2020 AIME II Problems/Problem 3

Revision as of 01:27, 8 June 2020 by Icematrix (talk | contribs) (Video Solution 2)

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) Then this problem turns into \[\frac{30}{x}\log_{2} 3 = \frac{3030}{x+3}\log_{2} 3\] Divide $\log_{2} 3$ from both sides. And we are left with $\frac{30}{x}=\frac{3030}{x+3}$.Solving this simple equation we get \[x = \tfrac{3}{100} \Rightarrow \boxed{103}\] ~mlgjeffdoge21

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AIME Problems and Solutions

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