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

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

2020 AIME I (Problems • Answer Key • Resources)
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2020 AIME I Problems/Problem 2

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Problem

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Solution 2

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