Difference between revisions of "2021 AMC 12A Problems/Problem 14"
Pi is 3.14 (talk | contribs) (→Video Solution by pi_is_3.14 (Using Logarithmic Manipulations)) |
|||
Line 5: | Line 5: | ||
<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 22:18, 12 February 2021
Contents
Problem
What is the value of
Solution
This equals ~JHawk0224
Solution 2
First, we can get rid of the exponents using properties of logarithms:
(Leaving the single in the exponent will come in handy later). Similarly,
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:
To evaluate the exponent of the in the first logarithm, we use the triangular numbers equation:
Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:
Thus,
-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.