# 10^x > x but more

Now, you might think,

Isn’t that kinda obvious?

You

No. No it’s not.

So we have to prove it.

Okay, here we go!

1. For x = 0, 10^0 = 1 > 0, so TRUE
2. For x ≥ 1, let’s do an induction-style proof. For x = 1, 10^1 = 10 > 1, so true. For x + 1, let the difference of the terms be f(x + 1) and g(x + 1), respectively. So f(x + 1) would be the difference between 10^x and 10^(x + 1), and g(x + 1) would be… 1. And since 10 * 10^x – 10^x is 9 * 10^x, and 9 * 10^x is always larger than 1, the statement will always be true. So TRUE
3. For 1 > x > 0, define a different number k, where it is less than 0. k^(any number you want whatever I don’t care as long as it’s positive) will always be less than 1. For positive numbers greater than 1, keep powering it and you won’t go under 1. So TRUE.
4. For x < 0, no matter what you do with the number 10 involving exponents, it will always be positive, so CHECK

So therefore it is TRUE.

Now, you were right, it was obvious, so let’s go one step higher.

Let’s prove that for any positive n, 2^n > n^2 (except n = 2, 3 and 4)

Proof by induction.

Let n = 1. 2^1 = 2, 1^2 = 1. So TRUE.

We have to prove that n + 1 also is true if n is true.

2^3 = 8, 3^2 = 9…

n = 4

2^4 = 16, 4^2 = 16… why

n = 5

2^5 = 32, 5^2 = 25

NOW let’s prove it for n + 1 as long as the term is true for n.

2^(n + 1) = 2 * 2^n, and (n + 1)^2 = n^2 + 2n + 1

Now from here, I did not create this answer. A Stank Overflow user named triple_sec created this part (or Googled it idk) and credit goes to them.

So remember how 2^n > n^2 for n = 5? So 2 * 2^n > 2 * n^2, and with basic knowledge (too lazy to prove lol) you’ll know that n^2 > 2n + 1 for n > 2, so:

2 * n^2 = n^2 + n^2 > n^2 + 2n + 1 and therefore n^2 > n + 1, which we just proved.

So it is proved true.

Q.E.D. * 2