Difference between revisions of "2019 AMC 12A Problems/Problem 23"
Sevenoptimus (talk | contribs) m (Slightly fixed formatting) |
(→Solution 1: Hopefully that makes it clearer. The solution was great, but some people might have a hard time following the logarithm manipulating.) |
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We now compute <math>k</math> from <math>a_3</math>. It is given that <math>a_3 = 3\,\heartsuit\,2 = 3^{\frac1{\log_7(2)}}</math>, so <math>k = \log_3(a_3) = \log_3\left(3^{\frac1{\log_7(2)}}\right) = \frac1{\log_7(2)} = \log_2(7)</math>. | We now compute <math>k</math> from <math>a_3</math>. It is given that <math>a_3 = 3\,\heartsuit\,2 = 3^{\frac1{\log_7(2)}}</math>, so <math>k = \log_3(a_3) = \log_3\left(3^{\frac1{\log_7(2)}}\right) = \frac1{\log_7(2)} = \log_2(7)</math>. | ||
− | Now, we must have <math>\log_{2019}(a_{2019}) = k = \log_2(7)</math>. | + | Now, we must have <math>\log_{2019}(a_{2019}) = k = \log_2(7)</math>. 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. |
+ | |||
+ | <math>\frac{\log{a_{2019}}}{\log{2019}} = \frac{\log{7}}{\log{2}}\implies | ||
+ | \frac{\log{a_{2019}}}{\log{7}} = \frac{\log{2019}}{\log{2}}\implies | ||
+ | \log_7(a_{2019}) =\log_2(2019)</math> | ||
+ | |||
+ | We conclude that <math>\log_7(a_{2019}) \approx \boxed{11}</math>, or choice <math>\boxed{\text{D}}</math>. | ||
==Solution 2== | ==Solution 2== |
Revision as of 11:22, 12 February 2019
Contents
Problem
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 ?
Solution 1
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 .
Solution 2
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.
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 log base 2019 of both sides gives us . Then, by changing to base 7 and after cancellation, we arrive at . Because and , our answer is .
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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