Why is $2^a > a^3$?

Question

I found this rather interesting and maybe, a bit too obvious for some people property about 2 raised to some power.
$2^a > a^3$, if $a=0,a=1 \text{ or } a\ge 10$ .($a \in N$)
I seem to get a bit of intuition as to why this happens(although the intuition is in "micro units of intuition"). Somebody please provide me some intuition regarding it the mathematical/logic way.

Answer 1

Prove it's true for $a = 0, a=1$. Then use induction on $a$ to prove it holds for $a\geq 10$, in which you'll first need to show it holds for $a = 10$. This will help you develop intuition about the conjecture, as well as provide a mathematical proof as to it's truth.

It is easy to see it holds for $a = 0$, since $2^0 = 1 > 0^3 = 0$, and $2^1 = 2\gt 1^3 = 1$.

Base case: $a = 10$

$$2^{10} = 1024 \gt 10^3 = 1000\large \checkmark$$

Inductive Hypothesis (IH): $a = k$. Assume that it is true that $$2^k \gt k^3\tag{IH}$$

Inductive Step: Use the Inductive hypothesis to prove that the conjecture then must hold for $a = k+1$: $$2^{a + 1} = 2 \cdot 2^a \gt (a+1)^3$$

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Added: For an inuitive understanding, let's use Wolfram alpha to plot the functions $y = 2^a$ vs. $y = a^3$ (note that both functions are increasing function for $a \geq 0$):

Notice the points of intersection, and the point, just below $a = 10$, where $y = 2^a$ overtakes $y = a^3$ "forever"!

Answer 2

Generally, powers always eventually get larger than polynomials. When $a$ is large, increasing $a$ doubles $2^a$, but doesn't increase $a^3$ nearly that much. $a^3$ roughly increases by $3a^2$, which is less than $a^3$. As others have suggested, you can turn this into an inductive proof: $2^{10}=1024\gt1000=10^3$. Now assume $2^k \gt k^3$ and $k \ge 10$. Then $2^{k+1}=2\cdot2^k \gt 2\cdot k^3 \gt (k+1)^3$ where I leave you the details of the last comparison.