Difference between revisions of "2019 AMC 12A Problems/Problem 23"
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Revision as of 19:36, 20 August 2019
Define binary operations and by for all real numbers and for which these expressions are defined. The sequence is defined recursively by and for all integers . To the nearest integer, what is ?
By definition, the recursion becomes . By the change of base formula, this reduces to . Thus, we have . Thus, for each positive integer , the value of must be some constant value .
We now compute from . It is given that , so .
Now, we must have . At this point, we simply switch some bases around. For those who are unfamiliar with logarithms, we can turn the logarithms into fractions which are less intimidating to work with.
We conclude that , or choice .
Using the recursive definition, or where and . Using logarithm rules, we can remove the exponent of the 3 so that . Therefore, , which is .
We claim that for all . We can prove this through induction.
Clearly, the base case where holds.
This can be simplified as .
Applying the diamond operation, we can simplify where . By using logarithm rules to remove the exponent of and after cancelling, .
Therefore, for all , completing the induction.
We have . Taking of both sides gives us . Then, by changing to base and after cancellation, we arrive at . Because and , our answer is .
We are given that Since we are asked to find , we directly apply Using the property that Now using the property that Once again applying the first property yields Rearranging the expression,
Now expressing in a similar expression as ,
Because of the fact that , we can cancel out the terms to get
Using the Chain Rule for Logarithm, , yields
Finally, substituting in , we have
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