Difference between revisions of "2020 AIME I Problems/Problem 2"
(→Problem) |
(→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 16:57, 12 March 2020
Note: Please do not post problems here until after the AIME.
Problem
There is a unique positive real number 
such that the three numbers 
, 
, and 
, in that order, form a geometric progression with positive common ratio. The number can be written as 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
Since these form a geometric series, 
is the common ratio. Rewriting this, we get 
by base change formula. Therefore, the common ratio is 2. Now 
. Therefore, 
.
~ 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. 
