Difference between revisions of "2021 AMC 12A Problems/Problem 14"
MRENTHUSIASM (talk | contribs) (Reformatted and made the boxes more consistent.) |
MRENTHUSIASM (talk | contribs) m (→Solution 3 (Properties of Logarithms)) |
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\sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\ | \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^{(1 + 2 + \dots + 20)} \\ | ||
− | &= \log_{5} 3^{\frac{20(20+1)}{2}} &&\hspace{10mm} | + | &= \log_{5} 3^{\frac{20(20+1)}{2}} &&\hspace{10mm}(*) \\ |
&= \log_{5} 3^{210}, \\ | &= \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 \\ | \sum_{k=1}^{100} \log_{9} 5^2 &= \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2 \\ | ||
Line 45: | Line 45: | ||
&= \log_{9} 5^{200}. | &= \log_{9} 5^{200}. | ||
\end{align*}</cmath> | \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> |
Revision as of 00:56, 26 June 2021
Contents
- 1 Problem
- 2 Solution 1 (Condensed Explanation of Logarithmic Identities)
- 3 Solution 2 (Detailed Explanation of Logarithmic Identities)
- 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 (Condensed Explanation of Logarithmic Identities)
This equals
~JHawk0224
Solution 2 (Detailed Explanation of Logarithmic Identities)
We will apply the following property of logarithms:
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 3 (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 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:
Since it follows that
By an extremely rough underestimation,
From here, it should be safe to guess that the answer is
As an extra guaranty, note that Therefore, we must have
~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.