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==
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==Solution 1==
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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>. 
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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>.
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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>.
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==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

Problem

Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a  \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\]for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsuit\, 2$ and \[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\]for all integers $n \geq 4$. To the nearest integer, what is $\log_{7}(a_{2019})$?

$\textbf{(A) } 8 \qquad  \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$

Solution 1

By definition, the recursion becomes $a_n = \left(n^{\frac1{\log_7(n-1)}}\right)^{\log_7(a_{n-1})}=n^{\frac{\log_7(a_{n-1})}{\log_7(n-1)}}$. By the change of base formula, this reduces to $a_n = n^{\log_{n-1}(a_{n-1})}$. Thus, we have $\log_n(a_n) = \log_{n-1}(a_{n-1})$. Thus, for each positive integer $m \geq 3$, the value of $\log_m(a_m)$ must be some constant value $k$.

We now compute $k$ from $a_3$. It is given that $a_3 = 3\,\heartsuit\,2 = 3^{\frac1{\log_7(2)}}$, so $k = \log_3(a_3) = \log_3\left(3^{\frac1{\log_7(2)}}\right) = \frac1{\log_7(2)} = \log_2(7)$.

Now, we must have $\log_{2019}(a_{2019}) = k = \log_2(7)$. Changing bases to $7$, this becomes $\frac{\log_7(a_{2019})}{\log_7(2019)} = \log_2(7)$, so $\log_7(a_{2019}) = \log_2(7) \cdot \log_7(2019) = \log_2(2019)$, where the last equality comes from the logarithmic chain rule. We conclude that $\log_7(a_{2019}) = \log_2(2019) \approx \boxed{11}$, or choice $\boxed{\text{D}}$.

Solution 2

Using the recursive definition, $a_4 = (4  \, \heartsuit \, 3) \, \diamondsuit\, (3 \, \heartsuit\, 2)$ or $a_4 = (4^{m})^{n}$ where $m = \frac{1}{\log_{7}(3)}$ and $n = \log_{7}(3^{\frac{1}{\log_{7}(2)}})$. Using logarithm rules, we can remove the exponent of the 3 so that $n = \frac{\log_{7}(3)}{\log_{7}(2)}$. Therefore, $a_4 = 4^{\frac{1}{\log_{7}(2)}}$, which is $4  \, \heartsuit \, 2$.

We claim that $a_n = n  \, \heartsuit \, 2$ for all $n \geq 3$. We can prove this through induction.

See Also

2019 AMC 12A (ProblemsAnswer KeyResources)
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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