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

Revision as of 22:07, 8 August 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 $\log_{2019}(a_{2019}) = k = \log_2(7)$ . 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.

$\frac{\log{a_{2019}}}{\log{2019}} = \frac{\log{7}}{\log{2}}\implies \frac{\log{a_{2019}}}{\log{7}} = \frac{\log{2019}}{\log{2}}\implies \log_7(a_{2019}) =\log_2(2019)$

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.

Clearly, the base case where $n = 3$ holds.

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

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 $\log_{2019}$ 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 $\boxed{\textbf{(D) } 11}$ .

Solution 3

We are given that $a_n=(n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}$ $$ (Error compiling LaTeX. Unknown error_msg)a_n=(n^{\frac{1}{

@@ Line 41: / Line 41: @@
 We are given that
-<cmath>a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}</cmath>
+<cmath>a_n=(n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}</cmath>
+<math></math>a_n=(n^{\frac{1}{
 ==See Also==

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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