For which $n,~~ 2^n>n^3~$ is true?

Question

For which $n,~~ 2^n>n^3~$ is true ?

It’s my school homework) I realized that I need to use induction for this task.

The base would be $n=10~~ (n=1~$ also works$)$.

I can’t really figure out the rest(.

Answer 1

Hint If you know that it works for $n=10$, and that you need to use induction, here is the hint for the inductive step $$(n+1)^3=n^3+3n^2+3n+1 < n^3+3n^2+3n^2+n^2< n^3+10n^2 \leq n^3+n^3$$ with the last step following from $n \ge 10$.

Answer 2

$$\frac{l_n}{l_{n-1}}=2$$ while

$$\frac{r_n}{r_{n-1}}=\left(\frac n{n-1}\right)^3$$ decreases towards $1$. The two ratios are equal when

$$n=\frac{\sqrt[3]2}{\sqrt[3]2-1}\approx 4.8$$

Hence as of $n=5$, the left sequence grows faster than the right one.

You find the exact crossing point by comparing

$$1,2,4,8,16,32,64,128,256,512,1024,\cdots$$

vs.

$$0,1,8,27,125,316,343,512,729,1000,\cdots$$

So for $n\ge10$, the left sequence is always larger.

Answer 3

Another approach:

$2^n>n^3$

Taking logarithm of both sides we get:

$n \log_{10}2>3 \log_{10}n$

$n. \frac {301}{1000}> 3\log_{10}n$

$\frac{n}{\log_{10} n}> \frac {3000}{301}≈9.9$

$\frac{n}{\log_{10} n}≥ 10=\frac{10}{\log_{10} 10}$

⇒$ n≥10$