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

(Solution)
Line 4: Line 4:
  
 
== 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>.
  
 
==See Also==
 
==See Also==

Revision as of 16:54, 12 March 2020

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

Problem

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 $\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\log_2{x} = \frac{2}{3} + \frac{2}{3}log_2{x} \implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}$. Therefore, $1 + 16 = \boxed{017}$.

See Also

2020 AIME I (ProblemsAnswer KeyResources)
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