For which positive integer n is $21n +127 \leq 3^n$?
Base case: \begin{gather} n=5\\ 21(5) + 127 \leq 3^5 = 243\\ 232 \leq 243 \end{gather} Inductive steps: \begin{gather} P(n+1) = 21(n+1) + 127 \leq 3^{n+1}\\\\ \text{Using the original base case:}\\ 21n+127 \leq 3^n\\ 3(21n+127) \leq 3^n\cdot3\\ 63n+381\leq3^{n+1}\\\\ \text{Since:}\\ 21(n+1) + 127 = 21n+ 148 \leq 63n+381\\ \therefore21(n+1)+127 \leq 3^{n+1} \end{gather}
Is this an acceptable way to proof the induction for this question? I am using the following logic in my argument wherby i split a constant into two and remove a part of it to fulfill the argument: \begin{gather} 3\leq5\\ 4\leq5\\ (3+1)\leq5\\ \therefore 3\leq5\\ \end{gather}
Your inductive proof is fine for showing the inequality is true for $n \ge 5$. To complete the question you need to show that the inequality fails for $1,2,3,4$. You can do this with a bit of computation. The fact that the inequality is so close at $n=5$ makes it seem likely, but you need to at least state it.