Prove via mathematical induction that $4n < 2^n$ for all $ n≥5$. 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 $(n+1)$.

Given $5\leq{n}$
Assume $4n<2^n$
Prove $4(n+1)<2^{n+1}$:


*

*$4(n+1)=4n+4$

*$4n+4<2^n+4$ assumption used here

*$2^n+4<2^n+32$

*$2^n+32=2^n+2^5$

*$2^n+2^5\leq2^n+2^n$ given fact used here

*$2^n+2^n=2^{n+1}$

