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

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 $43+11= \boxed{054}$ .

Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs $(a,b)$ , and not for the number of possible values of $b$ . Were the problem to ask for the number of possible values of $b$ , the values of $b^6$ under $2005$ would have to be subtracted, which would just be $2$ values: $2^6$ and $3^6$ .

Solution II

Let $a=log_a b$ . Then our equation becomes $a+\frac{1}{a}=5$ . Multiplying through by $a$ and solving the quadratic gives us $a=2$ or $a=3$ . Hence $a^2=b$ or $a^3=b$ .

For the first case $a^2=b$ , $a$ can range from 2 to 44, a total of 43 values. For the second case $a^3=b$ , $a$ can range from 2 to 12, a total of 11 values.

Thus the total number of possible values is $43+11=\boxed{54}$ .

2005 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2005 AIME II Problems/Problem 5

Contents

Problem

Solution

Solution II

See also