2021 AMC 12A Problems/Problem 14

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

We can prove it quickly using the Change of Base Formula: $\log_{p^n}{(q^n)}=\frac{\log_{p}{(q^n)}}{\log_{p}{(p^n)}}=\frac{n\log_{p}{q}}{n\log_{p}{p}}=\frac{\log_{p}{q}}{1}=\log_{p}{q}.$ 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: $\begin{align*} \left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)&=\left(\sum_{k=1}^{20} k\log_{5}{3}\right)\left(\sum_{k=1}^{100} \log_{3}{5}\right) \\ &= \left(\log_{5}{3}\cdot\sum_{k=1}^{20} k\right)\left(\log_{3}{5}\cdot\sum_{k=1}^{100} 1\right) \\ &= \left(\sum_{k=1}^{20} k\right)\left(\sum_{k=1}^{100} 1\right) \\ &= \left(\frac{21\cdot20}{2}\right)\left(100\right) \\ &= \boxed{\textbf{(E) }21,000}. \end{align*}$ ~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

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

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