Solving Exponential Inequality

Question

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

$$n^{100}\le2^{n^2}$$

Answer 1

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.

Answer 2

$$n^{100}\leq2^{n^2}$$ $$n\leq 2^{\frac{n^2}{100}}$$ 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: $$n=1 \vee n\geq 21$$

Answer 3

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.