I did the following base case $n = 5$ $\begin{align*} 4(5) &\lt 2 ^5\\ 20 &\lt 32 \end{align*}$ So true.
$\begin{align*} 4n &\lt 2^n\\ n &\lt 2^{n-2}\\ \log_2(n)+2 &\lt n \end{align*}$ But I don't think this is right. Where do I add in the $(n+1)$.
I did the following base case $n = 5$ $\begin{align*} 4(5) &\lt 2 ^5\\ 20 &\lt 32 \end{align*}$ So true.
$\begin{align*} 4n &\lt 2^n\\ n &\lt 2^{n-2}\\ \log_2(n)+2 &\lt n \end{align*}$ But I don't think this is right. Where do I add in the $(n+1)$.
The induction base is correct.
For the inductive step, we assume that the result holds for $n$, with $n\geq 5$; that is, are assuming that $4n\lt 2^n,\qquad n\geq 5.$ We want to prove that, under this assumption, $4(n+1)\lt 2^{n+1}$.
Hint the first. $4(n+1) = 4n+4 \lt 2^n+4$, with the last step using the induction hypothesis.
Hint the second. $2^{n+1} = 2\times 2^n = 2^n+2^n$.
Hint
So you want to show for $n>4$, $4n<2^n \implies 4n+4 < 2^n+2^n$
If 4n<2^n then 4n+4<2^n+4<2^n+2^n=2^{n+1} (the second inequality hold since $2^n\geq4 $ for $n\geq 2$). It follows that 4(n+1)<2^{n+1}.
Here is what I think might work:
Step 1: Base case. You've done this.
Step 2. Assume $4k<2^k$ for arbitrary k contained in the naturals. You can do this because your base case proves the assumption true for at least one k. That is, you've proved this true for k=5.
Step 3. Prove this holds for $n=k+1$.
$4k<2^{k}$ $8k<2^k*2$ $8k<2^{k+1}$
It is obvious that: $4k+1<8k<2^{K+1}$ for $K>4$
And therefore, by induction, that $4n<2^n$ for all n belonging to the naturals where n>4.
That is my attempt. Hope it works.
Given $5\leq{n}$
Assume $4n<2^n$
Prove $4(n+1)<2^{n+1}$:
assumption used here
given fact used here