Difference between revisions of "2019 AMC 12A Problems/Problem 23"

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We claim that <math>a_n = n  \, \heartsuit \, 2</math> for all <math>n \geq 3</math>. We can prove this through induction.
 
We claim that <math>a_n = n  \, \heartsuit \, 2</math> for all <math>n \geq 3</math>. We can prove this through induction.
  
<math>a_n = a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, ((n-1)  \, \heartsuit \, 2)</math>
+
<math>a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, ((n-1)  \, \heartsuit \, 2)</math>
  
 
This can be simplified as <math>a_n = ((n^{\log_{n-1}(7)})  \, \diamondsuit \, ((n-1)^{\log_{2}(7)}))</math>.
 
This can be simplified as <math>a_n = ((n^{\log_{n-1}(7)})  \, \diamondsuit \, ((n-1)^{\log_{2}(7)}))</math>.

Revision as of 20:49, 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.

$a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, ((n-1)  \, \heartsuit \, 2)$

This can be simplified as $a_n = ((n^{\log_{n-1}(7)})  \, \diamondsuit \, ((n-1)^{\log_{2}(7)}))$.

Applying the diamond operation, we can simplify $a_n = n^h$ where $h = \log_{n-1}(7) \cdot \log_{7}(n-1)^{\log_{2}(7)}$. By using logarithm rules to remove the exponent of $\log_{7}(n-1)$ and after cancelling, $h = \frac{1}{\log_{7}(2)}$.

Therefore, $a_n = n^{\frac{1}{\log_{7}(2)}} = n  \, \heartsuit \, 2$ for all $n \geq 3$, completing the induction.

We have $a_{2019} = 2019^{\log_{2}(7)}$. Taking log base 2019 of both sides gives us ${\log_{2019}(a_{2019})} = {\log_{2}(7)}$. Then, by changing to base 7 and after cancellation, we arrive at ${\log_{7}(a_{2019})} = {\log_{2}(2019)}$. Because $2^{11} = 2048$ and $2^{10} = 1024$, our answer is $\boxed{\textbf{(D)}11}$

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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