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==Problem== | ==Problem== | ||
− | What is the value of <cmath>\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?</cmath> | + | What is the value of <cmath>\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?</cmath> |
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− | ==Solution | + | <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> |
− | We will | + | |
− | <cmath>\log_{p^n}{ | + | ==Solution 1 (Properties of Logarithms)== |
− | + | We will apply the following logarithmic identity: | |
+ | <cmath>\log_{p^n}{q^n}=\log_{p}{q},</cmath> | ||
+ | which can be proven by the Change of Base Formula: <cmath>\log_{p^n}{q^n}=\frac{\log_{p}{q^n}}{\log_{p}{p^n}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.</cmath> | ||
Now, we simplify the expressions inside the summations: | Now, we simplify the expressions inside the summations: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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Using these results, we evaluate the original expression: | Using these results, we evaluate the original expression: | ||
<cmath>\begin{align*} | <cmath>\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(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)&=\left(\sum_{k=1}^{20} k\log_{5}{3}\right)\cdot\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(\log_{5}{3}\cdot\sum_{k=1}^{20} k\right)\cdot\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(\sum_{k=1}^{20} k\right)\cdot\left(\sum_{k=1}^{100} 1\right) \\ |
− | &= | + | &= \frac{21\cdot20}{2}\cdot100 \\ |
− | &= \boxed{\textbf{(E) }21,000}. | + | &= \boxed{\textbf{(E) }21{,}000}. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | ~MRENTHUSIASM | + | ~MRENTHUSIASM (Solution) |
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− | + | ~JHawk0224 (Proposal) | |
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+ | ==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> | ||
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: | ||
− | + | <cmath>\begin{align*} | |
− | <cmath>\ | + | \sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\ |
− | + | &= \log_{5} 3^{(1 + 2 + \dots + 20)} \\ | |
− | + | &= \log_{5} 3^{\frac{20(20+1)}{2}} &&\hspace{15mm}(*) \\ | |
− | + | &= \log_{5} 3^{210}, \\ | |
− | + | \sum_{k=1}^{100} \log_{9} 5^2 &= \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2 \\ | |
− | + | &= \log_{9} 5^{2(100)} \\ | |
− | <cmath>1 + 2 + \dots + n = \frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210</cmath> | + | &= \log_{9} 5^{200}. |
− | + | \end{align*}</cmath> | |
+ | In <math>(*),</math> we use the triangular numbers equation: <cmath>1 + 2 + \dots + n = \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: | 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> | |
− | <cmath>\log_{a} b\log_{x} y = \log_{a} y\log_{x} b</cmath> | ||
− | |||
Thus, | Thus, | ||
+ | <cmath>\begin{align*} | ||
+ | \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) }21{,}000}. | ||
+ | \end{align*}</cmath> | ||
+ | ~Joeya (Solution) | ||
− | + | ~MRENTHUSIASM (Reformatting) | |
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− | ==Solution | + | ==Solution 3 (Estimations and Answer Choices)== |
In <math>\sum_{k=1}^{20} \log_{5^k} 3^{k^2},</math> note that the addends are greater than <math>1</math> for all <math>k\geq2.</math> | In <math>\sum_{k=1}^{20} \log_{5^k} 3^{k^2},</math> note that the addends are greater than <math>1</math> for all <math>k\geq2.</math> | ||
In <math>\sum_{k=1}^{100} \log_{9^k} 25^k,</math> note that the addends are greater than <math>1</math> for all <math>k\geq1.</math> | In <math>\sum_{k=1}^{100} \log_{9^k} 25^k,</math> note that the addends are greater than <math>1</math> for all <math>k\geq1.</math> | ||
− | + | We have the inequality <cmath>\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} 1\right)=19\cdot100=1{,}900,</cmath> which eliminates choices <math>\textbf{(A)}, \textbf{(B)},</math> and <math>\textbf{(C)}.</math> We get the answer <math>\boxed{\textbf{(E) }21{,}000}</math> by either an educated guess or a continued approximation: | |
− | + | Observe that <math>\sum_{k=1}^{20} \log_{5^k} 3^{k^2} >> \sum_{k=2}^{20} 1 = 19</math> and <math>\sum_{k=1}^{100} \log_{9^k} 25^k\approx\sum_{k=1}^{100} \log_{9^k} 27^k = \sum_{k=1}^{100} \frac{3}{2} = 150.</math> Therefore, we obtain the following rough underestimation: <cmath>\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} \frac{3}{2}\right)=19\cdot150=2{,}850.</cmath> | |
+ | From here, it should be safe to guess that the answer is <math>\textbf{(E)}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 4 (Properties of Logarithms)== | ||
+ | Using the identity <cmath>\log_{p^n}{q^n}=\log_{p}{q},</cmath> simplify <cmath>\begin{align*} | ||
+ | \log_{5^k}{{3^k}^2}&=\log_{5^k}{(3^k)^k} \\ | ||
+ | &=\log_{5}{3^k} \\ | ||
+ | \end{align*}</cmath> | ||
+ | and | ||
+ | <cmath>\begin{align*} | ||
+ | \log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\ | ||
+ | &=\log_{3}{5}. | ||
+ | \end{align*}</cmath>. Now we have the product: <cmath>\left(\sum_{k=1}^{20} \log_{5} 3^{k}\right)\cdot\left(\sum_{k=1}^{100} \log_{3} 5\right)</cmath> | ||
+ | <cmath>\begin{align*} | ||
+ | \sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\ | ||
+ | &= \log_{5} 3^{(1 + 2 + \dots + 20)} \\ | ||
+ | &= \log_{5} 3^{\frac{20(20+1)}{2}} \\ | ||
+ | &= \log_{5} 3^{210} \\ | ||
+ | &= {210}\cdot\log_{5} {3}, \\ | ||
+ | \sum_{k=1}^{100}\log_{3} {5} &= {100}\cdot\log_{3} {5}. | ||
+ | \end{align*}</cmath> | ||
+ | With the reciprocal rule the logarithms cancel out leaving: <math>\boxed{\textbf{(E) }21{,}000}.</math> | ||
+ | |||
+ | ~[[User:PowerQualimit|PowerQualimit]] | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
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https://youtu.be/vgFPZ-hyd-I | https://youtu.be/vgFPZ-hyd-I | ||
− | ==Video Solution by TheBeautyofMath ( | + | ==Video Solution by TheBeautyofMath (Using Magical Ability)== |
https://youtu.be/ySWSHyY9TwI?t=999 | https://youtu.be/ySWSHyY9TwI?t=999 | ||
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== Video Solution by The Power of Logic == | == Video Solution by The Power of Logic == | ||
https://youtu.be/b7xEeR7HXkE | https://youtu.be/b7xEeR7HXkE | ||
+ | |||
+ | ==Video Solution (Logic and Simplification)== | ||
+ | https://youtu.be/F6w9zsiMZ8w | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=13|num-a=15}} | {{AMC12 box|year=2021|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:48, 23 November 2023
Contents
- 1 Problem
- 2 Solution 1 (Properties of Logarithms)
- 3 Solution 2 (Properties of Logarithms)
- 4 Solution 3 (Estimations and Answer Choices)
- 5 Solution 4 (Properties of Logarithms)
- 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 Video Solution (Logic and Simplification)
- 12 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)
~JHawk0224 (Proposal)
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 (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
Solution 4 (Properties of Logarithms)
Using the identity simplify
and
. Now we have the product:
With the reciprocal rule the logarithms cancel out leaving:
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
Video Solution (Logic and Simplification)
~Education, the Study of Everything
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.