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Difference between revisions of "2021 AMC 12A Problems/Problem 14"
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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==
 
==Solution 1==

Revision as of 16:47, 15 April 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 (Detailed Explanation of Solution 1)

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

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 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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