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

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==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><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>
==Solution==
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==Solution 1==
 
This equals
 
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>
 
<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>
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-Solution by Joeya
 
-Solution by Joeya
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 +
==Solution 3 (Detailed Explanation of Solution 1)==
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We use the following property of logarithms:
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<cmath>\log_{p^n}{(q^n)}=\log_{p}{q}.</cmath>
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Here is another AMC Problem that applies the property of logarithms shown in <math>(*)</math> above.
 +
 +
~MRENTHUSIASM
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==

Revision as of 15:46, 13 February 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

This equals \[\left(\sum_{k=1}^{20}k\log_5(3)\right)\left(\sum_{k=1}^{100}\log_9(25)\right)=\frac{20\cdot21}{2}\cdot\log_5(3)\cdot100\log_3(5)=\boxed{\textbf{(E)} 21000}\] ~JHawk0224

Solution 2

First, we can get rid of the $k$ exponents using properties of logarithms:

\[\left(\log_{5^k} 3^{k^2}\right) = k^2 * \frac{1}{k} * \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k\] (Leaving the single $k$ in the exponent will come in handy later). Similarly,

\[\left(\log_{9^k} 25^{k}\right) = k * \frac{1}{k} * \log_{9} 25 = \log_{9} 5^2\]

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:

\[\left(\sum_{k=1}^{20} \log_{5} 3^k\right) = \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} = \log_{5} 3^{(1 + 2 + \dots + 20)}\]

\[\left(\sum_{k=1}^{100} \log_{9} 5^2\right) = \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2= \log_{9} 5^{2(100)} = \log_{9} 5^{200}\]

To evaluate the exponent of the $3$ in the first logarithm, 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,

\[\left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) = \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right)\]

\[= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) = (200)(105) = \boxed{\textbf{(E)} 21000}\]

-Solution by Joeya

Solution 3 (Detailed Explanation of Solution 1)

We use the following property of logarithms: \[\log_{p^n}{(q^n)}=\log_{p}{q}.\]

Here is another AMC Problem that applies the property of logarithms shown in $(*)$ above.

~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

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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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