# Difference between revisions of "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

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:

$$\frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210$$

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