Prove that $2n-3\leq 2^{n-2}$ , for all $n \geq 5$ by mathematical induction

Prove that $2n-3\leq 2^{n-2}$ , for all $n \geq 5$ by mathematical induction

I have to prove by mathematical induction that: $2n-3\leq 2^{n-2}$ , for all $n \geq 5$

Thank you for the Review.

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For $n=5$ you have
$$ 8=2^{5-2} \geq 2\cdot5-3 = 7$$
So the statement is correct for $n=5$

Now assume $2^{n-2} \geq 2n-3$ is valid for some fixed $n$ then for $n+1$ you will get:

$$2^{n-1} = 2^{n-2} + 2^{n-2} \geq 2^{n-2} +2 \stackrel{assumption}{\geq}2n-3+2 = 2(n+1)-3$$

Thus
$$2^{n-2} \geq 2n-3 \Longrightarrow 2^{n-1} \geq 2(n+1)-3$$
which means the statement is correct for every $n\geq5$


(Answer to previous question)

For $n=6$ you have
$$2\cdot 6 = 12 < 2^{6-2} = 2^4 = 16$$

Thus your statement is wrong


However, if you assume $2n\leq2^{n-2}$ for all $n\geq 6$ now you have:
$$2(n+1) = 2 +2n \stackrel{assumption}{\leq} 2 +2^{n-2} \leq 2^{n-2}+2^{n-2} = 2^{n-1} $$
Thus
$$2n\leq2^{n-2}\Longrightarrow 2(n+1)\leq2^{n-1}$$
So this statement is correct for all $n\geq6$.

Another way: this is equivalent to showing that $n<2^{n-3}$ for $n \geq 6$. The base and assumption state are the same. The inductive step is
$$
n+1 < 2n<2^{n+1-3}=2 \cdot 2^{n-3}
$$
Divide through $2$ and you are back to your assumption step, hence the statement is proven.