2)/50}=-\frac{\ln 2}{50}$$
$$x\frac{-\ln 2}{50}=W\left(-\frac{\ln 2}{50}\right)$$
where $W$ is the Lambert W function.
$$x=-\frac{50}{\ln 2}W\left(-\frac{\ln 2}{50}\right)$$
$$n=\pm\sqrt{-\frac{50}{\ln 2}W\left(-\frac{\ln 2}{50}\right)}$$
Since $-\frac 1e<-\frac{\ln 2}{50}<0$, there are two values of the $W$ function here. We get four values for $n$:
$n\approx \pm 1.00705$ or $n\approx \pm 20.9496$
If we assume that $n$ is integral, the solution set is

$|n|\le 1$ or $|n|\ge 21$

If $n$ is real but not necessarily integral, replace the $1$ and $21$ with the approximate values above.
A: Hint
Take logarithms of both sides, the sign of the inequality should stay the same. Update with your thoughts how to proceed and I will be glad to post more hints as long as you show some effort to solve the problem.
A: $$n^{100}\leq2^{n^2}$$
$$n\leq 2^{\frac{n^2}{100}}$$
right-hand side grows faster, so if it's true for a constant $c$, then it's true for every number greater than $c$. $c=21$ and thus the inequality holds true for:
$$n=1 \vee n\geq 21$$
