Difference between revisions of "2020 AIME I Problems/Problem 2"

Latest revision as of 17:31, 29 January 2023

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 1

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: $4^{ar}=2^{ar^2} \implies 2^{2ar}=2^{ar^2} \implies 2ar=ar^2 \implies r=2$ 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 removing the exponent again gives: $3a-1=4a \implies a=-1$ Substituting this back into the first equation gives: $8^{-1}=2x \implies 2x=\frac{1}{8} \implies x=\frac{1}{16}$ 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

Solution 7

Again, by the Change of Base Formula, obtain that the common ratio is 2. If we let $y$ be the exponent of $\log_8 (2x)$ , then we have $8^y=2x;\:4^{2y}=x;\:2^{4y}=x.$ Wee can then divide the first equation by two to have the right side be $x$ . Also, $2^{4y}=\left(2^{4}\right)^y=16^y$ . Setting this equal to $\frac{8^y}{2}$ , we can divide the two equations to get $2^y=\frac12$ . Therefore, $y=-1$ . After that, we can raise $16$ to the $-1$ th power to get $x=16^{-1}\Rightarrow x=\frac1{16}$ . We then get our sum of $1+16=\boxed{\textbf{017}}$ .

~SweetMango77

Solution 8 (Official MAA)

By the Change of Base Formula the common ratio of the progression is $\frac{\log_2 x}{\log_4 x} = \frac{\hphantom{m}\log_2x\hphantom{m}}{\frac{\log_2x}{\log_24}} = 2.$ Hence $x$ must satisfy $2=\frac{\log_4 x}{\log_8 (2x)}= \frac{\log_2 x}{\log_2 4} \div \frac{\log_2(2x)}{\log_28} = \frac 32\cdot \frac{\log_2x}{1+\log_2x}.$ This is equivalent to $4 + 4\log_2x = 3\log_2x$ . Hence $\log_2x = -4$ and $x = \frac{1}{16}$ . The requested sum is $1+16 = 17$ .

Video Solutions

https://youtu.be/nPL7nUXnRbo (Unavailable)

Aarav Navani: https://youtu.be/4FvYVfhhTaQ

Mathematica De Estremo: https://youtu.be/FgrIgCyGVUI

https://youtu.be/mgRNqSDCvgM?t=281s (Unavailable)

