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

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== Problem ==
 
== Problem ==
 
 
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>
  
 
== Solution ==
 
== Solution ==
Note that the equation can be rewritten as: <math> \frac{\log b}{\log a}+6\frac{\log a}{\log b}=5 </math> or <math> \frac{(\log b)^2+6(\log a)^2}{\log a * \log b}=5 </math> Multiplying through by <math>\displaystyle \log a * \log b </math> and factoring yields <math> \displaystyle (\log b - 3\log a)(\log b - 2\log a)=0 </math>. Therefore, <math> \displaystyle \log b=3\log a </math> or <math>\displaystyle \log b=2\log a </math>. Therefore, 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 2 to 44 will work. For the case <math> b=a^3 </math>, note that <math> 12^3=1728 </math> while <math> 13^3=2197 </math>. Therefore, for this case, all values of <math>a</math> from 2 to 12 work. There are <math> 44-2+1=43 </math> possibilities for the square case and <math> 12-2+1=11 </math> possibilities for the cube case. Thus, the answer is <math> 43+11=054 </math>.
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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>.  
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*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^3 </math>, note that <math> 12^3=1728 </math> while <math> 13^3=2197 </math>. Therefore, for this case, all values of <math>a</math> from <math>2</math> to <math>12</math> work.  
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There are <math> 44-2+1=43 </math> possibilities for the square case and <math> 12-2+1=11 </math> possibilities for the cube case. Thus, the answer is <math> 43+11= \boxed{054}</math>.
  
 
== See also ==
 
== See also ==
*[[Logarithm]]
 
 
{{AIME box|year=2005|n=II|num-b=4|num-a=6}}
 
{{AIME box|year=2005|n=II|num-b=4|num-a=6}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 17:37, 25 July 2008

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AIME Problems and Solutions
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