Question

prove n>equal to 5, 4n<2^n

Answer 1

i would think proving by trial and error would be the easiest

Answer 2

proof by induction maybe?

Answer 3

Are you given anything about n, like n is in the set of integers or something?

Answer 4

I'll give it a shot. No promises on its validity, but hopefully if it isn't right it will at least give you an idea how to go about it.

The initial inequality is 4n < 2^n, which we want to prove is true for (assuming integer but I don't think it would change with Reals in this case) values of n >= 5.

Starting with the base case, n=5, we see that the inequality hold true: 20 < 32. Now we will propose an induction hypothesis by simply replacing n with (n+1) in the initial inequality. We should be looking at our I.H., the thing we want to prove, as 4(n+1) < 2^(n+1). Note: 2(2^n) = 2^(n+1). That suggests that we can multiply both sides of the initial inequality by 2 and then "prove" a much easier inequality which will consequently prove our I.H.

Multiplying both sides of 4n < 2^n by 2 leaves us with 8n < 2^(n+1). Now all we have to do is prove 8n > 4(n+1) and we should be almost done. 8n > 4n + 4 simplifies to 4n > 4. Since we were given the assumption that n >= 5, it naturally follows that 4n > 4 for all values of n > 5. I suppose, if your teacher is being especially rigorous, you may be required to prove 4n > 4 as well, but lets hope it doesn't come to that. Hint: 4(n+1) > 4 = 4n +4 > 4 = 4n > 0.

Unless I missed something, I think you are pretty much done. Your I.H. was proven when you demonstrated that 4(n+1) < 8n < 2^(n+1). Tombstone it, or Q.E.D. and on to the next one.