Difference between revisions of "2019 AMC 12A Problems/Problem 23"
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<math>\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12</math> | <math>\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12</math> | ||
− | ==Solution== | + | ==Solution 1== |
+ | |||
+ | By definition, the recursion becomes <math>a_n = \left(n^{\frac1{\log_7(n-1)}}\right)^{\log_7(a_{n-1})}=n^{\frac{\log_7(a_{n-1})}{\log_7(n-1)}}</math>. By the change of base formula, this reduces to <math>a_n = n^{\log_{n-1}(a_{n-1})}</math>. Thus, we have <math>\log_n(a_n) = \log_{n-1}(a_{n-1})</math>. Thus, for each positive integer <math>m \geq 3</math>, the value of <math>\log_m(a_m)</math> must be some constant value <math>k</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>. Changing bases to <math>7</math>, this becomes <math>\frac{\log_7(a_{2019})}{\log_7(2019)} = \log_2(7)</math>, so <math>\log_7(a_{2019}) = \log_2(7) \cdot \log_7(2019) = \log_2(2019)</math>, where the last equality comes from the logarithmic chain rule. We conclude that <math>\log_7(a_{2019}) = \log_2(2019) \approx \boxed{11}</math>, or choice <math>\boxed{\text{D}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
Using the recursive definition, <math>a_4 = (4 \, \heartsuit \, 3) \, \diamondsuit\, (3 \, \heartsuit\, 2)</math> or <math>a_4 = (4^{m})^{n}</math> where <math>m = \frac{1}{\log_{7}(3)}</math> and <math>n = \log_{7}(3^{\frac{1}{\log_{7}(2)}})</math>. Using logarithm rules, we can remove the exponent of the 3 so that <math>n = \frac{\log_{7}(3)}{\log_{7}(2)}</math>. Therefore, <math>a_4 = 4^{\frac{1}{\log_{7}(2)}}</math>, which is <math>4 \, \heartsuit \, 2</math>. | Using the recursive definition, <math>a_4 = (4 \, \heartsuit \, 3) \, \diamondsuit\, (3 \, \heartsuit\, 2)</math> or <math>a_4 = (4^{m})^{n}</math> where <math>m = \frac{1}{\log_{7}(3)}</math> and <math>n = \log_{7}(3^{\frac{1}{\log_{7}(2)}})</math>. Using logarithm rules, we can remove the exponent of the 3 so that <math>n = \frac{\log_{7}(3)}{\log_{7}(2)}</math>. Therefore, <math>a_4 = 4^{\frac{1}{\log_{7}(2)}}</math>, which is <math>4 \, \heartsuit \, 2</math>. |
Revision as of 20:02, 9 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 . Changing bases to , this becomes , so , where the last equality comes from the logarithmic chain rule. 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.
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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