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

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+ | 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> \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>. 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 a 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 a 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 == | == See Also == |

## Revision as of 23:13, 12 March 2007

## Problem

Determine the number of ordered pairs of integers such that and

## Solution

*This problem needs a solution. If you have a solution for it, please help us out by adding it.*

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 a from 2 to 44 will work. For the case , note that while . Therefore, for this case, all values of a from 2 to 12 work. There are possibilities for the square case and possibilities for the cube case. Thus, the answer is .