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/
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