Problem:
Determine the number of ordered pairs (a,b) of integers such that loga​b+ 6logb​a=5,2≤a≤2005, and 2≤b≤2005.
Solution:
Let x=loga​b. Because logb​a=1/loga​b, the given equation can be written as x+(6/x)=5, and because xî€ =0, this is equivalent to x2−5x+6=0, whose solutions are 2 and 3. If 2=x=loga​b, then a2=b. Now 442=1936 and 452=2025, so there are 44−1=43 ordered pairs (a,b) such that a2=b and a and b satisfy the given conditions. If 3=x=loga​b, then a3=b. Because 123=1728 and 133=2197, there are 12−1=11 ordered pairs (a,b) such that a3=b and a and b satisfy the given conditions. Thus there are 43+11=54​ of the requested ordered pairs.
The problems on this page are the property of the MAA's American Mathematics Competitions