Difference between revisions of "2020 AIME I Problems/Problem 2"
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==Solution 4 (Exponents > Logarithms)== | ==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> | 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} \ | + | 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 | ~IAmTheHazard | ||
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~Bowser498 | ~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>. | ||
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+ | ~[[User:Sweetmango77|SweetMango77]] | ||
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+ | ==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>. | ||
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+ | ==Video Solutions== | ||
https://youtu.be/nPL7nUXnRbo | 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.