# Difference between revisions of "2020 AIME II Problems/Problem 3"

## 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 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 $\log_a{b}=\dfrac{1}{\log_b{a},$ (Error compiling LaTeX. ! File ended while scanning use of \@genfrac.) $\log_{2^x} 3=\dfrac{1}{\log_{3} 2^x},$ 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 $\boxed{3+100=103}.$

## 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 $3 + 100 = 103$.

~IceMatrix