2005 AIME II Problems/Problem 5

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

Note that the equation can be rewritten as: $\frac{\log b}{\log a}+6\frac{\log a}{\log b}=5$ or $\frac{(\log b)^2+6(\log a)^2}{\log a * \log b}=5$ Multiplying through by $\displaystyle \log a * \log b$ and factoring yields $\displaystyle (\log b - 3\log a)(\log b - 2\log a)=0$ . Therefore, $\displaystyle \log b=3\log a$ or $\displaystyle \log b=2\log a$ . Therefore, 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 $43+11=054$ .

2005 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2005 AIME II Problems/Problem 5

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