Difference between revisions of "2019 AIME II Problems/Problem 6"
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<cmath>y^{54} = x</cmath> | <cmath>y^{54} = x</cmath> | ||
Plugging this in gives | Plugging this in gives | ||
− | <cmath>3\log _{b} (y | + | <cmath>3\log _{b} (y \cdot y^{27}) = 3\log _{b} y^{28} = 56</cmath> |
which gives | which gives | ||
<cmath>\log _{b} y = \dfrac{2}{3}</cmath> | <cmath>\log _{b} y = \dfrac{2}{3}</cmath> |
Revision as of 21:49, 5 June 2020
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 -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 |
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