leeper@ahutb.UUCP (leeper) (03/13/85)
This is a problem I made up when I was in high school. Define an "exponant dual" of the x to be some OTHER number y so that x^y = y^x For example an exponant dual of 2 is 4 since 2 is not 4 and 2^4 = 4^2 = 16 For what real numbers greater than one do there exist no exponant duals? For what real numbers greater than one does there exist one exponant dual? For what real numbers greater than one do there exist more than one exponant dual? Mark Leeper ...ihnp4!ahutb!leeper
matt@oddjob.UUCP (Matt Crawford) (03/13/85)
Nun! N chmmyr V pna fbyir! Vs k^l = l^k, gura ya(l)/l = ya(k)/k. Qenj gur tencu bs gur shapgvba s(k)=ya(k). Gur yvar guebhtu gur bevtva naq nal cbvag ba gur tencu unf fybcr ya(k)/k. Vs guvf yvar uvgf gjb cbvagf (k,s(k)) naq (l,s(l)) ba gur tencu gura gurl fngvfsl ya(l)/l = ya(k)/k. Nal yvar guebhtu gur bevtva juvpu vagrefrpgf gur tencu va gur svefg dhnqenag ng nyy jvyy vagrefrpg vg rknpgyl gjvpr (orpnhfr bs gur pbapnivgl) hayrff vg vf gur gnatrag yvar guebhtu (r,1). Guhf nyy k>1 unir bar "rkcbarag qhny" rkprcg sbe r. _____________________________________________________ Zngg Havirefvgl penjsbeq@nay-zpf.necn Penjsbeq bs Puvpntb vuac4!bqqwbo!zngg