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2020 AIME I Problems/Problem 2: Difference between revisions

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~ JHawk0224
~ JHawk0224
==Solution 2==
If we set <math>x=2^y</math>, we can obtain three terms of a geometric sequence through logarithm properties. The three terms are <cmath>\frac{y+1}{3}, \frac{y}{2}, y.</cmath> 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: <cmath>\frac{y^2+y}{3} = \frac{y^2}{4},</cmath> which can be solved to reveal <math>y = -4</math>. Therefore, <math>x = 2^{-4} = \frac{1}{16}</math>, so our answer is <math>\boxed{017}</math>.
-molocyxu


==See Also==
==See Also==

Revision as of 16:11, 12 March 2020

Note: Please do not post problems here until after the AIME.

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

See Also

2020 AIME I (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions

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