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

Revision as of 17:17, 12 March 2020

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 $\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\log_2{x} = \frac{2}{3} + \frac{2}{3}\log_2{x}$

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

~ JHawk0224

Solution 2

If we set $x=2^y$ , we can obtain three terms of a geometric sequence through logarithm properties. The three terms are $\frac{y+1}{3}, \frac{y}{2}, y.$ In a three-term geometric sequence, the middle term squared is equal to the product of the other two terms, so we obtain the following: $\frac{y^2+y}{3} = \frac{y^2}{4},$ which can be solved to reveal $y = -4$ . Therefore, $x = 2^{-4} = \frac{1}{16}$ , so our answer is $\boxed{017}$ .

-molocyxu

Solution 3

Let $r$ be the common ratio. We have $r = \frac{\log_4{(x)}}{\log_8{(2x)}} = \frac{\log_2{(x)}}{\log_4{(x)}}$ Hence we obtain $(\log_4{(x)})(\log_4{(x)}) = (\log_8{(2x)})(\log_2{(x)})$ Ideally we change everything to base $64$ and we can get: $(\log_{64}{(x^3)})(\log_{64}{(x^3)}) = (\log_{64}{(x^6)})(\log_{64}{(4x^2)})$ Now divide to get: $\frac{\log_{64}{(x^3)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}$ By change-of-base we obtain: $\log_{(4x^2)}{(x^3)} = \log_{(x^3)}{(x^6)} = 2$ Hence $(4x^2)^2 = x^3 \rightarrow 16x^4 = x^3 \rightarrow x = \frac{1}{16}$ and we have $1+16 = \boxed{017}$ as desired.

~skyscraper

@@ Line 19: / Line 19: @@
 Hence we obtain <cmath> (\log_4{(x)})(\log_4{(x)}) = (\log_8{(2x)})(\log_2{(x)})</cmath>
 Ideally we change everything to base <math>64</math> and we can get: <cmath> (\log_{64}{(x^3)})(\log_{64}{(x^3)}) = (\log_{64}{(x^6)})(\log_{64}{(4x^2)})</cmath>
-Now divide to get: <cmath>\frac{\log_{64}{(x^6)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}</cmath>
+Now divide to get: <cmath>\frac{\log_{64}{(x^3)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}</cmath>
 By change-of-base we obtain: <cmath>\log_{(4x^2)}{(x^3)} = \log_{(x^3)}{(x^6)} = 2</cmath>
 Hence <math>(4x^2)^2 = x^3 \rightarrow 16x^4 = x^3 \rightarrow x = \frac{1}{16}</math> and we have <math>1+16 = \boxed{017}</math> as desired.

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

Page

Toolbox

Search