2020 AIME I Problems/Problem 2

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

Solution 4 (Exponents > Logarithms)

Let $r$ be the common ratio, and let $a$ be the starting term ( $a=\log_{8}{(2x)}$ ). We then have: $\log_{8}{(2x)}=a, \log_{4}{(x)}=ar, \log_{2}{(x)}=ar^2$ Rearranging these equations gives: $8^a=2x, 4^{ar}=x, 2^{ar^2}=x$ Deal with the last two equations first: Setting them equal gives: $4^{ar}=2^{ar^2} \Rightarrow 2^{2ar}=2^{ar^2}$ Using LTE results in: $2ar=ar^2 \Rightarrow r=2$ Using this value of $r$ , substitute into the first and second equations (or the first and third, it doesn't really matter) to get: $8^a=2x, 4^{2a}=x$ Changing these to a common base gives: $2^{3a}=2x, 2^{4a}=x$ Dividing the first equation by 2 on both sides yields: $2^{3a-1}=x$ Setting these equations equal to each other and applying LTE again gives: $3a-1=4a \Rightarrow a=-1$ Substituting this back into the first equation gives: $8^{-1}=2x \Rightarrow 2x=\frac{1}{8} \Rightarrow x=\frac{1}{16}$ Therefore, $m+n=1+16=\boxed{017}$

~IAmTheHazard

Solution 5

We can relate the logarithms as follows:

$\frac{\log_4{x}}{\log_8{(2x)}}=\frac{\log_2{x}}{\log_4{x}}$ $\log_8{(2x)}\log_2{x}=\log_4{x}\log_4{x}$

Now we can convert all logarithm bases to $2$ using the identity $\log_a{b}=\log_{a^c}{b^c}$ :

$\log_2{\sqrt[3]{2x}}\log_2{x}=\log_2{\sqrt{x}}\log_2{\sqrt{x}}$

We can solve for $x$ as follows:

$\frac{1}{3}\log_2{(2x)}\log_2{x}=\frac{1}{4}\log_2{x}\log_2{x}$ $\frac{1}{3}\log_2{(2x)}=\frac{1}{4}\log_2{x}$ $\frac{1}{3}\log_2{2}+\frac{1}{3}\log_2{x}=\frac{1}{4}\log_2{x}$ We get $x=\frac{1}{16}$ . Verifying that the common ratio is positive, we find the answer of $\boxed{017}$ .

~QIDb602

Solution 6

If the numbers are in a geometric sequence, the middle term must be the geometric mean of the surrounding terms. We can rewrite the first two logarithmic expressions as $\frac{1+\log_2{x}}{3}$ and $\frac{1}{2}\log_2{x}$ , respectively. Therefore: $\frac{1}{2}\log_2{x}=\sqrt{\left(\frac{1+\log_2{x}}{3}\right)\left(\log_2{x}\right)}$ Let $n=\log_2{x}$ . We can rewrite the expression as: $\frac{n}{2}=\sqrt{\frac{n(n+1)}{3}}$ $\frac{n^2}{4}=\frac{n(n+1)}{3}$ $4n(n+1)=3n^2$ $4n^2+4n=3n^2$ $n^2+4n=0$ $n(n+4)=0$ $n=0 \text{ and } -4$ Zero does not work in this case, so we consider $n=-4$ : $\log_2{x}=-4 \rightarrow x=\frac{1}{16}$ . Therefore, $1+16=\boxed{017}$ .

~Bowser498

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