Difference between revisions of "2019 AIME II Problems/Problem 6"
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<cmath>x = (b^{\log \log x})^{54}</cmath> | <cmath>x = (b^{\log \log x})^{54}</cmath> | ||
<cmath>x = (\log x)^{54}</cmath> | <cmath>x = (\log x)^{54}</cmath> | ||
+ | Note by dolphin7 - you could also just rewrite the second equation in exponent form. | ||
Substituting this into the <math>\sqrt x</math> of the first equation, | Substituting this into the <math>\sqrt x</math> of the first equation, | ||
<cmath>3\log((\log x)^{27}\log x) = 56</cmath> | <cmath>3\log((\log x)^{27}\log x) = 56</cmath> |
Revision as of 14:01, 18 April 2021
Contents
Problem
In a Martian civilization, all logarithms whose bases are not specified as assumed to be base , for some fixed
. A Martian student writes down
and finds that this system of equations has a single real number solution
. Find
.
Solution 1
Using change of base on the second equation to base b,
Note by dolphin7 - you could also just rewrite the second equation in exponent form.
Substituting this into the
of the first equation,
We can manipulate this equation to be able to substitute a couple more times:
However, since we found that ,
is also equal to
. Equating these,
Solution 2
We start by simplifying the first equation to
Next, we simplify the second equation to
Substituting this into the first equation gives
Plugging this into
gives
-ktong
Solution 3
Apply change of base to to yield:
which can be rearranged as:
Apply log properties to
to yield:
Substituting
into the equation
yields:
So
Substituting this back in to
yields
So,
-Ghazt2002
Solution 4
1st equation:
2nd equation:
So now substitute
and
:
We also have that
This means that
, so
.
-Stormersyle
Solution 5 (Substitution)
Let
Then we have
which gives
Plugging this in gives
which gives
so
By substitution we have
which gives
Plugging in again we get
--Hi3142
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.