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 $(n+1)$.
 A: 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. $2^{n+1} = 2\times 2^n = 2^n+2^n$.
A: Hint
So you want to show for $n>4$,
$$4n<2^n \implies 4n+4 < 2^n+2^n$$
A: If $4n&lt2^n$ then $4n+4&lt2^n+4&lt2^n+2^n=2^{n+1}$ (the second inequality hold since $2^n\geq4 $ for $n\geq 2$). It follows that $4(n+1)&lt2^{n+1}$.
A: 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. 
A: 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

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

