Final answer:
An integer N is sought for which the inequality 2^n > n^4 holds true for all n larger than N. Through mathematical induction and experimentation, we conjecture and then prove that for a sufficiently large base case such a condition is consistently met.
Step-by-step explanation:
For the inequality 2^n > n^4 to hold true for all integers greater than a certain number N, we must find an integer N where this property starts to apply consistently. We can use mathematical induction to prove that a certain value of N works.
We first want to find a base case, a value of n, where 2^n > n^4 starts to hold true. After trying several values, we might conjecture that the inequality holds for all n ≥ k, where k is some integer. For a specific example, by testing integers we observe that 2^10 = 1024 and 10^4 = 10000, so n = 10 does not satisfy the inequality, but at n = 17, we have 2^17 = 131072 and 17^4 = 83521, so the condition is satisfied and it's a good candidate for our base case.
Let's assume that the inequality holds for some integer n = k; this is our induction hypothesis where 2^k > k^4. Now, we need to prove that 2^(k+1) > (k+1)^4. For mathematical induction, if we prove that the assumption for n = k implies the case for n = k+1, then the inequality will hold true for all n ≥ k.
For k sufficiently large, 2^(k+1) can be approached as multiplying 2^k by 2, while raising a base of k to the fourth power and adding one has smaller relative growth when compared to multiplying by 2. Hence, by the principle of mathematical induction, we prove that there exists an integer N such that for n > N, the inequality 2^n > n^4 is true.