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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 I ==
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== Solution 1 ==
 
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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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>.
 
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>.
  
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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>.
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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>. However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)
  
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==Solution II ==
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==Solution 2 ==
 
Let <math>k=\log_a b</math>. Then our equation becomes <math>k+\frac{6}{k}=5</math>. Multiplying through by <math>k</math> and solving the quadratic gives us <math>k=2</math> or <math>k=3</math>. Hence <math>a^2=b</math> or <math>a^3=b</math>.   
 
Let <math>k=\log_a b</math>. Then our equation becomes <math>k+\frac{6}{k}=5</math>. Multiplying through by <math>k</math> and solving the quadratic gives us <math>k=2</math> or <math>k=3</math>. Hence <math>a^2=b</math> or <math>a^3=b</math>.   
  
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Thus the total number of possible values is <math>43+11=\boxed{54}</math>.
 
Thus the total number of possible values is <math>43+11=\boxed{54}</math>.
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==Solution 3(similar to solution 2)==
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Using the change of base formula on the second equation to change to base <math>a</math>, we get <math>\log_a(b) + \frac{6 \log_a(a)}{\log_a(b)}</math>. If we substitute <math>x</math> for <math>\log_a(b)</math>, we get <math>x + \frac{6}{x}</math>. Multiplying by <math>x</math> on both sides and solving, we get <math>x=3,2</math>. Substituting back in, we get <math>\log_a(b) = 3,2</math>. That means <math>a^3 = b</math> or <math>a^2 = b</math>. Since <math>b \leq 2005</math>, we can see that for the cubed case, the maximum <math>a</math> can be without exceeding 2005 is 12(because <math>13^3 = 2197</math>) and for the squared case it can be a maximum of 44. Since <math>a \neq 1</math>, the number of values is <math>(44-1)+(12-1) = \boxed{54}</math>.
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~idk12345678
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:15, 4 April 2024

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 1

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 $3^6$. However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)

Solution 2

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 $43+11=\boxed{54}$.


Solution 3(similar to solution 2)

Using the change of base formula on the second equation to change to base $a$, we get $\log_a(b) + \frac{6 \log_a(a)}{\log_a(b)}$. If we substitute $x$ for $\log_a(b)$, we get $x + \frac{6}{x}$. Multiplying by $x$ on both sides and solving, we get $x=3,2$. Substituting back in, we get $\log_a(b) = 3,2$. That means $a^3 = b$ or $a^2 = b$. Since $b \leq 2005$, we can see that for the cubed case, the maximum $a$ can be without exceeding 2005 is 12(because $13^3 = 2197$) and for the squared case it can be a maximum of 44. Since $a \neq 1$, the number of values is $(44-1)+(12-1) = \boxed{54}$.

~idk12345678

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions

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