Difference between revisions of "2019 AIME II Problems/Problem 6"
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+ | ==Solution 4 (Easiest)== | ||
+ | From the first equation we have that <math>\log (\sqrt{x}\log x)=\frac{56}{3}</math>, so <math>\log(\sqrt x)+\log(\log x)=\frac{1}{2}\log x+\log(\log x)=\frac{56}{3}</math>. From the second equation we have that <math>x=(\log x)^{54}</math>, so now set <math>\log x=a</math> and <math>x=b^a</math>. Substituting, we have that <math>b^a=a^{54}</math>, so <math>b=a^{\frac{54}{a}}</math>. We also have that <math>\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}</math>, so <math>\frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}</math>. This means that <math>\frac{14}{27}a=\frac{56}{3}</math>, so <math>a=36</math>, and <math>b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}</math>. | ||
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+ | -Stormersyle | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=II|num-b=5|num-a=7}} | {{AIME box|year=2019|n=II|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:04, 23 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
-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 (Easiest)
From the first equation we have that , so
. From the second equation we have that
, so now set
and
. Substituting, we have that
, so
. We also have that
, so
. This means that
, so
, and
.
-Stormersyle
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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