Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 AMC 12A Problems/Problem 14"

m (→‎Solution 2)
(→‎Solution 4 (Properties of Logarithms))
 
(69 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
What is the value of<cmath>\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?</cmath><math>\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2,200\qquad \textbf{(E) }21,000</math>
+
What is the value of <cmath>\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?</cmath>
−
==Solution==
 
−
This equals
 
−
<cmath>\left(\sum_{k=1}^{20}k\log_5(3)\right)\left(\sum_{k=1}^{100}\log_9(25)\right)=\frac{20\cdot21}{2}\cdot\log_5(3)\cdot100\log_3(5)=\boxed{\textbf{(E)} 21000}</cmath>
 
−
~JHawk0224
 
  
−
==Solution 2==
+
<math>\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2{,}200\qquad \textbf{(E) }21{,}000</math>
−
First, we can get rid of the <math>k</math> exponents using properties of logarithms:
 
  
−
<cmath>\left(\log_{5^k} 3^{k^2}\right) = k^2 * \frac{1}{k} * \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k</cmath> (Leaving the single <math>k</math> in the exponent will come in handy later). Similarly,
+
==Solution 1 (Properties of Logarithms)==
 +
We will apply the following logarithmic identity:
 +
<cmath>\log_{p^n}{q^n}=\log_{p}{q},</cmath>
 +
which can be proven by the Change of Base Formula: <cmath>\log_{p^n}{q^n}=\frac{\log_{p}{q^n}}{\log_{p}{p^n}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.</cmath>
 +
Now, we simplify the expressions inside the summations:
 +
<cmath>\begin{align*}
 +
\log_{5^k}{{3^k}^2}&=\log_{5^k}{(3^k)^k} \\
 +
&=k\log_{5^k}{3^k} \\
 +
&=k\log_{5}{3},
 +
\end{align*}</cmath>
 +
and
 +
<cmath>\begin{align*}
 +
\log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\
 +
&=\log_{3}{5}.
 +
\end{align*}</cmath>
 +
Using these results, we evaluate the original expression:
 +
<cmath>\begin{align*}
 +
\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)&=\left(\sum_{k=1}^{20} k\log_{5}{3}\right)\cdot\left(\sum_{k=1}^{100} \log_{3}{5}\right) \\
 +
&= \left(\log_{5}{3}\cdot\sum_{k=1}^{20} k\right)\cdot\left(\log_{3}{5}\cdot\sum_{k=1}^{100} 1\right) \\
 +
&= \left(\sum_{k=1}^{20} k\right)\cdot\left(\sum_{k=1}^{100} 1\right) \\
 +
&= \frac{21\cdot20}{2}\cdot100 \\
 +
&= \boxed{\textbf{(E) }21{,}000}.
 +
\end{align*}</cmath>
 +
~MRENTHUSIASM (Solution)
  
−
<cmath>\left(\log_{9^k} 25^{k}\right) = k * \frac{1}{k} * \log_{9} 25 = \log_{9} 5^2</cmath>
+
~JHawk0224 (Proposal)
  
 +
==Solution 2 (Properties of Logarithms)==
 +
First, we can get rid of the <math>k</math> exponents using properties of logarithms: <cmath>\log_{5^k} 3^{k^2} = k^2 \cdot \frac{1}{k} \cdot \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k.</cmath> (Leaving the single <math>k</math> in the exponent will come in handy later). Similarly, <cmath>\log_{9^k} 25^{k} = k \cdot \frac{1}{k} \cdot \log_{9} 25 = \log_{9} 5^2.</cmath>
 
Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms:
 
Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms:
 +
<cmath>\begin{align*}
 +
\sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\
 +
&= \log_{5} 3^{(1 + 2 + \dots + 20)} \\
 +
&= \log_{5} 3^{\frac{20(20+1)}{2}} &&\hspace{15mm}(*) \\
 +
&= \log_{5} 3^{210}, \\
 +
\sum_{k=1}^{100} \log_{9} 5^2 &= \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2 \\
 +
&= \log_{9} 5^{2(100)} \\
 +
&= \log_{9} 5^{200}.
 +
\end{align*}</cmath>
 +
In <math>(*),</math> we use the triangular numbers equation: <cmath>1 + 2 + \dots + n = \frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210.</cmath>
 +
Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:
 +
<cmath>\log_{a} b\log_{x} y = \log_{a} y\log_{x} b.</cmath>
 +
Thus,
 +
<cmath>\begin{align*}
 +
\left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) &= \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right) \\
 +
&= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) \\
 +
&= (200)(105) \\
 +
&= \boxed{\textbf{(E) }21{,}000}.
 +
\end{align*}</cmath>
 +
~Joeya (Solution)
  
−
<cmath>\left(\sum_{k=1}^{20} \log_{5} 3^k\right) = \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} = \log_{5} 3^{(1 + 2 + \dots + 20)}</cmath>
+
~MRENTHUSIASM (Reformatting)
−
 
