Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AIME I Problems/Problem 2"

(→‎Solution)
(→‎Solution)
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
−
Since these form a geometric series, <math>\frac{\log_2{x}}{\log_4{x}}</math> is the common ratio. Rewriting this, we get <math>\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2</math> by base change formula. Therefore, the common ratio is 2. Now <math>\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\log_2{x} = \frac{2}{3} + \frac{2}{3}\log_2{x} \implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}</math>. Therefore, <math>1 + 16 = \boxed{017}</math>.
+
Since these form a geometric series, <math>\frac{\log_2{x}}{\log_4{x}}</math> is the common ratio. Rewriting this, we get <math>\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2</math> by base change formula. Therefore, the common ratio is 2. Now <math>\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\log_2{x} = \frac{2}{3} + \frac{2}{3}\log_2{x}  
 +
 
 +
\implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}</math>. Therefore, <math>1 + 16 = \boxed{017}</math>.
  
 
~ JHawk0224
 
~ JHawk0224

Revision as of 15:58, 12 March 2020

Note: Please do not post problems here until after the AIME.

Problem

There is a unique positive real number $x$ such that the three numbers $\log_8{2x}$, $\log_4{x}$, and $\log_2{x}$, in that order, form a geometric progression with positive common ratio. The number $x$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

Since these form a geometric series, $\frac{\log_2{x}}{\log_4{x}}$ is the common ratio. Rewriting this, we get $\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2$ by base change formula. Therefore, the common ratio is 2. Now $\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\log_2{x} = \frac{2}{3} + \frac{2}{3}\log_2{x}

\implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}$ (Error compiling LaTeX. Unknown error_msg). Therefore, $1 + 16 = \boxed{017}$.

~ JHawk0224

See Also

2020 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_2&oldid=119152"