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

[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.

 

[daon]

This may be easier: Let 2^k > c

Then 3^n > 2^(n+k) <=>
nLog(3) > (n+k)Log(2) <=>
n> kLog(2)/(Log(3)-Log(2))

Pick N = floor(kLog(2)/(Log(3)-Log(2)))+1

 

[toidayma]

Well, same as Daon's post, but may be shorter:
The problem is as to solve x for: 3^x > c*2^x
Since 2^x > 0 for every x, substitute both side with 2^x, we have: (3/2)^x > c
Then, we can always find an integer n > ln(c)/ln(3/2), or the ceiling of ln(c)/ln(3/2).