Problem:
Determine the number of ordered pairs (a,b) of integers such that logaβb+ 6logbβa=5,2β€aβ€2005, and 2β€bβ€2005.
Solution:
Let x=logaβb. Because logbβa=1/logaβb, the given equation can be written as x+(6/x)=5, and because xξ =0, this is equivalent to x2β5x+6=0, whose solutions are 2 and 3. If 2=x=logaβb, then a2=b. Now 442=1936 and 452=2025, so there are 44β1=43 ordered pairs (a,b) such that a2=b and a and b satisfy the given conditions. If 3=x=logaβb, then a3=b. Because 123=1728 and 133=2197, there are 12β1=11 ordered pairs (a,b) such that a3=b and a and b satisfy the given conditions. Thus there are 43+11=54β of the requested ordered pairs.
The problems on this page are the property of the MAA's American Mathematics Competitions