Difference between revisions of "2020 AIME I Problems/Problem 2"
Iamthehazard (talk | contribs) (Added a fourth solution (my own, which I used in the competition)) |
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~IAmTheHazard | ~IAmTheHazard | ||
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+ | ==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_4{\sqrt{x}}\log_4{\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 | ||
==See Also== | ==See Also== |
Revision as of 21:39, 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
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
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.