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Difference between revisions of "2020 AIME II Problems/Problem 3"
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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>, so <math>n=\frac{3}{100}</math> and the answer is <math>\boxed{103}</math>.
 
~rayfish
 
~rayfish
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=lPr4fYEoXi0 ~ CNCM
 
==See Also==
 
==See Also==

Revision as of 18:04, 7 June 2020

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, $(2^{\frac{3}{100}})^n=3^{20}$, so $n=\frac{3}{100}$ and the answer is $\boxed{103}$. ~rayfish

Video Solution

https://www.youtube.com/watch?v=lPr4fYEoXi0 ~ CNCM

See Also

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