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Difference between revisions of "2020 AIME II Problems/Problem 3"
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==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{30}{x}\log_{2} 3 = \frac{3030}{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{30}{x}=\frac{3030}{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>
 
~mlgjeffdoge21
 
~mlgjeffdoge21
  

Revision as of 09:43, 9 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

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 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 $3 + 100 = 103$.

Video Solution

https://youtu.be/lPr4fYEoXi0 ~ CNCM

Video Solution 2

https://www.youtube.com/watch?v=x0QznvXcwHY?t=528

~IceMatrix

See Also

2020 AIME II (Problems • Answer Key • Resources)
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Problem 2 		Followed by
Problem 4
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All AIME Problems and Solutions

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