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

m (No need an extra line before Sol 1.)
(Reformatted and made the boxes more consistent.)
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<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>
 
<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>
  
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==Solution 1==
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==Solution 1 (Condensed Explanation of Logarithmic Identities)==
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This equals
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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) }21,000}.</cmath>
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<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
 
~JHawk0224
  
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==Solution 2 (Detailed Explanation of Solution 1)==
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==Solution 2 (Detailed Explanation of Logarithmic Identities)==
 
We will apply the following property of logarithms:
 
We will apply the following property of logarithms:
 
<cmath>\log_{p^n}{q^n}=\log_{p}{q},</cmath>
 
<cmath>\log_{p^n}{q^n}=\log_{p}{q},</cmath>
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 3==
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==Solution 3 (Properties of Logarithms)==
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First, we can get rid of the <math>k</math> exponents using properties of logarithms:
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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>
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<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,
 
−
 
 
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<cmath>\left(\log_{9^k} 25^{k}\right) = k * \frac{1}{k} * \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*}
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<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>
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\sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\
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+
&= \log_{5} 3^{(1 + 2 + \dots + 20)} \\
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<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>
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&= \log_{5} 3^{\frac{20(20+1)}{2}} &&\hspace{10mm}\text{by Triangular Numbers: } 1 + 2 + \dots + n = \tfrac{n(n+1)}{2} \\
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&= \log_{5} 3^{210}, \\
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To evaluate the exponent of the <math>3</math> in the first logarithm, we use the triangular numbers equation:
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\sum_{k=1}^{100} \log_{9} 5^2 &= \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2 \\
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+
&= \log_{9} 5^{2(100)} \\
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<cmath>1 + 2 + \dots + n = \frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210</cmath>
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&= \log_{9} 5^{200}.
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\end{align*}</cmath>
 
Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:
 
Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:
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<cmath>\log_{a} b\log_{x} y = \log_{a} y\log_{x} b.</cmath>
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<cmath>\log_{a} b\log_{x} y = \log_{a} y\log_{x} b</cmath>
 
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Thus,
 
Thus,
 +
<cmath>\begin{align*}
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\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) \\
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&= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) \\
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&= (200)(105) \\
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&= \boxed{\textbf{(E) }21,000}.
 +
\end{align*}</cmath>
 +
~Joeya (Solution)
  
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<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>
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~MRENTHUSIASM (Reformatting)
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<cmath>= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) = (200)(105) = \boxed{\textbf{(E)} 21000}</cmath>
 
−
 
 
−
-Solution by Joeya
 
  
 
==Solution 4 (Estimations and Answer Choices)==
 
==Solution 4 (Estimations and Answer Choices)==
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https://youtu.be/vgFPZ-hyd-I
 
https://youtu.be/vgFPZ-hyd-I
  
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==Video Solution by TheBeautyofMath (using Magical Ability)==
+
==Video Solution by TheBeautyofMath (Using Magical Ability)==
 
https://youtu.be/ySWSHyY9TwI?t=999
 
https://youtu.be/ySWSHyY9TwI?t=999
  

Revision as of 00:38, 26 June 2021

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 (Condensed Explanation of Logarithmic Identities)

This equals \[\left(\sum_{k=1}^{20}k\log_5{3}\right)\left(\sum_{k=1}^{100}\log_9{25}\right)=\frac{20\cdot21}{2}\cdot\log_5{3}\cdot100\log_3{5}=\boxed{\textbf{(E) }21,000}.\] ~JHawk0224

Solution 2 (Detailed Explanation of Logarithmic Identities)

We will apply the following property of logarithms: \[\log_{p^n}{q^n}=\log_{p}{q},\] 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}{\left(3^k\right)^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 3 (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{10mm}\text{by Triangular Numbers: } 1 + 2 + \dots + n = \tfrac{n(n+1)}{2} \\ &= \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*} 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 4 (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:

Since $3^3=27\approx25,$ it follows that $9^{3/2}\approx25.$ By an extremely rough underestimation, \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)\approx\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} \frac{3}{2}\right)=19\cdot150=2,850.\] From here, it should be safe to guess that the answer is $\textbf{(E)}.$

As an extra guaranty, note that $\sum_{k=1}^{20} \log_{5^k} 3^{k^2} >> \sum_{k=2}^{20} 1 = 19.$ Therefore, we must have \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>>2,850.\]

~MRENTHUSIASM

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

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

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