2020 AIME I Problems/Problem 2

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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

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2020 AIME I Problems/Problem 2

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