Difference between revisions of "2020 AIME I Problems/Problem 2"
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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}} | 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>. | = 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== | ==Video Solutions== |
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 satisfy
This 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.