Difference between revisions of "2020 AIME II Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | Let <math>\log _{2^x}3^{20}=\log _{2^{x+3}}3^{2020}=n</math>. Based on the equation, we get <math>(2^x)^n=3^{20}</math> and <math>(2^{x+3})^n=3^{2020}</math>. Expanding the second equation, we get <math>8^n\cdot2^{xn}=3^{2020}</math>. Substituting the first equation in, we get <math>8^n\cdot3^{20}=3^{2020}</math>, so <math>8^n=3^{2000}</math>. Taking the 100th root, we get <math>8^{\frac{n}{100}}=3^{20}</math>. Therefore, <math>(2^{\frac{3}{100}})^n=3^{20}</math>, | + | Let <math>\log _{2^x}3^{20}=\log _{2^{x+3}}3^{2020}=n</math>. Based on the equation, we get <math>(2^x)^n=3^{20}</math> and <math>(2^{x+3})^n=3^{2020}</math>. Expanding the second equation, we get <math>8^n\cdot2^{xn}=3^{2020}</math>. Substituting the first equation in, we get <math>8^n\cdot3^{20}=3^{2020}</math>, so <math>8^n=3^{2000}</math>. Taking the 100th root, we get <math>8^{\frac{n}{100}}=3^{20}</math>. Therefore, <math>(2^{\frac{3}{100}})^n=3^{20}</math>, and using the our first equation(<math>2^{xn}=3^{20}</math>), we get <math>x=\frac{3}{100}</math> and the answer is <math>\boxed{103}</math>. |
~rayfish | ~rayfish | ||
==Easiest Solution== | ==Easiest Solution== | ||
− | Recall the identity <math>\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b </math> (which is easily proven using exponents or change of base) | + | Recall the identity <math>\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b </math> (which is easily proven using exponents or change of base). |
Then this problem turns into <cmath>\frac{20}{x}\log_{2} 3 = \frac{2020}{x+3}\log_{2} 3</cmath> | Then this problem turns into <cmath>\frac{20}{x}\log_{2} 3 = \frac{2020}{x+3}\log_{2} 3</cmath> | ||
Divide <math>\log_{2} 3</math> from both sides. And we are left with <math>\frac{20}{x}=\frac{2020}{x+3}</math>.Solving this simple equation we get <cmath>x = \tfrac{3}{100} \Rightarrow \boxed{103}</cmath> | Divide <math>\log_{2} 3</math> from both sides. And we are left with <math>\frac{20}{x}=\frac{2020}{x+3}</math>.Solving this simple equation we get <cmath>x = \tfrac{3}{100} \Rightarrow \boxed{103}</cmath> | ||
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==Solution 2== | ==Solution 2== | ||
− | Because <math>\log_a{b^c}=c\log_a{b},</math> we have that <math>20\log_{2^x} 3 = 2020\log_{2^{x+3}} 3,</math> or <math>\log_{2^x} 3 = 101\log_{2^{x+3}} 3.</math> Since <math>\log_a{b}=\dfrac{1}{\log_b{a},</math> <math>\log_{2^x} 3=\dfrac{1}{\log_{3} 2^x},</math> and <math>101\log_{2^{x+3}} 3=101\dfrac{1}{\log_{3}2^{x+3}},</math> thus resulting in <math>\log_{3}2^{x+3}=101\log_{3} 2^x,</math> or <math>\log_{3}2^{x+3}=\log_{3} 2^{101x}.</math> We remove the base 3 logarithm and the power of 2 to yield <math>x+3=101x,</math> or <math>x=\dfrac{3}{100}.</math> | + | Because <math>\log_a{b^c}=c\log_a{b},</math> we have that <math>20\log_{2^x} 3 = 2020\log_{2^{x+3}} 3,</math> or <math>\log_{2^x} 3 = 101\log_{2^{x+3}} 3.</math> Since <math>\log_a{b}=\dfrac{1}{\log_b{a}},</math> <math>\log_{2^x} 3=\dfrac{1}{\log_{3} 2^x},</math> and <math>101\log_{2^{x+3}} 3=101\dfrac{1}{\log_{3}2^{x+3}},</math> thus resulting in <math>\log_{3}2^{x+3}=101\log_{3} 2^x,</math> or <math>\log_{3}2^{x+3}=\log_{3} 2^{101x}.</math> We remove the base 3 logarithm and the power of 2 to yield <math>x+3=101x,</math> or <math>x=\dfrac{3}{100}.</math> |
Our answer is <math>\boxed{3+100=103}.</math> | Our answer is <math>\boxed{3+100=103}.</math> | ||
+ | ~ OreoChocolate | ||
+ | |||
==Solution 3 (Official MAA)== | ==Solution 3 (Official MAA)== | ||
Using the Change of Base Formula to convert the logarithms in the given equation to base <math>2</math> yields | Using the Change of Base Formula to convert the logarithms in the given equation to base <math>2</math> yields | ||
