Difference between revisions of "2005 AIME II Problems/Problem 5"
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== Problem == | == Problem == | ||
− | Determine the number of ordered | + | 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>. | ||
− | + | == See also == | |
− | + | *[[Logarithm]] | |
− | + | {{AIME box|year=2005|n=II|num-b=4|num-a=6}} | |
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− | == See | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Revision as of 16:00, 13 March 2007
Problem
Determine the number of ordered pairs of integers such that and
Solution
Note that the equation can be rewritten as: or Multiplying through by and factoring yields . Therefore, or . Therefore, either or . For the case , note that and . Thus, all values of from 2 to 44 will work. For the case , note that while . Therefore, for this case, all values of from 2 to 12 work. There are possibilities for the square case and possibilities for the cube case. Thus, the answer is .
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 |