Difference between revisions of "2019 AIME II Problems/Problem 6"
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<cmath>b=36^{\frac{3}{2}}=6^3=\boxed{216}</cmath> | <cmath>b=36^{\frac{3}{2}}=6^3=\boxed{216}</cmath> | ||
− | ==Solution | + | ==Solution 3== |
Apply change of base to <cmath>\log_{\log x}(x)=54</cmath> to yield: <cmath>\frac{\log_b(x)}{\log_b(\log_b(x))}=54</cmath> | Apply change of base to <cmath>\log_{\log x}(x)=54</cmath> to yield: <cmath>\frac{\log_b(x)}{\log_b(\log_b(x))}=54</cmath> | ||
which can be rearranged as: <cmath>\frac{\log_b(x)}{54}=\log_b(\log_b(x))</cmath> | which can be rearranged as: <cmath>\frac{\log_b(x)}{54}=\log_b(\log_b(x))</cmath> |
Revision as of 22:48, 22 March 2019
Contents
[hide]Problem 6
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,
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
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
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 |
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