<cmath>\frac{\log_2 3^{20}}{\log_2 2^x} = \frac{\log_2 3^{2020}}{\log_2 2^{x+3}}, \text{~ and then ~} | <cmath>\frac{\log_2 3^{20}}{\log_2 2^x} = \frac{\log_2 3^{2020}}{\log_2 2^{x+3}}, \text{~ and then ~} | ||
\frac{20\log_2 3}{x\cdot\log_2 2} = \frac{2020\log_2 3}{(x+3)\log_2 2}.</cmath>Canceling the logarithm factors then yields<cmath>\frac{20}x = \frac{2020}{x+3},</cmath>which has solution <math>x = \frac3{100}.</math> The requested sum is <math>3 + 100 = 103</math>. | \frac{20\log_2 3}{x\cdot\log_2 2} = \frac{2020\log_2 3}{(x+3)\log_2 2}.</cmath>Canceling the logarithm factors then yields<cmath>\frac{20}x = \frac{2020}{x+3},</cmath>which has solution <math>x = \frac3{100}.</math> The requested sum is <math>3 + 100 = 103</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | <math>\log_{2^x} 3^{20} = 2^{xy} = 3^{20}</math> | ||
+ | |||
+ | <math>\log_{2^{x+3}} 3^{2020} = (2^{x+3})^y = 3^{2020}</math> | ||
+ | |||
+ | <math>3^{2020} = (3^{20})^{101}</math> | ||
+ | |||
+ | <math>(2^{xy})^{101} = (2^{x+3})^y = 3^{2020}</math> | ||
+ | |||
+ | <math>(2^{xy})^{101} = (2^{x+3})^y \Rightarrow 2^{101xy} = 2^{xy+3y} \Rightarrow 101xy = xy + 3y \Rightarrow 101xy = y(x+3)</math> | ||
+ | |||
+ | <math>101x = x + 3</math> | ||
+ | |||
+ | <math>100x = 3</math> | ||
+ | |||
+ | <math>x = \frac{3}{100}</math> | ||
+ | |||
+ | <math>100 + 3 = \boxed{103}</math> ~Airplanes2007 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=ZCm0SOjTPVE | ||
+ | |||
+ | ~North America Math Contest Go Go Go | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/lPr4fYEoXi0 ~ CNCM | https://youtu.be/lPr4fYEoXi0 ~ CNCM | ||
+ | |||
==Video Solution 2== | ==Video Solution 2== | ||
https://www.youtube.com/watch?v=x0QznvXcwHY?t=528 | https://www.youtube.com/watch?v=x0QznvXcwHY?t=528 | ||
~IceMatrix | ~IceMatrix | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | |||
+ | https://youtu.be/-CkEF5nWOaI | ||
+ | |||
+ | ~avn | ||
+ | |||
+ | ==Video Solution 4== | ||
+ | |||
+ | https://www.youtube.com/watch?v=2TSNY2DDUbQ&t=3s ~ MathEx | ||
+ | |||
+ | == Video Solution 5 by OmegaLearn == | ||
+ | https://youtu.be/RdIIEhsbZKw?t=1648 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 04:23, 23 January 2023
Contents
Problem
The value of that satisfies
can be written as
, where
and
are relatively prime positive integers. Find
.
Solution
Let . Based on the equation, we get
and
. Expanding the second equation, we get
. Substituting the first equation in, we get
, so
. Taking the 100th root, we get
. Therefore,
, and using the our first equation(
), we get
and the answer is
.
~rayfish
Easiest Solution
Recall the identity (which is easily proven using exponents or change of base).
Then this problem turns into
Divide
from both sides. And we are left with
.Solving this simple equation we get
~mlgjeffdoge21
Solution 2
Because we have that
or
Since
and
thus resulting in
or
We remove the base 3 logarithm and the power of 2 to yield
or
Our answer is
~ OreoChocolate
Solution 3 (Official MAA)
Using the Change of Base Formula to convert the logarithms in the given equation to base yields
Canceling the logarithm factors then yields
which has solution
The requested sum is
.
Solution 4
~Airplanes2007
Video Solution
https://www.youtube.com/watch?v=ZCm0SOjTPVE
~North America Math Contest Go Go Go
Video Solution
https://youtu.be/lPr4fYEoXi0 ~ CNCM
Video Solution 2
https://www.youtube.com/watch?v=x0QznvXcwHY?t=528
~IceMatrix
Video Solution 3
~avn
Video Solution 4
https://www.youtube.com/watch?v=2TSNY2DDUbQ&t=3s ~ MathEx
Video Solution 5 by OmegaLearn
https://youtu.be/RdIIEhsbZKw?t=1648
~ pi_is_3.14
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.