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Difference between revisions of "2020 AIME II Problems/Problem 3"
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==Solution==
 
==Solution==
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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>, so <math>n=\frac{3}{100}</math> and the answer is <math>\boxed{103}</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==
 +
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>
 +
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>
 +
~mlgjeffdoge21
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 +
==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>
 +
 +
Our answer is <math>\boxed{3+100=103}.</math>
 +
~ OreoChocolate
 +
 +
==Solution 3 (Official MAA)==
 +
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 ~}
 +
\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>
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 +
<math>(2^{xy})^{101} = (2^{x+3})^y \Rightarrow 2^{101xy} = 2^{xy+3y} \Rightarrow 101xy = xy + 3y \Rightarrow 101xy = y(x+3)</math>
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 +
<math>101x = x + 3</math>
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<math>100x = 3</math>
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 +
<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==
 +
https://youtu.be/lPr4fYEoXi0 ~ CNCM
 +
 +
==Video Solution 2==
 +
https://www.youtube.com/watch?v=x0QznvXcwHY?t=528
 +
 +
~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==
 +
{{AIME box|year=2020|n=II|num-b=2|num-a=4}}
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 03:23, 23 January 2023

Problem

The value of $x$ that satisfies $\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Let $\log _{2^x}3^{20}=\log _{2^{x+3}}3^{2020}=n$. Based on the equation, we get $(2^x)^n=3^{20}$ and $(2^{x+3})^n=3^{2020}$. Expanding the second equation, we get $8^n\cdot2^{xn}=3^{2020}$. Substituting the first equation in, we get $8^n\cdot3^{20}=3^{2020}$, so $8^n=3^{2000}$. Taking the 100th root, we get $8^{\frac{n}{100}}=3^{20}$. Therefore, $(2^{\frac{3}{100}})^n=3^{20}$, and using the our first equation($2^{xn}=3^{20}$), we get $x=\frac{3}{100}$ and the answer is $\boxed{103}$. ~rayfish

Easiest Solution

Recall the identity $\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b$ (which is easily proven using exponents or change of base). Then this problem turns into \[\frac{20}{x}\log_{2} 3 = \frac{2020}{x+3}\log_{2} 3\] Divide $\log_{2} 3$ from both sides. And we are left with $\frac{20}{x}=\frac{2020}{x+3}$.Solving this simple equation we get \[x = \tfrac{3}{100} \Rightarrow \boxed{103}\] ~mlgjeffdoge21

Solution 2

Because $\log_a{b^c}=c\log_a{b},$ we have that $20\log_{2^x} 3 = 2020\log_{2^{x+3}} 3,$ or $\log_{2^x} 3 = 101\log_{2^{x+3}} 3.$ Since $\log_a{b}=\dfrac{1}{\log_b{a}},$ $\log_{2^x} 3=\dfrac{1}{\log_{3} 2^x},$ and $101\log_{2^{x+3}} 3=101\dfrac{1}{\log_{3}2^{x+3}},$ thus resulting in $\log_{3}2^{x+3}=101\log_{3} 2^x,$ or $\log_{3}2^{x+3}=\log_{3} 2^{101x}.$ We remove the base 3 logarithm and the power of 2 to yield $x+3=101x,$ or $x=\dfrac{3}{100}.$

Our answer is $\boxed{3+100=103}.$ ~ OreoChocolate

Solution 3 (Official MAA)

Using the Change of Base Formula to convert the logarithms in the given equation to base $2$ yields \[\frac{\log_2 3^{20}}{\log_2 2^x} = \frac{\log_2 3^{2020}}{\log_2 2^{x+3}}, \text{~ and then ~} \frac{20\log_2 3}{x\cdot\log_2 2} = \frac{2020\log_2 3}{(x+3)\log_2 2}.\]Canceling the logarithm factors then yields\[\frac{20}x = \frac{2020}{x+3},\]which has solution $x = \frac3{100}.$ The requested sum is $3 + 100 = 103$.

Solution 4

$\log_{2^x} 3^{20} = 2^{xy} = 3^{20}$

$\log_{2^{x+3}} 3^{2020} = (2^{x+3})^y = 3^{2020}$

$3^{2020} = (3^{20})^{101}$

$(2^{xy})^{101} = (2^{x+3})^y = 3^{2020}$

$(2^{xy})^{101} = (2^{x+3})^y \Rightarrow 2^{101xy} = 2^{xy+3y} \Rightarrow 101xy = xy + 3y \Rightarrow 101xy = y(x+3)$

$101x = x + 3$

$100x = 3$

$x = \frac{3}{100}$

$100 + 3 = \boxed{103}$ ~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

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

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

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