Comparing n^k and k^nDate: 04/03/2007 at 21:40:46 From: Heather Subject: Is a number larger with a bigger base or a bigger exponent? If n > k, then is k^n bigger, or is n^k? It seems that in most cases a larger exponent creates a larger number, but not always. My husband and I came up with this question over dinner tonight, and tested several numbers to find that: 2^3 is greater than 3^2 (2.5)^2 is less than 2^(2.5) 30.00001^30 is less than 30^(30.00001), etc. It seemed to be that the larger exponent returned the larger number in every case with a few exceptions. We could not, however, figure out the "rule" to know what the exceptions are. For example, we can tell graphically that 2^1000000 is greater than 1000000^2, but why? Is there a general rule about this? Thank you! Date: 04/04/2007 at 12:26:37 From: Doctor Peterson Subject: Re: Is a number larger with a bigger base or a bigger exponent? Hi, Heather. It turns out that it's not easy to determine which will be larger. This question was discussed briefly on the following page: Comparing x^y and y^x http://mathforum.org/library/drmath/view/61584.html Taking the suggestion made there, we can take the inequality a^b > b^a and raise each side to the 1/(ab) power, which will not change the inequality as long as we are only using positive numbers: (a^b)^[1/(ab)] > (b^a)^[1/(ab)] a^[b/(ab)] > b^[a/(ab)] a^(1/a) > b^(1/b) Thus the power with x as the base will be larger when x^(1/x) is larger. Look at the graph of the function y = x^(1/x): You'll see (if you trust me about some of the details) that this function increases until x = e (that is, 2.71828) and then starts decreasing slowly toward 1. If you are wondering about some pair of numbers a and b, you can just locate them on the x-axis and see which has a higher value of y; putting that one as the base will produce the larger power. Now, as long as both numbers are larger than e, putting the smaller one as the base and the larger one as the exponent will always produce the larger power. Therefore, let's call this the normal case. What we're looking for are special cases where a < b but a^b < b^a As mathematicians like to do, I'll give a name to such a pair of numbers; what strikes my fancy at the moment is to call it a "retrograde pair", meaning that the comparison goes backward from the usual. If you use only whole numbers, then if they are both larger than 2, they will behave normally. If a=1, then certainly 1^b < b^1 for any (larger) whole number b, so it is always retrograde in this trivial case. If a=2, then the only "retrograde" case will be when b=3: 2^3 < 3^2. When you extend to non-integers, you have more choices. For example, taking a = 1.5 and looking at the graph, I see that I will get a retrograde pair if b = 3, 4, 5, 6, and maybe 7. Let's try them: a b a^b b^a --- --- ------ ------- 1.5 3 3.375 5.1961 1.5 4 5.0625 8 1.5 5 7.5937 11.1803 1.5 6 11.3906 14.6969 1.5 7 17.0859 18.5202 1.5 8 25.6289 22.6274 And we were right: the pairs up through 7 are retrograde, but 1.5, 8 is normal. Since our function x^(1/x) is not easy to work with, it would probably be hard to come up with a simple way to test a pair of numbers, but at least the graph gives us a general sense of where they are, and a way to find "retrograde pairs". We know that one number has to be less than e, and the other has to be greater than that number and less than something on the other side of e that we can find using the graph. We also know that if the smaller number is less than 1, the pair is always retrograde. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 04/04/2007 at 23:43:25 From: Heather Subject: Thank you (Is a number larger with a bigger base or a bigger exponent?) Thank you for the quick and interesting response. That is quite fascinating. I guess we were close to the answer when we were testing 2.5 and 3, we just didn't know it. Thanks so much for taking the time to enlighten two fellow math lovers :) |
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