 
−
<cmath>\left(\sum_{k=1}^{100} \log_{9} 5^2\right) = \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2= \log_{9} 5^{2(100)} = \log_{9} 5^{200}</cmath>
 
  
−
To evaluate the exponent of the <math>3</math> in the first logarithm, we use the triangular numbers equation:
+
==Solution 3 (Estimations and Answer Choices)==
 +
In <math>\sum_{k=1}^{20} \log_{5^k} 3^{k^2},</math> note that the addends are greater than <math>1</math> for all <math>k\geq2.</math>
  
−
<cmath>\frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210</cmath>
+
In <math>\sum_{k=1}^{100} \log_{9^k} 25^k,</math> note that the addends are greater than <math>1</math> for all <math>k\geq1.</math>
  
−
Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:
+
We have the inequality <cmath>\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} 1\right)=19\cdot100=1{,}900,</cmath> which eliminates choices <math>\textbf{(A)}, \textbf{(B)},</math> and <math>\textbf{(C)}.</math> We get the answer <math>\boxed{\textbf{(E) }21{,}000}</math> by either an educated guess or a continued approximation:  
  
−
<cmath>\log_{a} b\log_{x} y = \log_{a} y\log_{x} b</cmath>
+
Observe that <math>\sum_{k=1}^{20} \log_{5^k} 3^{k^2} >> \sum_{k=2}^{20} 1 = 19</math> and <math>\sum_{k=1}^{100} \log_{9^k} 25^k\approx\sum_{k=1}^{100} \log_{9^k} 27^k = \sum_{k=1}^{100} \frac{3}{2} = 150.</math> Therefore, we obtain the following rough underestimation: <cmath>\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} \frac{3}{2}\right)=19\cdot150=2{,}850.</cmath>
 +
From here, it should be safe to guess that the answer is <math>\textbf{(E)}.</math>
  
−
Thus,
+
~MRENTHUSIASM
  
−
<cmath>\left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) = \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right)</cmath>
+
==Solution 4 (Properties of Logarithms)==
 +
Using the identity <cmath>\log_{p^n}{q^n}=\log_{p}{q},</cmath> simplify <cmath>\begin{align*}
 +
\log_{5^k}{{3^k}^2}&=\log_{5^k}{(3^k)^k} \\
 +
&=\log_{5}{3^k} \\
 +
\end{align*}</cmath>
 +
and
 +
<cmath>\begin{align*}
 +
\log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\
 +
&=\log_{3}{5}.
 +
\end{align*}</cmath>. Now we have the product: <cmath>\left(\sum_{k=1}^{20} \log_{5} 3^{k}\right)\cdot\left(\sum_{k=1}^{100} \log_{3} 5\right)</cmath>
 +
<cmath>\begin{align*}
 +
\sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\
 +
&= \log_{5} 3^{(1 + 2 + \dots + 20)} \\
 +
&= \log_{5} 3^{\frac{20(20+1)}{2}} \\
 +
&= \log_{5} 3^{210} \\
 +
&= {210}\cdot\log_{5} {3}, \\
 +
\sum_{k=1}^{100}\log_{3} {5} &= {100}\cdot\log_{3} {5}.
 +
\end{align*}</cmath>
 +
With the reciprocal rule the logarithms cancel out leaving: <math>\boxed{\textbf{(E) }21{,}000}.</math>
  
−
<cmath>= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) = (200)(105) = \boxed{\textbf{(E)} 21000}</cmath>
+
~[[User:PowerQualimit|PowerQualimit]]
−
 
 
−
-Solution by Joeya
 
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==
Line 43: Line 99:
 
== Video Solution by OmegaLearn (Using Logarithmic Manipulations) ==
 
== Video Solution by OmegaLearn (Using Logarithmic Manipulations) ==
 
https://youtu.be/vgFPZ-hyd-I
 
https://youtu.be/vgFPZ-hyd-I
 +
 +
==Video Solution by TheBeautyofMath (Using Magical Ability)==
 +
https://youtu.be/ySWSHyY9TwI?t=999
 +
 +
~IceMatrix
 +
 +
== Video Solution by The Power of Logic ==
 +
https://youtu.be/b7xEeR7HXkE
 +
 +
==Video Solution (Logic and Simplification)==
 +
https://youtu.be/F6w9zsiMZ8w
 +
 +
~Education, the Study of Everything
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2021|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2021|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:48, 23 November 2023

Problem

What is the value of \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?\]

$\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2{,}200\qquad \textbf{(E) }21{,}000$

Solution 1 (Properties of Logarithms)

