2020 AIME I Problems/Problem 2

Revision as of 18:08, 12 March 2020 by Skyscraper (talk | contribs) (Solution 3)


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 $m + n$.


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 $\boxed{017}$.


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: \[\frac{\log_{64}{(x^6)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}\] 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.


See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions

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