Induction with Inequalities

Suppose that we are trying to prove that $2^n > 10n + 9 $ $\forall n \geq 9$

Basis: $2^9 > 99$ is true.

Inductive Hypothesis: Assume that it is true that $n=k$ $\forall n \geq 9$

Then: $2^k > 10k + 9 $

Inductive step: Prove for $n=k+1$: $$2^{(k+1)} > 10(k+1) + 9 $$

This is where I get confused. Is it true that: $$2^{(k+1)} = 2^k*2 = 2^k +2^k > 2^k+2 > 10(k+1) +9$$

Is this how I string this inductive proof together? I feel like there is a gap in my understanding of whether or not this statement is true.

Answer

You have a flaw in your chain. All you can relate is using the inductive hypothesis that $$2^k>10k+9$$, so you have
$$2^{k+1}=2^k*2=2^k+2^k$$

now, you can add the inductive hypothesis to itself to get
$$2^k+2^k>10k+9+10k+9$$
Keep in mind, what we want it to be bigger than is $10(k+1)+9$, so all that is left is to show what we have is bigger than that. One way is to change the $9+9$ to $10+8$ in what we have, so we can then factor to get that magic $10(k+1)$ term:
$$2^k+2^k>10k+9+10k+9=10k+10+10k+8=10(k+1)+10k+8$$
So, we are left with needing that $10k+8\ge9$, which it is, since $k\ge 9$.