Difference between revisions of "2021 AMC 12A Problems/Problem 14"
MRENTHUSIASM (talk | contribs) m (Undo revision 160355 by MRENTHUSIASM (talk)) (Tag: Undo) |
MRENTHUSIASM (talk | contribs) (Moved the solutions around--prioritized comprehensive solutions, and group solutions with similar ideas together.) |
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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> | ||
− | ==Solution 1 ( | + | ==Solution 1 (Properties of Logarithms)== |
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We will apply the following logarithmic identity: | We will apply the following logarithmic identity: | ||
<cmath>\log_{p^n}{q^n}=\log_{p}{q},</cmath> | <cmath>\log_{p^n}{q^n}=\log_{p}{q},</cmath> | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution | + | ==Solution 2 (Properties of Logarithms)== |
First, we can get rid of the <math>k</math> exponents using properties of logarithms: <cmath>\log_{5^k} 3^{k^2} = k^2 \cdot \frac{1}{k} \cdot \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>\log_{9^k} 25^{k} = k \cdot \frac{1}{k} \cdot \log_{9} 25 = \log_{9} 5^2.</cmath> | First, we can get rid of the <math>k</math> exponents using properties of logarithms: <cmath>\log_{5^k} 3^{k^2} = k^2 \cdot \frac{1}{k} \cdot \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>\log_{9^k} 25^{k} = k \cdot \frac{1}{k} \cdot \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: | 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: | ||
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~MRENTHUSIASM (Reformatting) | ~MRENTHUSIASM (Reformatting) | ||
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+ | ==Solution 3 (Properties of Logarithms)== | ||
+ | This equals <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) }21{,}000}.</cmath> | ||
+ | ~JHawk0224 | ||
==Solution 4 (Estimations and Answer Choices)== | ==Solution 4 (Estimations and Answer Choices)== |
Revision as of 23:47, 25 August 2021
Contents
- 1 Problem
- 2 Solution 1 (Properties of Logarithms)
- 3 Solution 2 (Properties of Logarithms)
- 4 Solution 3 (Properties of Logarithms)
- 5 Solution 4 (Estimations and Answer Choices)
- 6 Video Solution by Punxsutawney Phil
- 7 Video Solution by Hawk Math
- 8 Video Solution by OmegaLearn (Using Logarithmic Manipulations)
- 9 Video Solution by TheBeautyofMath (Using Magical Ability)
- 10 Video Solution by The Power of Logic
- 11 See also
Problem
What is the value of
Solution 1 (Properties of Logarithms)
We will apply the following logarithmic identity:
which can be proven by the Change of Base Formula:
Now, we simplify the expressions inside the summations:
and
Using these results, we evaluate the original expression:
~MRENTHUSIASM
Solution 2 (Properties of Logarithms)
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:
In
we use the triangular numbers equation:
Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:
Thus,
~Joeya (Solution)
~MRENTHUSIASM (Reformatting)
Solution 3 (Properties of Logarithms)
This equals
~JHawk0224
Solution 4 (Estimations and Answer Choices)
In note that the addends are greater than
for all
In note that the addends are greater than
for all
We have the inequality which eliminates choices
and
We get the answer
by either an educated guess or a continued approximation:
Observe that and
Therefore, we obtain the following rough underestimation:
From here, it should be safe to guess that the answer is
~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)
Video Solution by TheBeautyofMath (Using Magical Ability)
https://youtu.be/ySWSHyY9TwI?t=999
~IceMatrix
Video Solution by The Power of Logic
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.