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

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Using the identity <cmath>\log_{p^n}{q^n}=\log_{p}{q},</cmath> simplify <cmath>\begin{align*}
 
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^k}{{3^k}^2}&=\log_{5^k}{(3^k)^k} \\
−
&=k\log_{5^k}{3^k} \\
+
&=\log_{5}{3^k} \\
−
&=k\log_{5}{3},
 
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
and  
 
and  
Line 79: Line 78:
 
\log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\
 
\log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\
 
&=\log_{3}{5}.
 
&=\log_{3}{5}.
−
\end{align*}</cmath>. Now we have the sum: <cmath>\left(\sum_{k=1}^{20} \log_{5} 3^{k}\right)\cdot\left(\sum_{k=1}^{100} \log_{3} 5\right)</cmath>  
+
\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*}
 
<cmath>\begin{align*}
 
\sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\
 
\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^{(1 + 2 + \dots + 20)} \\
−
&= \log_{5} 3^{\frac{20(20+1)}{2}} &&\hspace{15mm}(*) \\
+
&= \log_{5} 3^{\frac{20(20+1)}{2}} \\
−
&= \log_{5} 3^{210}
+
&= \log_{5} 3^{210} \\
 
&= {210}\cdot\log_{5} {3}, \\
 
&= {210}\cdot\log_{5} {3}, \\
−
\sum_{k=1}^{100} \log_{9} 5^2 &= \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2 \\
+
\sum_{k=1}^{100}\log_{3} {5} &= {100}\cdot\log_{3} {5}.
−
&= {100}\cdot\log_{3} {5} \\.
 
 
\end{align*}</cmath>  
 
\end{align*}</cmath>  
−
<cmath>\begin{align*}
+
With the reciprocal rule the logarithms cancel out leaving: <math>\boxed{\textbf{(E) }21{,}000}.</math>
−
\{210}\cdot\log_{5} {3}\cdot\{100}\cdot\log_{3} {5}\
+
 
−
&={21000}\cdot\log_{5} {3}\cdot\log_{3} {5}
+
~[[User:PowerQualimit|PowerQualimit]]
−
\end{align*}</cmath>
 
−
With the reciprocal rule the logarithms cancel out leaving: <math>\boxed{\textbf{(E) }21{,}000}</math>
 
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==

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
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All AMC 12 Problems and Solutions

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