Comparing Numbers With Huge Exponents

Date: 05/23/2002 at 16:12:24
From: Hunter Hill
Subject: Huge Exponents

Which is larger? 1999^1999 or 2000^1998

I am stumped, and do not know how to solve this. Please help!

Date: 05/23/2002 at 17:49:54
From: Doctor Ian
Subject: Re: Huge Exponents

Hi Hunter,

Let's look at some smaller versions of the same problem:

  n      n^n     (n+1)^(n-1)
  -     -----    -----------
  2         4              3 
  3        27             16 
  4       256            125 
  5      3125           1296 
  6     46656          16807 

This is hardly a proof, but it suggests that the term with the
higher exponent will be larger.  If you can prove (e.g., by 
induction) that n^n grows faster than (n+1)^(n-1), then it's got 
to hold for the special case of n=1999.

Does this help? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/24/2002 at 15:19:56
From: Hunter Hill
Subject: Huge Exponents

Doctor Ian,

I appreciate the help. However, in order for me to get full credit 
for this extra credit problem, I need to show my work and explain it 
at a college algebra level. If possible, please assist. Thanks for 
your help.

-Hunter Hill

Date: 05/24/2002 at 17:17:37
From: Doctor Ian
Subject: Re: Huge Exponents

Hi Hunter,

We're trying to show that n^n grows faster than (n+1)^(n-1).  One 
way to do that is to show that the ratio 

       n^n
   ----------- 
   (n+1)^(n-1)

increases with increasing n, i.e., that if we choose some k to be 
the value of n, then 

      k^k        (k+1)^(k+1)
   ----------- < -----------
   (k+1)^(k-1)    (k+2)^k
   
       ^             ^
       |             |
    ratio for     ratio for
    some k        next k

Since we know that all the terms are positive, we can 
cross-multiply to get

  k^k * (k+2)^k  <  (k+1)^(k+1) * (k+1)^(k-1)

If we can show that this inequality holds, then we've shown that 
if k^k is larger than (k+1)^(k-1) for any any value of k, it 
holds for every larger value of k.  And we've already shown that 
it holds for k=2. 

Since 1999 is larger than 2, it must be true for k=1999 as well. 
 
Can you take it from here?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/