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2005 AIME II Problems/Problem 5

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Problem

Determine the number of ordered pairs (a,b) of integers such that \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, and 2 \leq b \leq 2005.

Solution

The equation can be rewritten as \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a  \log b}=5 Multiplying through by \log a \log b and factoring yields (\log b - 3\log a)(\log b - 2\log a)=0. Therefore, \log b=3\log a or \log b=2\log a, so either b=a^3 or b=a^2.

  • For the case b=a^2, note that 44^2=1936 and 45^2=2025. Thus, all values of a from 2 to 44 will work.
  • For the case b=a^3, note that 12^3=1728 while 13^3=2197. Therefore, for this case, all values of a from 2 to 12 work.

There are 44-2+1=43 possibilities for the square case and 12-2+1=11 possibilities for the cube case. Thus, the answer is 43+11= \boxed{054}.

See also

2005 AIME II (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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