Proove that $2n-3\leq 2^{n-2}$ for all $n\geq 5$

The following is the proof asked in https://goo.gl/h5uyhK


Prove that $$2n-3\leq 2^{n-2}\text{ for all } n\geq 5$$

Solution (by Mathematical Induction)

1. Base case: If $n=5$, the left side is 7 and the right side is $2^{5-2}=2^3=8$. So, the inequality holds when $n=5$.

2. Inductive hypothesis: Suppose the inequality holds for all values of $n$ up to some $k\geq5$, $P(k):2k-3\leq 2^{k-2}$.

3. Inductive step: Let $n=k+1$. Now we have to prove that the relation also holds for $k+1$ by using the induction hypothesis. This means that we have to prove

$$P(k+1):2(k+1)-3\leq 2^{(k+1)-2}$$
Then our left side is
$$\begin{align} P(k+1):2(k+1)-3 &= 2k-3+2 \\ &\leq 2^{k-2} +2 \text{, by our inductive hypothesis}\\ &\leq 2^{k-2} + 2^{k-2} \text{ since } 2\leq 2^{k-2}, \forall k\geq 5 \\ &\leq 2\cdot 2^{k-2} \\ &\leq 2^{k-2+1} \\ &=2^{k-1} \end{align}$$
Thus
$$2(k+1)-3\leq 2^{k-1} \Longrightarrow 2k-3 \leq 2^{k-2}$$

which is our right side. So, the theorem holds for $n=k+1$.

By the principle of mathematical induction, the theorem holds for all $n\geq 5$.