Difference between revisions of "2021 AMC 12A Problems/Problem 14"
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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> | <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 | ||
| + | |||
| + | ==Solution 2== | ||
| + | First, we can get rid of the <math>k</math> exponents using properties of logarithms: | ||
| + | |||
| + | <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, | ||
| + | |||
| + | <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: | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | To evaluate the exponent of the <math>3</math> in the first logarithm, we use the triangular numbers equation: | ||
| + | |||
| + | <cmath>\frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210</cmath> | ||
| + | |||
| + | Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify: | ||
| + | |||
| + | <cmath>\log_{a} b\log_{x} y = \log_{a} y\log_{x} b</cmath> | ||
| + | |||
| + | Thus, | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <cmath>= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) = (200)(105) = \boxed{\textbf{(E)} 21000}</cmath> | ||
| + | |||
| + | -Solution by Joeya | ||
| + | |||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
Revision as of 21:18, 12 February 2021
Contents
Problem
What is the value of![\[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?\]](http://latex.artofproblemsolving.com/9/9/3/9934e0771c30ad96de59aa26766c6befd3570ccd.png)
Solution
This equals
![\[\left(\sum_{k=1}^{20}k\log_5(3)\right)\left(\sum_{k=1}^{100}\log_9(25)\right)=\frac{20\cdot21}{2}\cdot\log_5(3)\cdot100\log_3(5)=\boxed{\textbf{(E)} 21000}\]](http://latex.artofproblemsolving.com/1/e/e/1ee3550086ed338d704398aad18c9252b730baea.png)
~JHawk0224
Solution 2
First, we can get rid of the 
exponents using properties of logarithms:
![\[\left(\log_{5^k} 3^{k^2}\right) = k^2 * \frac{1}{k} * \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k\]](http://latex.artofproblemsolving.com/a/d/c/adcecdbcabfa0caa2424049ec4bc365d37cdc1f0.png)
(Leaving the single in the exponent will come in handy later). Similarly,
![\[\left(\log_{9^k} 25^{k}\right) = k * \frac{1}{k} * \log_{9} 25 = \log_{9} 5^2\]](http://latex.artofproblemsolving.com/7/e/8/7e847ba486e60a60869387116689ef6528c61686.png)
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:
![\[\left(\sum_{k=1}^{20} \log_{5} 3^k\right) = \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} = \log_{5} 3^{1 + 2 + \dots + 20}\]](http://latex.artofproblemsolving.com/1/c/0/1c0416aa2f7a45416d54f48caccc5c92995ba851.png)
![\[\left(\sum_{k=1}^{100} \log_{9} 5^2\right) = \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2= \log_{9} 5^{2(100)} = \log_{9} 5^{200}\]](http://latex.artofproblemsolving.com/f/c/2/fc2b2b505d799b7714eb0d46d1d948923c43703e.png)
To evaluate the exponent of the 
in the first logarithm, we use the triangular numbers equation:
![\[\frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210\]](http://latex.artofproblemsolving.com/3/7/3/37309e5f9c247e1c275c9d325b5da167e073e849.png)
Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:
![\[\log_{a} b\log_{x} y = \log_{a} y\log_{x} b\]](http://latex.artofproblemsolving.com/a/d/c/adc75a189e981c4f6c35974f8278ebf707320396.png)
Thus,
![\[\left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) = \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right)\]](http://latex.artofproblemsolving.com/1/8/f/18f3ec7c003f8ccefb245d1d0f2a52a49eb4bde0.png)
![\[= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) = (200)(105) = \boxed{\textbf{(E)} 21000}\]](http://latex.artofproblemsolving.com/b/2/5/b25094a894a8d80e0dd89240c3399d1ff181510c.png)
-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)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
