Difference between revisions of "2020 AIME I Problems/Problem 2"
Skyscraper (talk | contribs) (→Solution 3) |
Skyscraper (talk | contribs) (→Solution 3) |
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==Solution 3== | ==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> | |
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Hence we obtain <cmath> (\log_4{(x)})(\log_4{(x)}) = (\log_8{(2x)})(\log_2{(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> | 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> |
Revision as of 18:08, 12 March 2020
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
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.