Difference between revisions of "2020 AIME I Problems/Problem 2"
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== Problem == | == Problem == | ||
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== Solution == | == Solution == | ||
− | 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} \implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}</math>. Therefore, <math>1 + 16 = \boxed{017}</math>. | + | 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> |
+ | |||
+ | <math>\implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}</math>. Therefore, <math>1 + 16 = \boxed{017}</math>. | ||
~ 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 | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let <math>r</math> be the common ratio. We have <cmath> r = \frac{\log_4{(x)}}{\log_8{(2x)}} = \frac{\log_2{(x)}}{\log_4{(x)}}</cmath> | ||
+ | Hence we obtain <cmath> (\log_4{(x)})(\log_4{(x)}) = (\log_8{(2x)})(\log_2{(x)})</cmath> | ||
+ | Ideally we change everything to base <math>64</math> and we can get: <cmath> (\log_{64}{(x^3)})(\log_{64}{(x^3)}) = (\log_{64}{(x^6)})(\log_{64}{(4x^2)})</cmath> | ||
+ | Now divide to get: <cmath>\frac{\log_{64}{(x^3)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}</cmath> | ||
+ | By change-of-base we obtain: <cmath>\log_{(4x^2)}{(x^3)} = \log_{(x^3)}{(x^6)} = 2</cmath> | ||
+ | Hence <math>(4x^2)^2 = x^3 \rightarrow 16x^4 = x^3 \rightarrow x = \frac{1}{16}</math> and we have <math>1+16 = \boxed{017}</math> as desired. | ||
+ | |||
+ | ~skyscraper | ||
+ | |||
+ | ==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} \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 | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | We can relate the logarithms as follows: | ||
+ | |||
+ | <cmath>\frac{\log_4{x}}{\log_8{(2x)}}=\frac{\log_2{x}}{\log_4{x}}</cmath> | ||
+ | <cmath>\log_8{(2x)}\log_2{x}=\log_4{x}\log_4{x}</cmath> | ||
+ | |||
+ | Now we can convert all logarithm bases to <math>2</math> using the identity <math>\log_a{b}=\log_{a^c}{b^c}</math>: | ||
+ | |||
+ | <cmath>\log_2{\sqrt[3]{2x}}\log_2{x}=\log_2{\sqrt{x}}\log_2{\sqrt{x}}</cmath> | ||
+ | |||
+ | We can solve for <math>x</math> as follows: | ||
+ | |||
+ | <cmath>\frac{1}{3}\log_2{(2x)}\log_2{x}=\frac{1}{4}\log_2{x}\log_2{x}</cmath> | ||
+ | <cmath>\frac{1}{3}\log_2{(2x)}=\frac{1}{4}\log_2{x}</cmath> | ||
+ | <cmath>\frac{1}{3}\log_2{2}+\frac{1}{3}\log_2{x}=\frac{1}{4}\log_2{x}</cmath> | ||
+ | We get <math>x=\frac{1}{16}</math>. Verifying that the common ratio is positive, we find the answer of <math>\boxed{017}</math>. | ||
+ | |||
+ | ~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 | ||
+ | |||
+ | https://youtu.be/4FvYVfhhTaQ | ||
+ | |||
+ | https://youtu.be/FgrIgCyGVUI | ||
+ | |||
+ | https://youtu.be/mgRNqSDCvgM?t=281s | ||
==See Also== | ==See Also== |
Latest revision as of 23:20, 1 February 2021
Contents
Problem
There is a unique positive real number such that the three numbers , , and , in that order, form a geometric progression with positive common ratio. The number can be written as , where and are relatively prime positive integers. Find .
Solution
Since these form a geometric series, is the common ratio. Rewriting this, we get by base change formula. Therefore, the common ratio is 2. Now
. Therefore, .
~ JHawk0224
Solution 2
If we set , we can obtain three terms of a geometric sequence through logarithm properties. The three terms are 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: which can be solved to reveal . Therefore, , so our answer is .
-molocyxu
Solution 3
Let be the common ratio. We have Hence we obtain Ideally we change everything to base and we can get: Now divide to get: By change-of-base we obtain: Hence and we have as desired.
~skyscraper
Solution 4 (Exponents > Logarithms)
Let be the common ratio, and let be the starting term (). We then have: Rearranging these equations gives: Deal with the last two equations first: Setting them equal gives: Using this value of , substitute into the first and second equations (or the first and third, it doesn't really matter) to get: Changing these to a common base gives: Dividing the first equation by 2 on both sides yields: Setting these equations equal to each other and removing the exponent again gives: Substituting this back into the first equation gives: Therefore,
~IAmTheHazard
Solution 5
We can relate the logarithms as follows:
Now we can convert all logarithm bases to using the identity :
We can solve for as follows:
We get . Verifying that the common ratio is positive, we find the answer of .
~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 and , respectively. Therefore: Let . We can rewrite the expression as: Zero does not work in this case, so we consider : . Therefore, .
~Bowser498
Solution 7
Again, by the Change of Base Formula, obtain that the common ratio is 2. If we let be the exponent of , then we have Wee can then divide the first equation by two to have the right side be . Also, . Setting this equal to , we can divide the two equations to get . Therefore, . After that, we can raise to the th power to get . We then get our sum of .
Solution 8 (Official MAA)
By the Change of Base Formula the common ratio of the progression isHence must satisfyThis is equivalent to . Hence and . The requested sum is .
Video Solutions
https://youtu.be/mgRNqSDCvgM?t=281s
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.