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

Revision as of 22:51, 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

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

@@ Line 13: / Line 13: @@
 ==Video Solution==
 https://youtu.be/lPr4fYEoXi0 ~ CNCM
+==Video Solution 2==
+https://www.youtube.com/watch?v=x0QznvXcwHY
 ==See Also==
 {{AIME box|year=2020|n=II|num-b=2|num-a=4}}
 [[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

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

Page

Toolbox

Search