Prove by Induction that $2^n \geq n+12$ for all $n\geq4$

Question

Prove by induction that $$2^n \geq n+12$$ for all $n\geq4.$

Not sure how to proceed with this question. If someone could answer it in a step by step induction process (base step, induction hypothesis, induction step), I'd very much appreciate it.

Answer 1

Well you've already got the good idea of using induction. So your start is to pick the lowest $n$ you can ($4$) in this case, and seeing if it works for that.

Assuming it does, why don't we try extending it, see if we can somehow prove it works for $n+1$?

Doing that will complete the proof.

Answer 2

For $n=4$ it's obvious.

Let $2^n\geq n+12$ for any $n\geq4$.

Thus, $$2^{n+1}=2\cdot2^n\geq2(n+12)=n+1+12+n+11>(n+1)+12$$ and by induction we are done!

Answer 3

Define the functions $f(n) = 2^n$ and $g(n) = n + 12$. If we look at the difference between consecutive $n$ for the functions, i.e.:

$$f(n+1) - f(n) = 2^{n+1} - 2^n = 2^n$$ $$g(n+1)-g(n) = (n+1) + 12 - (n+12) = 1$$

We can clearly see that $f(n)$ grows with $2^n$ between each term as $n$ increments, and $g(n)$ grows by $1$ between each term as $n$ increments.

Solving for $n$ in $2^n = n + 12$ we find that they are equal at $n = 4$. However $f(n)$ will grow exponentially faster than $g(x)$; so after $n=4$, $f(x)$ will always be larger than $g(x)$.