# Thread: Proving 3^n has higher growth rate than c*2^n?

1. ## Proving 3^n has higher growth rate than c*2^n?

Does anyone know how to formally prove that 3^n has a higher growth rate than c*2^n? In other words no matter how large c is, you can always find some natural number n that makes 3^n larger? It's quite obvious this is true but I don't know the steps to proving it. I used L'Hopital's rule in comparing these two as n -> ∞, but don't know how to isolate the n so that one side has n and the other has c.
Thanks.

2. ## suggestion

Originally Posted by Selena
Does anyone know how to formally prove that 3^n has a higher growth rate than c*2^n? In other words no matter how large c is, you can always find some natural number n that makes 3^n larger? It's quite obvious this is true but I don't know the steps to proving it. I used L'Hopital's rule in comparing these two as n -> ∞, but don't know how to isolate the n so that one side has n and the other has c.
Thanks.
if you need to prove 3^n>c*2^n for any natural number you could always go by the easiest(infact its most effective as well in many cases) way that is proof by induction.the steps you need are:
1)first prove it for 1 assuming it to be true for n.
2)then prove it for 2 .
3)Now prove it for n+1.
4)while proving for n+1, substitute values you got for n which was assumed to be true earlier.
5)if this results into the relation being true for n+1 then you are done.
speaking frankly i never understood how this is a proof. perhaps it shows that since for every n and n+1 the relation is true therefore it is true for all natural numbers. natural numbers progress uniformly. so perhaps like that!