We will apply the following logarithmic identity: \[\log_{p^n}{q^n}=\log_{p}{q},\] which can be proven by the Change of Base Formula: \[\log_{p^n}{q^n}=\frac{\log_{p}{q^n}}{\log_{p}{p^n}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.\] Now, we simplify the expressions inside the summations: \begin{align*} \log_{5^k}{{3^k}^2}&=\log_{5^k}{(3^k)^k} \\ &=k\log_{5^k}{3^k} \\ &=k\log_{5}{3}, \end{align*} and \begin{align*} \log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\ &=\log_{3}{5}. \end{align*} Using these results, we evaluate the original expression: \begin{align*} \left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)&=\left(\sum_{k=1}^{20} k\log_{5}{3}\right)\cdot\left(\sum_{k=1}^{100} \log_{3}{5}\right) \\ &= \left(\log_{5}{3}\cdot\sum_{k=1}^{20} k\right)\cdot\left(\log_{3}{5}\cdot\sum_{k=1}^{100} 1\right) \\ &= \left(\sum_{k=1}^{20} k\right)\cdot\left(\sum_{k=1}^{100} 1\right) \\ &= \frac{21\cdot20}{2}\cdot100 \\ &= \boxed{\textbf{(E) }21{,}000}. \end{align*} ~MRENTHUSIASM (Solution)

~JHawk0224 (Proposal)

Solution 2 (Properties of Logarithms)

First, we can get rid of the $k$ exponents using properties of logarithms: \[\log_{5^k} 3^{k^2} = k^2 \cdot \frac{1}{k} \cdot \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k.\] (Leaving the single $k$ in the exponent will come in handy later). Similarly, \[\log_{9^k} 25^{k} = k \cdot \frac{1}{k} \cdot \log_{9} 25 = \log_{9} 5^2.\] Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms: \begin{align*} \sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\ &= \log_{5} 3^{(1 + 2 + \dots + 20)} \\ &= \log_{5} 3^{\frac{20(20+1)}{2}} &&\hspace{15mm}(*) \\ &= \log_{5} 3^{210}, \\ \sum_{k=1}^{100} \log_{9} 5^2 &= \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2 \\ &= \log_{9} 5^{2(100)} \\ &= \log_{9} 5^{200}. \end{align*} In $(*),$ we use the triangular numbers equation: \[1 + 2 + \dots + n = \frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210.\] Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify: \[\log_{a} b\log_{x} y = \log_{a} y\log_{x} b.\] Thus, \begin{align*} \left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) &= \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right) \\ &= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) \\ &= (200)(105) \\ &= \boxed{\textbf{(E) }21{,}000}. \end{align*} ~Joeya (Solution)

~MRENTHUSIASM (Reformatting)

Solution 3 (Estimations and Answer Choices)

In $\sum_{k=1}^{20} \log_{5^k} 3^{k^2},$ note that the addends are greater than $1$ for all $k\geq2.$

In $\sum_{k=1}^{100} \log_{9^k} 25^k,$ note that the addends are greater than $1$ for all $k\geq1.$

We have the inequality \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} 1\right)=19\cdot100=1{,}900,\] which eliminates choices $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}.$ We get the answer $\boxed{\textbf{(E) }21{,}000}$ by either an educated guess or a continued approximation:

Observe that $\sum_{k=1}^{20} \log_{5^k} 3^{k^2} >> \sum_{k=2}^{20} 1 = 19$ and $\sum_{k=1}^{100} \log_{9^k} 25^k\approx\sum_{k=1}^{100} \log_{9^k} 27^k = \sum_{k=1}^{100} \frac{3}{2} = 150.$ Therefore, we obtain the following rough underestimation: \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} \frac{3}{2}\right)=19\cdot150=2{,}850.\] From here, it should be safe to guess that the answer is $\textbf{(E)}.$

~MRENTHUSIASM

Solution 4 (Properties of Logarithms)

Using the identity \[\log_{p^n}{q^n}=\log_{p}{q},\] simplify \begin{align*} \log_{5^k}{{3^k}^2}&=\log_{5^k}{(3^k)^k} \\ &=\log_{5}{3^k} \\ \end{align*} and \begin{align*} \log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\ &=\log_{3}{5}. \end{align*}. Now we have the product: \[\left(\sum_{k=1}^{20} \log_{5} 3^{k}\right)\cdot\left(\sum_{k=1}^{100} \log_{3} 5\right)\] \begin{align*} \sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\ &= \log_{5} 3^{(1 + 2 + \dots + 20)} \\ &= \log_{5} 3^{\frac{20(20+1)}{2}} \\ &= \log_{5} 3^{210} \\ &= {210}\cdot\log_{5} {3}, \\ \sum_{k=1}^{100}\log_{3} {5} &= {100}\cdot\log_{3} {5}. \end{align*} With the reciprocal rule the logarithms cancel out leaving: $\boxed{\textbf{(E) }21{,}000}.$

~PowerQualimit

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=FD9BE7hpRvg&t=322s

Video Solution by Hawk Math

https://www.youtube.com/watch?v=AjQARBvdZ20

Video Solution by OmegaLearn (Using Logarithmic Manipulations)

https://youtu.be/vgFPZ-hyd-I

Video Solution by TheBeautyofMath (Using Magical Ability)

https://youtu.be/ySWSHyY9TwI?t=999

~IceMatrix

Video Solution by The Power of Logic

https://youtu.be/b7xEeR7HXkE

Video Solution (Logic and Simplification)

https://youtu.be/F6w9zsiMZ8w

~Education, the Study of Everything

See also

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_14&oldid=205477"