Difference between revisions of "2020 AIME I Problems/Problem 2"
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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 |
Revision as of 20:59, 20 September 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
See here for a video solution:
Another video solution:
Video solution (*)
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 (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
.
See here for a video solution:
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.