@@ Line 3: / Line 3: @@
 There is a unique positive real number <math>x</math> such that the three numbers <math>\log_8{2x}</math>, <math>\log_4{x}</math>, and <math>\log_2{x}</math>, in that order, form a geometric progression with positive common ratio.  The number <math>x</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
-== Solution ==
+== Solution 1==
 Since these form a geometric series, <math>\frac{\log_2{x}}{\log_4{x}}</math> is the common ratio. Rewriting this, we get <math>\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2</math> by base change formula. Therefore, the common ratio is 2. Now <math>\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}</math>
@@ Line 27: / Line 27: @@
 ==Solution 4 (Exponents > Logarithms)==
 Let <math>r</math> be the common ratio, and let <math>a</math> be the starting term (<math>a=\log_{8}{(2x)}</math>). We then have: <cmath>\log_{8}{(2x)}=a, \log_{4}{(x)}=ar, \log_{2}{(x)}=ar^2</cmath> Rearranging these equations gives: <cmath>8^a=2x, 4^{ar}=x, 2^{ar^2}=x</cmath>
-Deal with the last two equations first: Setting them equal gives: <cmath>4^{ar}=2^{ar^2} \Rightarrow 2^{2ar}=2^{ar^2}</cmath> Using LTE results in: <cmath>2ar=ar^2 \Rightarrow r=2</cmath> Using this value of <math>r</math>, substitute into the first and second equations (or the first and third, it doesn't really matter) to get: <cmath>8^a=2x, 4^{2a}=x</cmath> Changing these to a common base gives: <cmath>2^{3a}=2x, 2^{4a}=x</cmath> Dividing the first equation by 2 on both sides yields: <cmath>2^{3a-1}=x</cmath> Setting these equations equal to each other and applying LTE again gives: <cmath>3a-1=4a \Rightarrow a=-1</cmath> Substituting this back into the first equation gives: <cmath>8^{-1}=2x \Rightarrow 2x=\frac{1}{8} \Rightarrow x=\frac{1}{16}</cmath> Therefore, <math>m+n=1+16=\boxed{017}</math>
+Deal with the last two equations first: Setting them equal gives: <cmath>4^{ar}=2^{ar^2} \implies 2^{2ar}=2^{ar^2} \implies 2ar=ar^2 \implies r=2</cmath> Using this value of <math>r</math>, substitute into the first and second equations (or the first and third, it doesn't really matter) to get: <cmath>8^a=2x, 4^{2a}=x</cmath> Changing these to a common base gives: <cmath>2^{3a}=2x, 2^{4a}=x</cmath> Dividing the first equation by 2 on both sides yields: <cmath>2^{3a-1}=x</cmath> Setting these equations equal to each other and removing the exponent again gives: <cmath>3a-1=4a \implies a=-1</cmath> Substituting this back into the first equation gives: <cmath>8^{-1}=2x \implies 2x=\frac{1}{8} \implies x=\frac{1}{16}</cmath> Therefore, <math>m+n=1+16=\boxed{017}</math>
 ~IAmTheHazard
@@ Line 50: / Line 50: @@
 ~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 <math>\frac{1+\log_2{x}}{3}</math> and <math>\frac{1}{2}\log_2{x}</math>, respectively. Therefore:
+<cmath>\frac{1}{2}\log_2{x}=\sqrt{\left(\frac{1+\log_2{x}}{3}\right)\left(\log_2{x}\right)}</cmath>
+Let <math>n=\log_2{x}</math>. We can rewrite the expression as:
+<cmath>\frac{n}{2}=\sqrt{\frac{n(n+1)}{3}}</cmath>
+<cmath>\frac{n^2}{4}=\frac{n(n+1)}{3}</cmath>
+<cmath>4n(n+1)=3n^2</cmath>
+<cmath>4n^2+4n=3n^2</cmath>
+<cmath>n^2+4n=0</cmath>
+<cmath>n(n+4)=0</cmath>
+<cmath>n=0 \text{ and } -4</cmath>
+Zero does not work in this case, so we consider <math>n=-4</math>: <math>\log_2{x}=-4 \rightarrow x=\frac{1}{16}</math>. Therefore, <math>1+16=\boxed{017}</math>.
+~Bowser498
+==Solution 7 ==
+Again, by the Change of Base Formula, obtain that the common ratio is 2. If we let <math>y</math> be the exponent of <math>\log_8 (2x)</math>, then we have <math>8^y=2x;\:4^{2y}=x;\:2^{4y}=x.</math> Wee can then divide the first equation by two to have the right side be <math>x</math>. Also, <math>2^{4y}=\left(2^{4}\right)^y=16^y</math>. Setting this equal to <math>\frac{8^y}{2}</math>, we can divide the two equations to get <math>2^y=\frac12</math>. Therefore, <math>y=-1</math>. After that, we can raise <math>16</math> to the <math>-1</math>th power to get <math>x=16^{-1}\Rightarrow x=\frac1{16}</math>. We then get our sum of <math>1+16=\boxed{\textbf{017}}</math>.
+~[[User:Sweetmango77|SweetMango77]]
+==Solution 8 (Official MAA)==
+By the Change of Base Formula the common ratio of the progression is<cmath>\frac{\log_2 x}{\log_4 x} = \frac{\hphantom{m}\log_2x\hphantom{m}}{\frac{\log_2x}{\log_24}}
+= 2.</cmath>Hence <math>x</math> must satisfy<cmath>2=\frac{\log_4 x}{\log_8 (2x)}= \frac{\log_2 x}{\log_2 4} \div \frac{\log_2(2x)}{\log_28} = \frac 32\cdot \frac{\log_2x}{1+\log_2x}.</cmath>This is equivalent to <math>4 + 4\log_2x = 3\log_2x</math>. Hence <math>\log_2x = -4</math> and <math>x = \frac{1}{16}</math>. The requested sum is <math>1+16 = 17</math>.
+==Video Solutions==
+https://youtu.be/nPL7nUXnRbo (Unavailable)
+Aarav Navani: https://youtu.be/4FvYVfhhTaQ
+Mathematica De Estremo: https://youtu.be/FgrIgCyGVUI
+https://youtu.be/mgRNqSDCvgM?t=281s (Unavailable)
 ==See Also==

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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