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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>
  
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==Solution 1==
+
== Solution 1 ==
 +
First note that by log properties <math>a\diamondsuit b = 7^{(\log_7a)(\log_7b)}</math> and <math>a \heartsuit b = 7^{\frac{\log_7a}{\log_7b}} = 7^{\log_ba}</math>.
 +
 
 +
Now, define <math>b_n = \log_7(a_n)</math>. Thus <math>b_3 = \log_7(3\heartsuit 2) = \log_7(7^{\log_23}) = \log_23</math>.
 +
 
 +
Taking logs of both sides of the recursion and using the definition of <math>\diamondsuit</math> gives <math>\log_7(a_n) = \log_7(7^{\log_7n\heartsuit (n-1)\log_7a_{n-1}})</math>.
 +
 
 +
The logs and the exponent cancel to <math> \log_7((n\heartsuit (n-1))^{\log_7(a_{n-1})})</math>, and by the definition of <math>\heartsuit</math>, ths is <math>\log_7(7^{(\log_{n-1}n)\log_7(a_{n-1})})</math>, which quickly simplifies to <math>\log_7(a_{n-1})\log_{n-1}n</math> <math> = b_{n-1}\log_{n-1}n</math>.
 +
 
 +
Thus <math>b_n = b_{n-1}\log_{n-1}n</math>. From this, we have <math>b_4 = b_3\log_34 = \log_23\log_34 = \log_24</math>, <math>b_5 = \log_45\log_24 = \log_25</math>, and in general, <math>b_n = \log_2n</math>.
 +
 
 +
Finally, <math>\log_7(a_{2019}) = b_{2019}= \log_22019</math>.
 +
Since <math>2^{11} = 2048</math> and <math>2019</math> is slightly less than <math>2048</math>, <math>\log_22019 \approx \boxed{\text{(D) }11}</math>.
 +
 
 +
- NamelyOrange
 +
 
 +
 
 +
==Solution 2==
  
 
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>.   
 
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 conclude that <math>\log_7(a_{2019}) = \log_2(2019) \approx \boxed{11}</math>, or choice <math>\boxed{\text{D}}</math>.
 
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==
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==Solution 3==
  
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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})^{k}</math> where <math>m = \frac{1}{\log_{7}(3)}</math> and <math>k = \log_{7}(3^{\frac{1}{\log_{7}(2)}})</math>. Using logarithm rules, we can remove the exponent of the 3 so that <math>k = \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>.
  
 
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.
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<math>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>.
  
 
Applying the diamond operation, we can simplify <math>a_n = n^h</math> where <math>h = \log_{n-1}(7) \cdot \log_{7}(n-1)^{\log_{2}(7)}</math>. By using logarithm rules to remove the exponent of <math>\log_{7}(n-1)</math> and after cancelling, <math>h = \frac{1}{\log_{7}(2)}</math>.
 
Applying the diamond operation, we can simplify <math>a_n = n^h</math> where <math>h = \log_{n-1}(7) \cdot \log_{7}(n-1)^{\log_{2}(7)}</math>. By using logarithm rules to remove the exponent of <math>\log_{7}(n-1)</math> and after cancelling, <math>h = \frac{1}{\log_{7}(2)}</math>.
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We have <math>a_{2019} = 2019^{\log_{2}(7)}</math>. Taking <math>\log_{2019}</math> of both sides gives us <math>{\log_{2019}(a_{2019})} = {\log_{2}(7)}</math>. Then, by changing to base <math>7</math> and after cancellation, we arrive at <math>{\log_{7}(a_{2019})} = {\log_{2}(2019)}</math>. Because <math>2^{11} = 2048</math> and <math>2^{10} = 1024</math>, our answer is <math>\boxed{\textbf{(D) } 11}</math>.
 
We have <math>a_{2019} = 2019^{\log_{2}(7)}</math>. Taking <math>\log_{2019}</math> of both sides gives us <math>{\log_{2019}(a_{2019})} = {\log_{2}(7)}</math>. Then, by changing to base <math>7</math> and after cancellation, we arrive at <math>{\log_{7}(a_{2019})} = {\log_{2}(2019)}</math>. Because <math>2^{11} = 2048</math> and <math>2^{10} = 1024</math>, our answer is <math>\boxed{\textbf{(D) } 11}</math>.
  
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+
== Solution 4 ==
−
== Solution 3 ==
 
  
 
We are given that  
 
We are given that  
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<cmath>\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7a_{n-1})</cmath>
 
<cmath>\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7a_{n-1})</cmath>
  
−
Now expressing <math>\log_7a_(n-1)</math> in a similar expression as <math>\log_7a_n</math>,
+
Now expressing <math>\log_7a_{n-1}</math> in a similar expression as <math>\log_7a_n</math>,
  
 
<cmath>\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7n-1)(\log_{n-2}7)(\log_7a_{n-2})</cmath>
 
<cmath>\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7n-1)(\log_{n-2}7)(\log_7a_{n-2})</cmath>
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~ Nafer
 
~ Nafer
 +
 +
=== Video Solution by Richard Rusczyk ===
 +
 +
https://artofproblemsolving.com/videos/amc/2019amc12a/495
 +
 +
~ dolphin7
  
 
==See Also==
 
==See Also==

Latest revision as of 09:10, 29 September 2024

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 $2^{11} = 2048$ and $2019$ is slightly less than $2048$, $\log_22019 \approx \boxed{\text{(D) }11}$.

- 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

See Also

2019 AMC 12A (Problems • Answer Key • Resources)
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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