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Difference between revisions of "2005 AIME II Problems/Problem 5"

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Determine the number of [[ordered pair]]s <math> (a,b) </math> of [[integer]]s such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math>
 
Determine the number of [[ordered pair]]s <math> (a,b) </math> of [[integer]]s such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math>
  
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== Solution ==
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== Solution I ==
 
The equation can be rewritten as <math> \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a  \log b}=5 </math> Multiplying through by <math>\log a \log b </math> and factoring yields <math>(\log b - 3\log a)(\log b - 2\log a)=0 </math>. Therefore, <math>\log b=3\log a </math> or <math>\log b=2\log a </math>, so either <math> b=a^3 </math> or <math> b=a^2 </math>.  
 
The equation can be rewritten as <math> \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a  \log b}=5 </math> Multiplying through by <math>\log a \log b </math> and factoring yields <math>(\log b - 3\log a)(\log b - 2\log a)=0 </math>. Therefore, <math>\log b=3\log a </math> or <math>\log b=2\log a </math>, so either <math> b=a^3 </math> or <math> b=a^2 </math>.  
 
*For the case <math> b=a^2 </math>, note that <math> 44^2=1936 </math> and <math> 45^2=2025 </math>. Thus, all values of <math>a</math> from <math>2</math> to <math>44</math> will work.  
 
*For the case <math> b=a^2 </math>, note that <math> 44^2=1936 </math> and <math> 45^2=2025 </math>. Thus, all values of <math>a</math> from <math>2</math> to <math>44</math> will work.  
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Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs <math>(a,b)</math>, and not for the number of possible values of <math>b</math>.  Were the problem to ask for the number of possible values of <math>b</math>, the values of <math>b^6</math> under <math>2005</math> would have to be subtracted, which would just be <math>2</math> values:  <math>2^6</math> and <math>3^6</math>.
 
Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs <math>(a,b)</math>, and not for the number of possible values of <math>b</math>.  Were the problem to ask for the number of possible values of <math>b</math>, the values of <math>b^6</math> under <math>2005</math> would have to be subtracted, which would just be <math>2</math> values:  <math>2^6</math> and <math>3^6</math>.
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==Solution II ==
 
==Solution II ==

Revision as of 00:27, 3 March 2012

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 I

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 $k=\log_a b$. Then our equation becomes $k+\frac{6}{k}=5$. Multiplying through by $k$ and solving the quadratic gives us $k=2$ or $k=3$. 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 $43+11=\boxed{54}$.

See also

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
All AIME Problems and Solutions
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