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

First note that by log properties $a\diamondsuit b = 7^{(\log_7a)(\log_7b)}$ and $a \heartsuit b = 7^{\frac{\log_7a}{\log_7b}} = 7^{\log_ba}$ .

Now, define $b_n = \log_7(a_n)$ . Thus $b_3 = \log_7(3\heartsuit 2) = \log_7(7^{\log_23}) = \log_23$ .

Taking logs of both sides of the recursion and using the definition of $\diamondsuit$ gives $\log_7(a_n) = \log_7(7^{\log_7n\heartsuit (n-1)\log_7a_{n-1}})$ .

The logs and the exponent cancel to $\log_7((n\heartsuit (n-1))^{\log_7(a_{n-1})})$ , and by the definition of $\heartsuit$ , ths is $\log_7(7^{(\log_{n-1}n)\log_7(a_{n-1})})$ , which quickly simplifies to $\log_7(a_{n-1})\log_{n-1}n$ $= b_{n-1}\log_{n-1}n$ .

Thus $b_n = b_{n-1}\log_{n-1}n$ . From this, we have $b_4 = b_3\log_34 = \log_23\log_34 = \log_24$ , $b_5 = \log_45\log_24 = \log_25$ , and in general, $b_n = \log_2n$ .

Finally, $\log_7(a_{2019}) = b_{2019}= \log_22019$ . Since $2^{11} = 2048$ , $\log_22019 \approx \boxed{\text{(D) }11}$ .

- NamelyOrange

Solution 2

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 3

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 4

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

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