2019 AMC 12A Problems/Problem 23

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 $a_4 = (4^{m})^{k}$ where $m = \frac{1}{\log_{7}(3)}$ and $k = \log_{7}(3^{\frac{1}{\log_{7}(2)}})$. Using logarithm rules, we can remove the exponent of the 3 so that $k = \frac{\log_{7}(3)}{\log_{7}(2)}$. 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 \[a_n=(n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] \[a_n=(n^{\frac{1}{\log_7(n-1)}})^{\log_7(a_{n-1})}\] Since we are asked to find $\log_7(a_{2019})$, we directly apply \[\log_7(a_n)=\log_7(n^{\frac{1}{\log_7(n-1)}})^{\log_7(a_{n-1})}\] Using the property that $\log_ab^c=c\log_ab$ \[\log_7(a_n)=(\log_7a_{n-1})(\log_7(n^{\frac{1}{\log_7(n-1)}}))\] Now using the property that $\frac{1}{\log_ab}=\log_ba$ \[\log_7(a_n)=(\log_7a_{n-1})(\log_7n^{\log_{n-1}7})\] Once again applying the first property yields \[\log_7(a_n)=(\log_7a_{n-1})(\log_{n-1}7)(\log_7n)\] Rearranging the expression, \[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7a_{n-1})\]

Now expressing $\log_7a_{n-1}$ in a similar expression as $\log_7a_n$,

\[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7n-1)(\log_{n-2}7)(\log_7a_{n-2})\] \[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7n-1)(\log_{n-2}7)(\log_7n-2)(\log_{n-3}7)...(\log_74)(\log_37)(\log_7a_3)\]

Because of the fact that $(\log_ab)(\log_ba)=1$, we can cancel out the terms to get

\[\log_7(a_n)=(\log_7n)(\log_37)(\log_7a_3)\] \[\log_7(a_n)=(\log_7n)(\log_37)(\log_7(3^{\frac{1}{\log_72}}))\] \[\log_7(a_n)=(\log_7n)(\log_37)(\log_27)(\log_73)\] \[\log_7(a_n)=(\log_27)(\log_7n)\]

Using the Chain Rule for Logarithm, $(\log_ab)(\log_bc)=\log_ac$, yields

\[\log_7(a_n)=(\log_2n)\] Finally, substituting in $n=2019$, we have \[\log_7(a_{2019})=(\log_22019)\] \[\log_7(a_{2019})\approx11\boxed{\mathrm{(D)}}\]

~ Nafer

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2019amc12a/495

~ dolphin7

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS