Difference between revisions of "2020 AIME I Problems/Problem 2"
Skyscraper (talk | contribs) (→Solution 3) |
Iamthehazard (talk | contribs) (Added a fourth solution (my own, which I used in the competition)) |
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+ | ==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> | ||
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+ | ~IAmTheHazard | ||
==See Also== | ==See Also== |
Revision as of 18:32, 12 March 2020
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 LTE results in: 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 applying LTE again gives: Substituting this back into the first equation gives: Therefore,
~IAmTheHazard
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.