Difference between revisions of "2021 AMC 12A Problems/Problem 14"

m (Solution 2 (Detailed Explanation of Logarithmic Identities))
 
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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><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>
+
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>
==Solution==
 
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)} 21000}</cmath>
 
~JHawk0224
 
  
==Solution 2==
+
<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>
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,
+
==Solution 1 (Condensed Explanation of Logarithmic Identities)==
 +
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
  
<cmath>\left(\log_{9^k} 25^{k}\right) = k * \frac{1}{k} * \log_{9} 25 = \log_{9} 5^2</cmath>
+
==Solution 2 (Detailed Explanation of Logarithmic Identities)==
 +
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:
 +
<cmath>\begin{align*}
 +
\log_{5^k}{{3^k}^2}&=\log_{5^k}{\left(3^k\right)^k} \\
 +
&=k\log_{5^k}{3^k} \\
 +
&=k\log_{5}{3},
 +
\end{align*}</cmath>
 +
and
 +
<cmath>\begin{align*}
 +
\log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\
 +
&=\log_{3}{5}.
 +
\end{align*}</cmath>
 +
Using these results, we evaluate the original expression:
 +
<cmath>\begin{align*}
 +
\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)\cdot\left(\log_{3}{5}\cdot\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}.
 +
\end{align*}</cmath>
 +
~MRENTHUSIASM
  
 +
==Solution 3 (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*}
 +
\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)} \\
 +
&= \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:
 +
<cmath>\log_{a} b\log_{x} y = \log_{a} y\log_{x} b.</cmath>
 +
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)
  
<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>
+
~MRENTHUSIASM (Reformatting)
  
<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>
+
==Solution 4 (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>
  
To evaluate the exponent of the <math>3</math> in the first logarithm, we use the triangular numbers equation:
+
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>
  
<cmath>\frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210</cmath>
+
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:
  
Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:
+
Since <math>3^3=27\approx25,</math> it follows that <math>9^{3/2}\approx25.</math> By an extremely 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)\approx\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>
  
<cmath>\log_{a} b\log_{x} y = \log_{a} y\log_{x} b</cmath>
+
As an extra guaranty, note that <math>\sum_{k=1}^{20} \log_{5^k} 3^{k^2} >> \sum_{k=2}^{20} 1 = 19.</math> Therefore, we must have <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)>>2,850.</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
 
  
 +
~MRENTHUSIASM
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==
Line 44: Line 80:
 
== Video Solution by OmegaLearn (Using Logarithmic Manipulations) ==
 
== Video Solution by OmegaLearn (Using Logarithmic Manipulations) ==
 
https://youtu.be/vgFPZ-hyd-I
 
https://youtu.be/vgFPZ-hyd-I
 +
 +
==Video Solution by TheBeautyofMath (Using Magical Ability)==
 +
https://youtu.be/ySWSHyY9TwI?t=999
 +
 +
~IceMatrix
 +
 +
== Video Solution by The Power of Logic ==
 +
https://youtu.be/b7xEeR7HXkE
  
 
==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 01:00, 26 June 2021

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 1 (Condensed Explanation of Logarithmic Identities)

This equals \[\left(\sum_{k=1}^{20}k\log_5{3}\right)\left(\sum_{k=1}^{100}\log_9{25}\right)=\frac{20\cdot21}{2}\cdot\log_5{3}\cdot100\log_3{5}=\boxed{\textbf{(E) }21,000}.\] ~JHawk0224

Solution 2 (Detailed Explanation of Logarithmic Identities)

We will apply the following logarithmic identity: \[\log_{p^n}{q^n}=\log_{p}{q},\] which can be proven by the Change of Base Formula: \[\log_{p^n}{q^n}=\frac{\log_{p}{q^n}}{\log_{p}{p^n}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.\] Now, we simplify the expressions inside the summations: \begin{align*} \log_{5^k}{{3^k}^2}&=\log_{5^k}{\left(3^k\right)^k} \\ &=k\log_{5^k}{3^k} \\ &=k\log_{5}{3}, \end{align*} and \begin{align*} \log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\ &=\log_{3}{5}. \end{align*} Using these results, we evaluate the original expression: \begin{align*} \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)\cdot\left(\log_{3}{5}\cdot\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}. \end{align*} ~MRENTHUSIASM

Solution 3 (Properties of Logarithms)

First, we can get rid of the $k$ exponents using properties of logarithms: \[\log_{5^k} 3^{k^2} = k^2 \cdot \frac{1}{k} \cdot \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k.\] (Leaving the single $k$ in the exponent will come in handy later). Similarly, \[\log_{9^k} 25^{k} = k \cdot \frac{1}{k} \cdot \log_{9} 25 = \log_{9} 5^2.\] 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: \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}} &&\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)} \\ &= \log_{9} 5^{200}. \end{align*} In $(*),$ we use the triangular numbers equation: \[1 + 2 + \dots + n = \frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210.\] 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, \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*} ~Joeya (Solution)

~MRENTHUSIASM (Reformatting)

Solution 4 (Estimations and Answer Choices)

In $\sum_{k=1}^{20} \log_{5^k} 3^{k^2},$ note that the addends are greater than $1$ for all $k\geq2.$

In $\sum_{k=1}^{100} \log_{9^k} 25^k,$ note that the addends are greater than $1$ for all $k\geq1.$

We have the inequality \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} 1\right)=19\cdot100=1,900,\] which eliminates choices $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}.$ We get the answer $\boxed{\textbf{(E) }21,000}$ by either an educated guess or a continued approximation:

Since $3^3=27\approx25,$ it follows that $9^{3/2}\approx25.$ By an extremely rough underestimation, \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)\approx\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} \frac{3}{2}\right)=19\cdot150=2,850.\] From here, it should be safe to guess that the answer is $\textbf{(E)}.$

As an extra guaranty, note that $\sum_{k=1}^{20} \log_{5^k} 3^{k^2} >> \sum_{k=2}^{20} 1 = 19.$ Therefore, we must have \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>>2,850.\]

~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)

https://youtu.be/vgFPZ-hyd-I

Video Solution by TheBeautyofMath (Using Magical Ability)

https://youtu.be/ySWSHyY9TwI?t=999

~IceMatrix

Video Solution by The Power of Logic

https://youtu.be/b7xEeR7HXkE

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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

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